subject:"Re\: \[PHP\] \" \" in a Variable"



Terion Miller wrote:
 Hi Guys! Well I tried the INNER JOIN and still can not get it to echo
 the AdminID so I know it isn't working, (what kind of things should I
 think about that could make it not work)  so far the only query that
 did work and return the AdminID was my original I believe it was
 referred to as hosed query, yet then it was explained to me that
 query while picking up the AdminID was then returning all the rows
 from workorders anyways because I needed the INNER JOIN, so since that
 isn't working, I'm thinking my best and fastest route ( I have until
 Monday on this project and this is just one bit of it OUCH)  is to use
 my hosed query then somehow use the resulting AdminID to fetch the
 orders from the workorders table, question is , would that be a sub
 query, or do I just make the query results for AdminID a variable to
 use in another query?
 Thanks guys and ps. I'm a she not a he, funny that coders are
 primarily always assumed to be guys...lol
 Facebook Me:
 http://www.facebook.com/profile.php?id=1542024891ref=name
 http://www.facebook.com/profile.php?id=1542024891ref=name
So the subquery that I gave you doesn't work?  Run it and then do a
print_r($row); and post what you get.

-Shawn


 Terion

 On Wed, Jan 28, 2009 at 7:39 PM, Shawn McKenzie nos...@mckenzies.net
 mailto:nos...@mckenzies.net wrote:

 Chris wrote:
 
  The main problem is that you've never explained what you want
 to get
  from the query.  The replies have used your code as an example
 and I'm
  pretty sure that's not what you want.  Unless I totally
 mis-understand
  what you want, you have 2 options:
 
  1. Use the 2 queries that I gave you in a previous post.
  2. Use a subquery:
 
  $sql = SELECT * FROM workorders WHERE AdminID = (SELECT
 AdminID FROM
  admin WHERE UserName = ' .
  mysql_real_escape_string($_SESSION['user']) . ');
 
  3 - fix the join ;)
 

 Yes, however, I'm going out on a limb here because we don't really
 know
 what he wants - he is only getting admin.AdminID, workorders.AdminID
 returned in all of the queries I've seen.  I'm assuming that he wants
 some of the workorder details.

 --
 Thanks!
 -Shawn
 http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

2009-01-29 Thread Terion Miller

Hi Again

Here is the query and code I tried:
$sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM admin
WHERE UserName = '
. mysql_real_escape_string($_SESSION['user']) . ');


  $result2 = mysql_query ($sql);
  $row2 = mysql_fetch_assoc ($result2);
   $printrow = print_r($row2);



Here is my print variables--- Nothing printed with the print_r:

$sqlSELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM admin
WHERE UserName = 'tmiller')$result2Resource id #6$row2$printrow1


So the subquery that I gave you doesn't work?  Run it and then do a
 print_r($row); and post what you get.

 -Shawn

 
  Terion
 
  On Wed, Jan 28, 2009 at 7:39 PM, Shawn McKenzie nos...@mckenzies.net
  mailto:nos...@mckenzies.net wrote:
 
  Chris wrote:
  
   The main problem is that you've never explained what you want
  to get
   from the query.  The replies have used your code as an example
  and I'm
   pretty sure that's not what you want.  Unless I totally
  mis-understand
   what you want, you have 2 options:
  
   1. Use the 2 queries that I gave you in a previous post.
   2. Use a subquery:
  
   $sql = SELECT * FROM workorders WHERE AdminID = (SELECT
  AdminID FROM
   admin WHERE UserName = ' .
   mysql_real_escape_string($_SESSION['user']) . ');
  
   3 - fix the join ;)
  
 
  Yes, however, I'm going out on a limb here because we don't really
  know
  what he wants - he is only getting admin.AdminID, workorders.AdminID
  returned in all of the queries I've seen.  I'm assuming that he wants
  some of the workorder details.
 
  --
  Thanks!
  -Shawn
  http://www.spidean.com
 
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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Terion Miller wrote:
 Hi Again

 Here is the query and code I tried:
 $sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM
 admin WHERE UserName = '
 . mysql_real_escape_string($_SESSION['user']) . ');


   $result2 = mysql_query ($sql);
   $row2 = mysql_fetch_assoc ($result2);
$printrow = print_r($row2);



 Here is my print variables--- Nothing printed with the print_r:

 $sql  SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM
 admin WHERE UserName = 'tmiller')
 $result2  Resource id #6
 $row2 
 $printrow 1

O.K., easy:

1. There is no UserName in admin that = $_SESSION['user']
2. Or, the AdminID that is returned for $_SESSION['user'] is not present
in workorders

That's why I suggested the 2 query approach the first time so that you
could echo out the AdminID after the first query and then go into the db
and make sure it exists in workorders.  Or, if none were returned, then
echo out $_SESSION['user'] and make sure it exists in admin.

-Shawn

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

2009-01-29 Thread Terion Miller

Well upon looking it was number 2, I had no orders in the workorder table,
but here is an oddity I wonder if someone can explain if I run this simple
query:

 $query =  SELECT * FROM admin, workorders
  WHERE   admin.UserName =   '.$_SESSION['user'].' ;

It returns adminID = 7 (my number is 20) username= tmiller (this is correct)


it's the only query I can get to return anything but it's the wrong adminID
can someone explain what can be making that happen

On Thu, Jan 29, 2009 at 2:48 PM, Shawn McKenzie nos...@mckenzies.netwrote:

 Terion Miller wrote:
  Hi Again
 
  Here is the query and code I tried:
  $sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM
  admin WHERE UserName = '
  . mysql_real_escape_string($_SESSION['user']) . ');
 
 
$result2 = mysql_query ($sql);
$row2 = mysql_fetch_assoc ($result2);
 $printrow = print_r($row2);
 
 
 
  Here is my print variables--- Nothing printed with the print_r:
 
  $sql  SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM
  admin WHERE UserName = 'tmiller')
  $result2  Resource id #6
  $row2 
  $printrow 1
 
 O.K., easy:

 1. There is no UserName in admin that = $_SESSION['user']
 2. Or, the AdminID that is returned for $_SESSION['user'] is not present
 in workorders

 That's why I suggested the 2 query approach the first time so that you
 could echo out the AdminID after the first query and then go into the db
 and make sure it exists in workorders.  Or, if none were returned, then
 echo out $_SESSION['user'] and make sure it exists in admin.

 -Shawn

Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Terion Miller wrote:
 Well upon looking it was number 2, I had no orders in the workorder table,
 but here is an oddity I wonder if someone can explain if I run this simple
 query:
 
  $query =  SELECT * FROM admin, workorders
   WHERE   admin.UserName =   '.$_SESSION['user'].' ;
 
 It returns adminID = 7 (my number is 20) username= tmiller (this is correct)
 
 
 it's the only query I can get to return anything but it's the wrong adminID
 can someone explain what can be making that happen
 

You cannot return related values from 2 tables without joining the
tables in the query which we've shown is not going to work because you
need to get AdminID from admin before you can join it to workorders.

What your query says is, give me all fields for all rows from admin,
where UserName = the session user, and give me the same number of rows
from workorders arbitrarily or just the first however many rows.  I
would be willing to bet that the first record in workorders has AdminID = 7.

This works, obviously:

$query =  SELECT * FROM admin WHERE admin.UserName =
'.$_SESSION['user'].' ;

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Shawn McKenzie wrote:
 Terion Miller wrote:
 Well upon looking it was number 2, I had no orders in the workorder table,
 but here is an oddity I wonder if someone can explain if I run this simple
 query:

  $query =  SELECT * FROM admin, workorders
   WHERE   admin.UserName =   '.$_SESSION['user'].' ;

 It returns adminID = 7 (my number is 20) username= tmiller (this is correct)


 it's the only query I can get to return anything but it's the wrong adminID
 can someone explain what can be making that happen

 
 You cannot return related values from 2 tables without joining the
 tables in the query which we've shown is not going to work because you
 need to get AdminID from admin before you can join it to workorders.
 
 What your query says is, give me all fields for all rows from admin,
 where UserName = the session user, and give me the same number of rows
 from workorders arbitrarily or just the first however many rows.  I
 would be willing to bet that the first record in workorders has AdminID = 7.
 
 This works, obviously:
 
 $query =  SELECT * FROM admin WHERE admin.UserName =
 '.$_SESSION['user'].' ;
 

I also bet that now that you know there are no valid records for your
query in the workorders table, that the join queries that others have
provided will work great!

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Db fields?

2009-01-28 Thread Andrew Ballard

On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller webdev.ter...@gmail.com wrote:
 Not sure if I'm wording this right, what I am trying to do is look in two
 tables, match the ID to use to pull information

 Here's my code but it's not right, although it is picking up the user from
 the session, I will also post what my variable debugging lists:

$query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
 workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
$result = mysql_query ($query);
$row = mysql_fetch_assoc ($result);

echo $row['AdminID'];

if ($row['ViewMyOrders'] == NO) {
header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you do
 not have access to that page.);
}

 *Also tried this to pull just this persons orders:*

$sql = SELECT workorders.WorkOrderID , workorders.AdminID,
 admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
 admin.AdminID ;
$result = mysql_query ($sql);

 Thanks for looking, t.


Your first version gives you a Cartesian product containing more rows
than you are expecting. (All rows from the workorders table joined
with the row in the admin table where the username matches.) The
second version returns all rows where the AdminIDs match, but for all
users. You need to combine them:

$sql =
SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
FROM workorders, admin
WHERE workorders.AdminID = admin.AdminID
  AND admin.UserName = ' . mysql_real_escape_string($username) . ';


Although I believe the preferred syntax (at least, I think it's the
preferred) is

$sql =
SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
FROM workorders
 INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';


Andrew

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Re: [PHP] Making a Variable from different tables with Matching Db fields?

Well I'm stuck I have the AdminID but now I can't seem to use it to pull
workorders with that AdminID . I couldn't get your block to work Andrew :(

I think I'm just not using it right now that I have it...lol


On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com wrote:

 On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller webdev.ter...@gmail.com
 wrote:
  Not sure if I'm wording this right, what I am trying to do is look in two
  tables, match the ID to use to pull information
 
  Here's my code but it's not right, although it is picking up the user
 from
  the session, I will also post what my variable debugging lists:
 
 $query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
  workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
 $result = mysql_query ($query);
 $row = mysql_fetch_assoc ($result);
 
 echo $row['AdminID'];
 
 if ($row['ViewMyOrders'] == NO) {
 header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you do
  not have access to that page.);
 }
 
  *Also tried this to pull just this persons orders:*
 
 $sql = SELECT workorders.WorkOrderID , workorders.AdminID,
  admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
  admin.AdminID ;
 $result = mysql_query ($sql);
 
  Thanks for looking, t.
 

 Your first version gives you a Cartesian product containing more rows
 than you are expecting. (All rows from the workorders table joined
 with the row in the admin table where the username matches.) The
 second version returns all rows where the AdminIDs match, but for all
 users. You need to combine them:

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders, admin
 WHERE workorders.AdminID = admin.AdminID
   AND admin.UserName = ' . mysql_real_escape_string($username) . ';


 Although I believe the preferred syntax (at least, I think it's the
 preferred) is

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders
  INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
 WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';


 Andrew

Re: [PHP] Making a Variable from different tables with Matching Db fields?

2009-01-28 Thread Christoph Boget

 Well I'm stuck I have the AdminID but now I can't seem to use it to pull
 workorders with that AdminID . I couldn't get your block to work Andrew :(
 I think I'm just not using it right now that I have it...lol

Because there is ambiguity w/r/t the columns you are selecting, you'll
need to use aliases.

http://dev.mysql.com/doc/refman/5.1/en/identifiers.html

In your code, when you are referencing the column, do so using the
alias.  That should solve your problem.

thnx,
Christoph


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Re: [PHP] Making a Variable from different tables with Matching Db fields?

I just read it 3 times and I don't understand it.

On Wed, Jan 28, 2009 at 3:22 PM, Christoph Boget
christoph.bo...@gmail.comwrote:

  Well I'm stuck I have the AdminID but now I can't seem to use it to pull
  workorders with that AdminID . I couldn't get your block to work Andrew
 :(
  I think I'm just not using it right now that I have it...lol

 Because there is ambiguity w/r/t the columns you are selecting, you'll
 need to use aliases.

 http://dev.mysql.com/doc/refman/5.1/en/identifiers.html

 In your code, when you are referencing the column, do so using the
 alias.  That should solve your problem.

 thnx,
 Christoph

Re: [PHP] Making a Variable from different tables with Matching Db fields?

2009-01-28 Thread Andrew Ballard

On Wed, Jan 28, 2009 at 4:22 PM, Christoph Boget
christoph.bo...@gmail.com wrote:
 Well I'm stuck I have the AdminID but now I can't seem to use it to pull
 workorders with that AdminID . I couldn't get your block to work Andrew :(
 I think I'm just not using it right now that I have it...lol

 Because there is ambiguity w/r/t the columns you are selecting, you'll
 need to use aliases.

 http://dev.mysql.com/doc/refman/5.1/en/identifiers.html

 In your code, when you are referencing the column, do so using the
 alias.  That should solve your problem.

 thnx,
 Christoph



You're right about the ambiguity. I didn't even pay attention to his
SELECT list. Actually, though, in this case aliases aren't even
necessary. Since workorders.AdminID and admin.AdminID are going to be
the same value since that's a join condition, so the OP really only
needs to choose one and use it.

Andrew

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Re: [PHP] Making a Variable from different tables with Matching Db fields?

2009-01-28 Thread Andrew Ballard

On Wed, Jan 28, 2009 at 4:31 PM, Terion Miller webdev.ter...@gmail.com wrote:
 On Wed, Jan 28, 2009 at 3:22 PM, Christoph Boget christoph.bo...@gmail.com
 wrote:

  Well I'm stuck I have the AdminID but now I can't seem to use it to pull
  workorders with that AdminID . I couldn't get your block to work Andrew
  :(
  I think I'm just not using it right now that I have it...lol

 Because there is ambiguity w/r/t the columns you are selecting, you'll
 need to use aliases.

 http://dev.mysql.com/doc/refman/5.1/en/identifiers.html

 In your code, when you are referencing the column, do so using the
 alias.  That should solve your problem.

 thnx,
 Christoph

 I just read it 3 times and I don't understand it.


Try this... (just make sure to assign a valid $username value.)

$sql =
SELECT workorders.WorkOrderID , workorders.AdminID
FROM workorders
INNER JOIN
  admin
ON  workorders.AdminID = admin.AdminID
WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';

Andrew

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Re: [PHP] Making a Variable from different tables with Matching Db fields?

Terion Miller wrote:
 Well I'm stuck I have the AdminID but now I can't seem to use it to pull
 workorders with that AdminID . I couldn't get your block to work Andrew :(
 
 I think I'm just not using it right now that I have it...lol
 
 
 On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com wrote:
 
 On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller webdev.ter...@gmail.com
 wrote:
 Not sure if I'm wording this right, what I am trying to do is look in two
 tables, match the ID to use to pull information

 Here's my code but it's not right, although it is picking up the user
 from
 the session, I will also post what my variable debugging lists:

$query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
 workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
$result = mysql_query ($query);
$row = mysql_fetch_assoc ($result);

echo $row['AdminID'];

if ($row['ViewMyOrders'] == NO) {
header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you do
 not have access to that page.);
}

 *Also tried this to pull just this persons orders:*

$sql = SELECT workorders.WorkOrderID , workorders.AdminID,
 admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
 admin.AdminID ;
$result = mysql_query ($sql);

 Thanks for looking, t.

 Your first version gives you a Cartesian product containing more rows
 than you are expecting. (All rows from the workorders table joined
 with the row in the admin table where the username matches.) The
 second version returns all rows where the AdminIDs match, but for all
 users. You need to combine them:

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders, admin
 WHERE workorders.AdminID = admin.AdminID
   AND admin.UserName = ' . mysql_real_escape_string($username) . ';


 Although I believe the preferred syntax (at least, I think it's the
 preferred) is

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders
  INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
 WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';


 Andrew

 

I think I see what you're trying to do:

$query = SELECT AdminID FROM admin WHERE UserName = '
. mysql_real_escape_string($_SESSION['user']) . ';
$result = mysql_query($query);
$admins = mysql_fetch_assoc($result);

$query = SELECT * FROM workorders WHERE AdminID = '
. $admins['AdminID'] . ';
$result = mysql_query($query);
$workorders = mysql_fetch_assoc($result);

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Db fields?

2009-01-28 Thread Christoph Boget

  Because there is ambiguity w/r/t the columns you are selecting, you'll
  need to use aliases.
  http://dev.mysql.com/doc/refman/5.1/en/identifiers.html
  In your code, when you are referencing the column, do so using the
  alias.  That should solve your problem.
  I just read it 3 times and I don't understand it.
 Try this... (just make sure to assign a valid $username value.)
 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID
 FROM workorders
 INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
 WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';

Andrew is right.  If you use the above query, that will probably get you
going.  But in general, aliases allow you to reference a column as
something other than what it is named in the table's schema.  It would
allow you to do something along the lines of:

SELECT
  Employee.Name AS EmployeeName,
  Employer.Name AS EmployerName
FROM
  blah blah blah.

In the above example, the column Name is ambiguous.  By defining an
alias, it removes the ambiguity and allows mysql to return the value of
both columns.  In your code, you would just reference it using:

$rowData['EmployeeName']
$rowData['EmployerName']

I hope that helps explain aliases a bit.

thnx,
Christoph


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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Shawn McKenzie wrote:
 Terion Miller wrote:
 Well I'm stuck I have the AdminID but now I can't seem to use it to pull
 workorders with that AdminID . I couldn't get your block to work Andrew :(

 I think I'm just not using it right now that I have it...lol


 On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com wrote:

 On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller webdev.ter...@gmail.com
 wrote:
 Not sure if I'm wording this right, what I am trying to do is look in two
 tables, match the ID to use to pull information

 Here's my code but it's not right, although it is picking up the user
 from
 the session, I will also post what my variable debugging lists:

$query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
 workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
$result = mysql_query ($query);
$row = mysql_fetch_assoc ($result);

echo $row['AdminID'];

if ($row['ViewMyOrders'] == NO) {
header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you do
 not have access to that page.);
}

 *Also tried this to pull just this persons orders:*

$sql = SELECT workorders.WorkOrderID , workorders.AdminID,
 admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
 admin.AdminID ;
$result = mysql_query ($sql);

 Thanks for looking, t.

 Your first version gives you a Cartesian product containing more rows
 than you are expecting. (All rows from the workorders table joined
 with the row in the admin table where the username matches.) The
 second version returns all rows where the AdminIDs match, but for all
 users. You need to combine them:

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders, admin
 WHERE workorders.AdminID = admin.AdminID
   AND admin.UserName = ' . mysql_real_escape_string($username) . ';


 Although I believe the preferred syntax (at least, I think it's the
 preferred) is

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders
  INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
 WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';


 Andrew

 
 I think I see what you're trying to do:
 
 $query = SELECT AdminID FROM admin WHERE UserName = '
 . mysql_real_escape_string($_SESSION['user']) . ';
 $result = mysql_query($query);
 $admins = mysql_fetch_assoc($result);
 
 $query = SELECT * FROM workorders WHERE AdminID = '
 . $admins['AdminID'] . ';
 $result = mysql_query($query);
 $workorders = mysql_fetch_assoc($result);
 

Well maybe not.  Has anyone noticed that all the proposed selects
including the OPs are only returning AdminID and WorkOrderID?  But in
the OPs code he's trying to use $row['ViewMyOrders']!

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

On Wed, Jan 28, 2009 at 3:43 PM, Shawn McKenzie nos...@mckenzies.netwrote:

 Shawn McKenzie wrote:
  Terion Miller wrote:
  Well I'm stuck I have the AdminID but now I can't seem to use it to pull
  workorders with that AdminID . I couldn't get your block to work Andrew
 :(
 
  I think I'm just not using it right now that I have it...lol
 
 
  On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com
 wrote:
 
  On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller 
 webdev.ter...@gmail.com
  wrote:
  Not sure if I'm wording this right, what I am trying to do is look in
 two
  tables, match the ID to use to pull information
 
  Here's my code but it's not right, although it is picking up the user
  from
  the session, I will also post what my variable debugging lists:
 
 $query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
  workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
 $result = mysql_query ($query);
 $row = mysql_fetch_assoc ($result);
 
 echo $row['AdminID'];
 
 if ($row['ViewMyOrders'] == NO) {
 header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you
 do
  not have access to that page.);
 }
 
  *Also tried this to pull just this persons orders:*
 
 $sql = SELECT workorders.WorkOrderID , workorders.AdminID,
  admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
  admin.AdminID ;
 $result = mysql_query ($sql);
 
  Thanks for looking, t.
 
  Your first version gives you a Cartesian product containing more rows
  than you are expecting. (All rows from the workorders table joined
  with the row in the admin table where the username matches.) The
  second version returns all rows where the AdminIDs match, but for all
  users. You need to combine them:
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders, admin
  WHERE workorders.AdminID = admin.AdminID
AND admin.UserName = ' . mysql_real_escape_string($username) . ';
 
 
  Although I believe the preferred syntax (at least, I think it's the
  preferred) is
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders
   INNER JOIN
admin
  ON  workorders.AdminID = admin.AdminID
  WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';
 
 
  Andrew
 
 
  I think I see what you're trying to do:
 
  $query = SELECT AdminID FROM admin WHERE UserName = '
  . mysql_real_escape_string($_SESSION['user']) . ';
  $result = mysql_query($query);
  $admins = mysql_fetch_assoc($result);
 
  $query = SELECT * FROM workorders WHERE AdminID = '
  . $admins['AdminID'] . ';
  $result = mysql_query($query);
  $workorders = mysql_fetch_assoc($result);
 

 Well maybe not.  Has anyone noticed that all the proposed selects
 including the OPs are only returning AdminID and WorkOrderID?  But in
 the OPs code he's trying to use $row['ViewMyOrders']!

 --
 Thanks!
 -Shawn
 http://www.spidean.com


I have to get only the work orders associated with the adminID, I get the
pages but no orders.  and if I print my variables I am grabbing the right
adminID but it's not then going and grabbing the work orders with it.  I'm
not up on the correct phrasing, been doing this about 2 months.

Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Terion Miller wrote:
 On Wed, Jan 28, 2009 at 3:43 PM, Shawn McKenzie nos...@mckenzies.netwrote:
 
 Shawn McKenzie wrote:
 Terion Miller wrote:
 Well I'm stuck I have the AdminID but now I can't seem to use it to pull
 workorders with that AdminID . I couldn't get your block to work Andrew
 :(
 I think I'm just not using it right now that I have it...lol


 On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com
 wrote:
 On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller 
 webdev.ter...@gmail.com
 wrote:
 Not sure if I'm wording this right, what I am trying to do is look in
 two
 tables, match the ID to use to pull information

 Here's my code but it's not right, although it is picking up the user
 from
 the session, I will also post what my variable debugging lists:

$query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
 workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
$result = mysql_query ($query);
$row = mysql_fetch_assoc ($result);

echo $row['AdminID'];

if ($row['ViewMyOrders'] == NO) {
header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry, you
 do
 not have access to that page.);
}

 *Also tried this to pull just this persons orders:*

$sql = SELECT workorders.WorkOrderID , workorders.AdminID,
 admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
 admin.AdminID ;
$result = mysql_query ($sql);

 Thanks for looking, t.

 Your first version gives you a Cartesian product containing more rows
 than you are expecting. (All rows from the workorders table joined
 with the row in the admin table where the username matches.) The
 second version returns all rows where the AdminIDs match, but for all
 users. You need to combine them:

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders, admin
 WHERE workorders.AdminID = admin.AdminID
   AND admin.UserName = ' . mysql_real_escape_string($username) . ';


 Although I believe the preferred syntax (at least, I think it's the
 preferred) is

 $sql =
 SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
 FROM workorders
  INNER JOIN
   admin
 ON  workorders.AdminID = admin.AdminID
 WHERE admin.UserName = ' . mysql_real_escape_string($username) . ';


 Andrew

 I think I see what you're trying to do:

 $query = SELECT AdminID FROM admin WHERE UserName = '
 . mysql_real_escape_string($_SESSION['user']) . ';
 $result = mysql_query($query);
 $admins = mysql_fetch_assoc($result);

 $query = SELECT * FROM workorders WHERE AdminID = '
 . $admins['AdminID'] . ';
 $result = mysql_query($query);
 $workorders = mysql_fetch_assoc($result);

 Well maybe not.  Has anyone noticed that all the proposed selects
 including the OPs are only returning AdminID and WorkOrderID?  But in
 the OPs code he's trying to use $row['ViewMyOrders']!

 --
 Thanks!
 -Shawn
 http://www.spidean.com

 
 I have to get only the work orders associated with the adminID, I get the
 pages but no orders.  and if I print my variables I am grabbing the right
 adminID but it's not then going and grabbing the work orders with it.  I'm
 not up on the correct phrasing, been doing this about 2 months.
 
Well, try what I posted (needs some error checking).  Where does
ViewMyOrders come from?  admin table?  It would be even easier if you
put the AdminID in the SESSION also :-)

There also seems to be some design flaws.  Why query the database for
orders if the user is not allowed to view their orders?

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

On Wed, Jan 28, 2009 at 4:00 PM, Shawn McKenzie nos...@mckenzies.netwrote:

 Terion Miller wrote:
  On Wed, Jan 28, 2009 at 3:43 PM, Shawn McKenzie nos...@mckenzies.net
 wrote:
 
  Shawn McKenzie wrote:
  Terion Miller wrote:
  Well I'm stuck I have the AdminID but now I can't seem to use it to
 pull
  workorders with that AdminID . I couldn't get your block to work
 Andrew
  :(
  I think I'm just not using it right now that I have it...lol
 
 
  On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com
  wrote:
  On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller 
  webdev.ter...@gmail.com
  wrote:
  Not sure if I'm wording this right, what I am trying to do is look
 in
  two
  tables, match the ID to use to pull information
 
  Here's my code but it's not right, although it is picking up the
 user
  from
  the session, I will also post what my variable debugging lists:
 
 $query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
  workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
 $result = mysql_query ($query);
 $row = mysql_fetch_assoc ($result);
 
 echo $row['AdminID'];
 
 if ($row['ViewMyOrders'] == NO) {
 header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry,
 you
  do
  not have access to that page.);
 }
 
  *Also tried this to pull just this persons orders:*
 
 $sql = SELECT workorders.WorkOrderID , workorders.AdminID,
  admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
  admin.AdminID ;
 $result = mysql_query ($sql);
 
  Thanks for looking, t.
 
  Your first version gives you a Cartesian product containing more rows
  than you are expecting. (All rows from the workorders table joined
  with the row in the admin table where the username matches.) The
  second version returns all rows where the AdminIDs match, but for all
  users. You need to combine them:
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders, admin
  WHERE workorders.AdminID = admin.AdminID
AND admin.UserName = ' . mysql_real_escape_string($username) .
 ';
 
 
  Although I believe the preferred syntax (at least, I think it's the
  preferred) is
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders
   INNER JOIN
admin
  ON  workorders.AdminID = admin.AdminID
  WHERE admin.UserName = ' . mysql_real_escape_string($username) .
 ';
 
 
  Andrew
 
  I think I see what you're trying to do:
 
  $query = SELECT AdminID FROM admin WHERE UserName = '
  . mysql_real_escape_string($_SESSION['user']) . ';
  $result = mysql_query($query);
  $admins = mysql_fetch_assoc($result);
 
  $query = SELECT * FROM workorders WHERE AdminID = '
  . $admins['AdminID'] . ';
  $result = mysql_query($query);
  $workorders = mysql_fetch_assoc($result);
 
  Well maybe not.  Has anyone noticed that all the proposed selects
  including the OPs are only returning AdminID and WorkOrderID?  But in
  the OPs code he's trying to use $row['ViewMyOrders']!
 
  --
  Thanks!
  -Shawn
  http://www.spidean.com
 
 
  I have to get only the work orders associated with the adminID, I get the
  pages but no orders.  and if I print my variables I am grabbing the right
  adminID but it's not then going and grabbing the work orders with it.
  I'm
  not up on the correct phrasing, been doing this about 2 months.
 
 Well, try what I posted (needs some error checking).  Where does
 ViewMyOrders come from?  admin table?  It would be even easier if you
 put the AdminID in the SESSION also :-)

 There also seems to be some design flaws.  Why query the database for
 orders if the user is not allowed to view their orders?

 --
 Thanks!
 -Shawn
 http://www.spidean.com

 --
 PHP General Mailing List (http://www.php.net/)
 To unsubscribe, visit: http://www.php.net/unsub.php

 No the user is allowed to view them, or that is what I'm trying to do
exactly , now I have it returning some orders but they don't belong to the
correct AdminID , I'm getting closer, I appreciate everyone's help in the
right direction!!
Terion

Re: [PHP] Making a Variable from different tables with Matching Dbfields?

On Wed, Jan 28, 2009 at 4:12 PM, Terion Miller webdev.ter...@gmail.comwrote:



 On Wed, Jan 28, 2009 at 4:00 PM, Shawn McKenzie nos...@mckenzies.netwrote:

 Terion Miller wrote:
  On Wed, Jan 28, 2009 at 3:43 PM, Shawn McKenzie nos...@mckenzies.net
 wrote:
 
  Shawn McKenzie wrote:
  Terion Miller wrote:
  Well I'm stuck I have the AdminID but now I can't seem to use it to
 pull
  workorders with that AdminID . I couldn't get your block to work
 Andrew
  :(
  I think I'm just not using it right now that I have it...lol
 
 
  On Wed, Jan 28, 2009 at 2:26 PM, Andrew Ballard aball...@gmail.com
  wrote:
  On Wed, Jan 28, 2009 at 3:18 PM, Terion Miller 
  webdev.ter...@gmail.com
  wrote:
  Not sure if I'm wording this right, what I am trying to do is look
 in
  two
  tables, match the ID to use to pull information
 
  Here's my code but it's not right, although it is picking up the
 user
  from
  the session, I will also post what my variable debugging lists:
 
 $query = SELECT  admin.AdminID, workorders.AdminID FROM admin,
  workorders WHERE admin.UserName =   '.$_SESSION['user'].' ;
 $result = mysql_query ($query);
 $row = mysql_fetch_assoc ($result);
 
 echo $row['AdminID'];
 
 if ($row['ViewMyOrders'] == NO) {
 header (Location: Welcome.php?AdminID=$AdminIDmsg=Sorry,
 you
  do
  not have access to that page.);
 }
 
  *Also tried this to pull just this persons orders:*
 
 $sql = SELECT workorders.WorkOrderID , workorders.AdminID,
  admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
  admin.AdminID ;
 $result = mysql_query ($sql);
 
  Thanks for looking, t.
 
  Your first version gives you a Cartesian product containing more
 rows
  than you are expecting. (All rows from the workorders table joined
  with the row in the admin table where the username matches.) The
  second version returns all rows where the AdminIDs match, but for
 all
  users. You need to combine them:
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders, admin
  WHERE workorders.AdminID = admin.AdminID
AND admin.UserName = ' . mysql_real_escape_string($username) .
 ';
 
 
  Although I believe the preferred syntax (at least, I think it's the
  preferred) is
 
  $sql =
  SELECT workorders.WorkOrderID , workorders.AdminID, admin.AdminID
  FROM workorders
   INNER JOIN
admin
  ON  workorders.AdminID = admin.AdminID
  WHERE admin.UserName = ' . mysql_real_escape_string($username) .
 ';
 
 
  Andrew
 
  I think I see what you're trying to do:
 
  $query = SELECT AdminID FROM admin WHERE UserName = '
  . mysql_real_escape_string($_SESSION['user']) . ';
  $result = mysql_query($query);
  $admins = mysql_fetch_assoc($result);
 
  $query = SELECT * FROM workorders WHERE AdminID = '
  . $admins['AdminID'] . ';
  $result = mysql_query($query);
  $workorders = mysql_fetch_assoc($result);
 
  Well maybe not.  Has anyone noticed that all the proposed selects
  including the OPs are only returning AdminID and WorkOrderID?  But in
  the OPs code he's trying to use $row['ViewMyOrders']!
 
  --
  Thanks!
  -Shawn
  http://www.spidean.com
 
 
  I have to get only the work orders associated with the adminID, I get
 the
  pages but no orders.  and if I print my variables I am grabbing the
 right
  adminID but it's not then going and grabbing the work orders with it.
  I'm
  not up on the correct phrasing, been doing this about 2 months.
 
 Well, try what I posted (needs some error checking).  Where does
 ViewMyOrders come from?  admin table?  It would be even easier if you
 put the AdminID in the SESSION also :-)

 There also seems to be some design flaws.  Why query the database for
 orders if the user is not allowed to view their orders?

 --
 Thanks!
 -Shawn
 http://www.spidean.com

 --
 PHP General Mailing List (http://www.php.net/)
 To unsubscribe, visit: http://www.php.net/unsub.php

 No the user is allowed to view them, or that is what I'm trying to do
 exactly , now I have it returning some orders but they don't belong to the
 correct AdminID , I'm getting closer, I appreciate everyone's help in the
 right direction!!
 Terion


Here are my variables when I reveal them, I am picking up the right adminID
I can't figure out why it's returning random orders though:
$querySELECT admin.AdminID, workorders.AdminID FROM admin, workorders WHERE
admin.UserName = 'tmiller' $resultResource id #5$rowkeyvalue[WorkOrderID]
44[AdminID]7$SortByWorkOrderID DESC$Page2$PerPage30$StartPage30
$sqlSELECT workorders.WorkOrderID, workorders.AdminID, admin.AdminID FROM
workorders, admin WHERE workorders.AdminID = admin.AdminID $Total3

Re: [PHP] Making a Variable from different tables with Matching Dbfields?



 Here are my variables when I reveal them, I am picking up the right
 adminID I can't figure out why it's returning random orders though:
 $querySELECT admin.AdminID, workorders.AdminID FROM admin,
 workorders WHERE admin.UserName = 'tmiller' 
 $result   Resource id #5
 $row  
 key   value
 [WorkOrderID] 44
 [AdminID] 7

 $SortBy   WorkOrderID DESC
 $Page 2
 $PerPage  30
 $StartPage30
 $sql  SELECT workorders.WorkOrderID, workorders.AdminID,
 admin.AdminID FROM workorders, admin WHERE workorders.AdminID =
 admin.AdminID 
 $Total3


Because your queries are hosed.  You want to populate $row with what? 
If you want all the fields in workorders, then this works great:

SELECT * FROM workorders WHERE AdminID = 7

-Shawn

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?




Here are my variables when I reveal them, I am picking up the right adminID
I can't figure out why it's returning random orders though:
$querySELECT admin.AdminID, workorders.AdminID FROM admin, workorders WHERE
admin.UserName = 'tmiller' 


Please please please trim your posts to relevant stuff. I had to go 
through 4 pages of previous attempts to try and find what you posted. 
Trim it down to what you need to post.


It's picking random stuff because you're not joining the tables properly.

You have:

select
  admin.AdminID, workorers.AdminID
from
  admin, workorders
WHERE
  admin.userName='tmiller';

You haven't told the db how to join the two tables together, so it's 
doing a cross or cartesian join (useful in some cases, but not here). 
What that means is for each row in 'admin', it will show every row in 
'workorders' and vice versa.


What you want is to only show rows that match up:

select
  admin.AdminID, workorers.AdminID
from
  admin inner join workorders using (AdminID)
WHERE
  admin.userName='tmiller';


which means

for every row in admin, make sure there is a matching row (based on the 
adminid) in workorders.


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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

I'm not sure what you mean by trim the posts, please explain so I can spare
folks from redundant text.

Your  post made perfect sense to me about the INNER JOIN , I looked it up
but it is not returning the AdminID, maybe my syntax is wrong?

$query =  SELECT admin.AdminID , workorders.AdminID
 FROM admin
 INNER JOIN
 workorders
 ON  AdminID(admin, workorders)
 WHERE admin.UserName =   '.$_SESSION['user'].' ;

$result = mysql_query ($query);
$row = mysql_fetch_assoc ($result);

echo $row['AdminID'];

thanks again!!
terion


On Wed, Jan 28, 2009 at 4:28 PM, Chris dmag...@gmail.com wrote:


  Here are my variables when I reveal them, I am picking up the right
 adminID
 I can't figure out why it's returning random orders though:
 $querySELECT admin.AdminID, workorders.AdminID FROM admin, workorders
 WHERE
 admin.UserName = 'tmiller' 


 Please please please trim your posts to relevant stuff. I had to go through
 4 pages of previous attempts to try and find what you posted. Trim it down
 to what you need to post.

 It's picking random stuff because you're not joining the tables properly.

 You have:

 select
  admin.AdminID, workorers.AdminID
 from
  admin, workorders
 WHERE
  admin.userName='tmiller';

 You haven't told the db how to join the two tables together, so it's doing
 a cross or cartesian join (useful in some cases, but not here). What that
 means is for each row in 'admin', it will show every row in 'workorders' and
 vice versa.

 What you want is to only show rows that match up:

 select
  admin.AdminID, workorers.AdminID
 from
  admin inner join workorders using (AdminID)
 WHERE
  admin.userName='tmiller';


 which means

 for every row in admin, make sure there is a matching row (based on the
 adminid) in workorders.

 --
 Postgresql  php tutorials
 http://www.designmagick.com/

Re: [PHP] Making a Variable from different tables with Matching Dbfields?


Terion Miller wrote:
I'm not sure what you mean by trim the posts, please explain so I can 
spare folks from redundant text. 


Before you post, remove any text that you're not referencing (in this 
case I removed my suggestion).


Your  post made perfect sense to me about the INNER JOIN , I looked it 
up but it is not returning the AdminID, maybe my syntax is wrong?


$query =  SELECT admin.AdminID , workorders.AdminID
 FROM admin
 INNER JOIN
 workorders
 ON  AdminID(admin, workorders)
 WHERE admin.UserName =   '.$_SESSION['user'].' ;


The syntax is wrong.

inner join workorders using (adminid)
^^
only works if both tables are using the same field name, the db will 
expand it into the syntax below. Here you just need to specify the 
fieldname to join on (ie the common one).


or

inner join workorders on (admin.adminID=workorders.adminID)
^^
if the field names are not named the same, eg:

select * from
comments
inner join users on (comments.user_id=users.id)

You have to use the full tablename (or alias) and the field name.

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?



  Your  post made perfect sense to me about the INNER JOIN , I looked it up
 but it is not returning the AdminID, maybe my syntax is wrong?

 $query =  SELECT admin.AdminID , workorders.AdminID
 FROM admin
 INNER JOIN
 workorders
 ON  AdminID(admin, workorders)
 WHERE admin.UserName =   '.$_SESSION['user'].' ;


 The syntax is wrong.

 inner join workorders using (adminid)
 ^^
 only works if both tables are using the same field name, the db will expand
 it into the syntax below. Here you just need to specify the fieldname to
 join on (ie the common one).

 or

 inner join workorders on (admin.adminID=workorders.adminID)
 ^^
 if the field names are not named the same, eg:

 select * from
 comments
 inner join users on (comments.user_id=users.id)

 Well I tried both ways and still cannot get it to pick up the AdminID,

$querySELECT admin.AdminID , workorders.AdminID FROM admin INNER JOIN
workorders on AdminID WHERE admin.UserName = 'tmiller' $result$row
$SortByWorkOrderID DESC$Page1$PerPage30$StartPage0$sqlSELECT
admin.AdminID , workorders.AdminID FROM admin INNER JOIN workorders ON
AdminID(admin, workorders) WHERE admin.UserName = 'tmiller' $Total0

Re: [PHP] Making a Variable from different tables with Matching Dbfields?




Well I tried both ways and still cannot get it to pick up the AdminID,

$query	SELECT admin.AdminID , workorders.AdminID FROM admin INNER JOIN 
workorders on AdminID WHERE admin.UserName = 'tmiller' 

$result 
$row
$SortBy WorkOrderID DESC
$Page   1
$PerPage30
$StartPage  0
$sql	SELECT admin.AdminID , workorders.AdminID FROM admin INNER JOIN 
workorders ON AdminID(admin, workorders) WHERE admin.UserName = 'tmiller' 

$Total  0


That is not the right syntax still.

It needs to be:

select
 admin.AdminID, workorers.AdminID
from
 admin inner join workorders using (AdminID)
WHERE
 admin.userName='tmiller';


or


select
 admin.AdminID, workorers.AdminID
from
 admin inner join workorders ON (admin.AdminID=workorders.AdminID)
WHERE
 admin.userName='tmiller';

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?



Terion Miller wrote:



 Your  post made perfect sense to me about the INNER JOIN , I
 looked it up but it is not returning the AdminID, maybe my
 syntax is wrong?

 $query =  SELECT admin.AdminID , workorders.AdminID
 FROM admin
 INNER JOIN
 workorders
 ON  AdminID(admin, workorders)
 WHERE admin.UserName =   '.$_SESSION['user'].' ;


 The syntax is wrong.

 inner join workorders using (adminid)
 ^^
 only works if both tables are using the same field name, the db
 will expand it into the syntax below. Here you just need to
 specify the fieldname to join on (ie the common one).

 or

 inner join workorders on (admin.adminID=workorders.adminID)
 ^^
 if the field names are not named the same, eg:

 select * from
 comments
 inner join users on (comments.user_id=users.id http://users.id)

 Well I tried both ways and still cannot get it to pick up the AdminID,

The main problem is that you've never explained what you want to get
from the query.  The replies have used your code as an example and I'm
pretty sure that's not what you want.  Unless I totally mis-understand
what you want, you have 2 options:

1. Use the 2 queries that I gave you in a previous post.
2. Use a subquery:

$sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM admin 
WHERE UserName = ' 
. mysql_real_escape_string($_SESSION['user']) . ');

-Shawn





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Re: [PHP] Making a Variable from different tables with Matching Dbfields?




The main problem is that you've never explained what you want to get
from the query.  The replies have used your code as an example and I'm
pretty sure that's not what you want.  Unless I totally mis-understand
what you want, you have 2 options:

1. Use the 2 queries that I gave you in a previous post.
2. Use a subquery:

$sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM admin WHERE UserName = ' 
. mysql_real_escape_string($_SESSION['user']) . ');


3 - fix the join ;)

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Re: [PHP] Making a Variable from different tables with Matching Dbfields?

Chris wrote:
 
 The main problem is that you've never explained what you want to get
 from the query.  The replies have used your code as an example and I'm
 pretty sure that's not what you want.  Unless I totally mis-understand
 what you want, you have 2 options:

 1. Use the 2 queries that I gave you in a previous post.
 2. Use a subquery:

 $sql = SELECT * FROM workorders WHERE AdminID = (SELECT AdminID FROM
 admin WHERE UserName = ' .
 mysql_real_escape_string($_SESSION['user']) . ');
 
 3 - fix the join ;)
 

Yes, however, I'm going out on a limb here because we don't really know
what he wants - he is only getting admin.AdminID, workorders.AdminID
returned in all of the queries I've seen.  I'm assuming that he wants
some of the workorder details.

-- 
Thanks!
-Shawn
http://www.spidean.com

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Re: [PHP] Remote File Variable Injection Safety?

2009-01-07 Thread Stuart

2009/1/7 Daniel Kolbo kolb0...@umn.edu:
 suppose there is a file at http://otherhost.com/remote.php that looks like
 this:

 ?php
 if (!isset($safe_flag))
 {
   die(hacking attempt);
 }
 echo You are in;
 ?

 Suppose i executed the following php file at http://myhost.com/local.php

 ?php
 require_once(http://otherhost.com/remote.php;);
 ?

 Is there any way to get local.php to display You are in, by only modifying
 local.php?  That is, is there a way to set $safe_flag on the remote host as
 one requests a file from the remote host from within local.php?

 I have genuine, academic, non-belligerent intentions when asking this
 question.

Doing this is evil and should be avoided if at all possible. However,
assuming you really need to do it this way...

The best way to validate inclusion is to check the value of
$_SERVER['REMOTE_ADDR'] in the remote script and only allow known IPs.
This is not foolproof but will kill off casual attempts to get the
code.

Alternatively if you change the test for $safe_flag to
$_GET['safe_flag'] and add ?safe_flag=1 to the end of the URL in the
require call that should also work, but is easily copied. You could
randomise safe_flag and the value to make it more difficult, but
checking the IP is far better IMHO.

-Stuart

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Re: [PHP] Remote File Variable Injection Safety?

2009-01-07 Thread ceo


If register_globals is on (ewww!) at otherhost.com, then ?safe_flag on the 
URL will get in.



This is one of the reasons why register_globals should be OFF.



NOTE:

The code you gave does not describe the circumstances whereby $safe_flag is 
set.  There could be all manner of other issues around that code that we 
cannot address without seeing more.



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Re: [PHP] Remote File Variable Injection Safety?

2009-01-07 Thread Daniel Kolbo


c...@l-i-e.com wrote:

If register_globals is on (ewww!) at otherhost.com, then ?safe_flag on the 
URL will get in.

This is one of the reasons why register_globals should be OFF.

NOTE:
The code you gave does not describe the circumstances whereby $safe_flag is 
set.  There could be all manner of other issues around that code that we 
cannot address without seeing more.
  
There is no other code.  This is an example I created to help me 
understand finer points of PHP.

thanks,
dK

Re: [PHP] Passing a variable which is an aggregate function

2008-10-10 Thread Jim Lucas

The Doctor wrote:
 What I am trying to do is to
 pass an aggregate variable which is the sum of
 some smaller variable.
 
 This var also need to finish in the form of $.YY .
 
 Pointers please.

Most of the above can be done using number_format().  The dollar sign you will
have to add yourself.

http://php.net/number_format

-- 
Jim Lucas

   Some men are born to greatness, some achieve greatness,
   and some have greatness thrust upon them.

Twelfth Night, Act II, Scene V
by William Shakespeare


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Re: [PHP] Passing a variable which is an aggregate function

2008-10-10 Thread Andrew Ballard

On Fri, Oct 10, 2008 at 1:33 PM, The Doctor [EMAIL PROTECTED] wrote:
 What I am trying to do is to
 pass an aggregate variable which is the sum of
 some smaller variable.

 This var also need to finish in the form of $.YY .

 Pointers please.

Would you care to explain a little further what you are trying to do.
I can't tell from your post what you mean by an aggregate variable
or the sum of some smaller variable. What are you passing this
to/from? Are you looking for something like this?

http://www.php.net/manual/en/function.array-sum.php

Andrew

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Re: [PHP] Passing a variable which is an aggregate function

2008-10-10 Thread The Doctor

On Fri, Oct 10, 2008 at 01:48:01PM -0400, Andrew Ballard wrote:
 On Fri, Oct 10, 2008 at 1:33 PM, The Doctor [EMAIL PROTECTED] wrote:
  What I am trying to do is to
  pass an aggregate variable which is the sum of
  some smaller variable.
 
  This var also need to finish in the form of $.YY .
 
  Pointers please.
 
 Would you care to explain a little further what you are trying to do.
 I can't tell from your post what you mean by an aggregate variable
 or the sum of some smaller variable. What are you passing this
 to/from? Are you looking for something like this?
 
 http://www.php.net/manual/en/function.array-sum.php
 
 Andrew
 
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 believed to be clean.
 


Certainly:

Here is a relevant sample of the script:

form action=https://esqa.moneris.com/HPPDP/index.php; method=post
input type=hidden name=ps_store_id value=CL535tore1
input type=hidden name=hpp_key value=hpYHLXQ5G6BY
input type=hidden name=charge_total value=$var
input type=hidden name=subject value=PD Solutions Order Form
input type=hidden name=recipients value=[EMAIL PROTECTED]
input type=hidden name=required value=realname,email
input type=hidden name=mail_options value=NoEmpty
input type=hidden name=title value=PD Solutions Order
input type=hidden name=cc_visitor value=1
input type=hidden name=bad_url 
value=https://secure.nl2k.ab.ca/pdsolutions/;
INPUT TYPE=HIDDEN NAME=env_report VALUE=REMOTE_HOST,REMOTE_ADDR







TABLE border=0 cellpadding=0 cellspacing=0 width=90%



tr bgcolor=#EBEBEB

td align=centerfont face=Arial,Helvetica size=2 
color=blackstrongItem/strong/tdtdnbsp;/td

td align=centerfont face=Arial,Helvetica size=2 color=blackstrongDate  
Time/strong/td

td align=centerfont face=Arial,Helvetica size=2 
color=blackstrongQuantity/strong/td

td align=centerfont face=Arial,Helvetica size=2 
color=blackstrongPrice/strong/td

/tr



tr
td align=leftfont face=Arial,Helvetica size=2Pest Control Products 
Act/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 27-2008 12.00 
Noon/td

td align=leftinput value=0 type=text size=2 name=PCPAOCT27/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left
font face=Arial,Helvetica size=2Sprayer Circuits/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 27-2008 3.00pm/td

td align=leftinput value=0 type=text size=2 name=SCOCT27/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=leftfont face=Arial,Helvetica size=2IPM 
Fundamentals/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 27-2008 6.00pm/td

td align=leftinput value=0 type=text size=2 name=IPMOCT27/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left
font face=Arial,Helvetica size=2Acute  Chronic 
Toxicity/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 27-2008 8.00pm/td

td align=leftinput value=0 type=text size=2 name=ACTOCT27/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left

font face=Arial,Helvetica size=2Developing New 
Pesticides/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 28-2008 10.00am/td

td align=leftinput value=0 type=text size=2 name=DNPOCT28 = 
$60.00/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left
font face=Arial,Helvetica size=2Rinsate Management/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 28-2008 12.00 
noon/td

td align=leftinput value=0 type=text size=2 name=RMOCT28/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left
font face=Arial,Helvetica size=2Developing an Environmental 
Plan/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 28-2008 3.00pm/td

td align=leftinput value=0 type=text size=2 name=DEPOCT28/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr


trtd align=left
font face=Arial,Helvetica size=2Sprayer Calibration/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 28-2008 7.00pm/td

td align=leftinput value=0 type=text size=2 name=SCOCT28/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left

font face=Arial,Helvetica size=2Provincial Pesticide 
Legislation/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 29-2008 10.00am/td

td align=leftinput value=0 type=text size=2 name=PPLOCT29/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left

font face=Arial,Helvetica size=2Perception of Risk/td/atdnbsp;/td

td align=leftfont face=Arial,Helvetica size=2October 29-2008 12.00 
noon/td

td align=leftinput value=0 type=text size=2 name=POROCT29/td

td align=rightfont face=Arial,Helvetica size=2$60.00/td/tr



trtd align=left

font face=Arial,Helvetica size=2Environmental 
Monitoring/td/atdnbsp;/td

Re: [PHP] Static method variable

2008-09-18 Thread Thijs Lensselink


Quoting Christoph Boget [EMAIL PROTECTED]:


Perhaps I'm misunderstanding what a static method variable is supposed
to do.  I thought the value would be static for an class' instance but
it appears it is static across all instances of the class.  Consider:

class StaticTest
{
  public function __construct()
  {
  }

  public function test( $newVal )
  {
static $retval = '';

if( $retval == '' )
{
  $retval = $newVal;
}
echo $retval . 'br';
  }
}

$one = new StaticTest();
$one-test( 'joe' );
$two = new StaticTest();
$two-test( 'bob' );

Should it be working that way?

thnx,
Chris



That's exactly how a static class var functions. It's available cross  
all instances of the class.




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Re: [PHP] Static method variable

2008-09-18 Thread Micah Gersten

If you're looking for a persistent variable in one class instance, then
you need a member variable.  If you want it persistent across all class
instances, you want a static variable.

Thank you,
Micah Gersten
onShore Networks
Internal Developer
http://www.onshore.com



Christoph Boget wrote:
 Perhaps I'm misunderstanding what a static method variable is supposed
 to do.  I thought the value would be static for an class' instance but
 it appears it is static across all instances of the class.  Consider:

 class StaticTest
 {
   public function __construct()
   {
   }

   public function test( $newVal )
   {
 static $retval = '';

 if( $retval == '' )
 {
   $retval = $newVal;
 }
 echo $retval . 'br';
   }
 }

 $one = new StaticTest();
 $one-test( 'joe' );
 $two = new StaticTest();
 $two-test( 'bob' );

 Should it be working that way?

 thnx,
 Chris

   

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Re: [PHP] Static method variable

2008-09-18 Thread Jochem Maas


Christoph Boget schreef:

Perhaps I'm misunderstanding what a static method variable is supposed
to do.  I thought the value would be static for an class' instance but
it appears it is static across all instances of the class.  Consider:

class StaticTest
{
  public function __construct()
  {
  }

  public function test( $newVal )
  {
static $retval = '';

if( $retval == '' )
{
  $retval = $newVal;
}
echo $retval . 'br';
  }
}

$one = new StaticTest();
$one-test( 'joe' );
$two = new StaticTest();
$two-test( 'bob' );

Should it be working that way?


yes. it's a function variable that retains state. everyone misread your
opst so far ... this has nothing to do with static class scope, although
it does work in the same way.

you can do the exact same thing with a normal function:

function test( $newVal )
{
static $retval;

if(!isset($retval))
$retval = $newVal;

echo $retval . \n;
}

test('joe');
test('bob');





thnx,
Chris




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Re: [PHP] Re: Passing variable to a page in a frameset


 Frames?!

 As a fellow Wisconsinite and a web developer, I'm going to have to ask you
 to leave the state.  Minnesota can have you.

 :P


 Jay


 PS -  No, but seriously, frames?!?!


There's nothing wrong with a Frame every once in a while!  Granted they
aren't used much anymore, but sometimes they can be useful!

RE: [PHP] Re: Passing variable to a page in a frameset

2008-08-15 Thread Boyd, Todd M.

 -Original Message-
 From: Dan Shirah [mailto:[EMAIL PROTECTED]
 Sent: Friday, August 15, 2008 9:20 AM
 To: Jay Moore
 Cc: php-general@lists.php.net
 Subject: Re: [PHP] Re: Passing variable to a page in a frameset

  Frames?!

  As a fellow Wisconsinite and a web developer, I'm going to have to
 ask you
  to leave the state.  Minnesota can have you.

  :P

  Jay

  PS -  No, but seriously, frames?!?!

 There's nothing wrong with a Frame every once in a while!  Granted
they
 aren't used much anymore, but sometimes they can be useful!

I believe frames and framesets are being phased out of XHTML altogether.
It's all about DIVs nowadays. I think there's an XHTML specification for
something like frames, but it's just as ugly.

Todd Boyd
Web Programmer

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Re: [PHP] Re: Passing variable to a page in a frameset



On Aug 15, 2008, at 9:20 AM, Dan Shirah wrote:



Frames?!


As a fellow Wisconsinite and a web developer, I'm going to have to  
ask you

to leave the state.  Minnesota can have you.

:P


Jay


PS -  No, but seriously, frames?!?!



There's nothing wrong with a Frame every once in a while!  Granted  
they

aren't used much anymore, but sometimes they can be useful!


I know, I know. I'm not a fan of frames either, and have never even  
thought of using them until now. The problem is, our library catalog  
is on a different server and provides results for a consortium of 30  
libraries. So, what we've been doing is including a search box for the  
catalog on web pages. Which works fine, but it's easy for people to  
get caught up in the catalog with no good way to get back to the site  
you came from. So, I created a frame page that has a small button in  
the top frame that takes you back to where you came from. So, the site  
itself isn't using frames, only when you search the catalog. Of  
course, that's if I can get it working...


- jody

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RE: [PHP] Re: Passing variable to a page in a frameset

2008-08-15 Thread Warren Vail

Actually you may want to check back with basic html at the target
parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]
 
 

 -Original Message-
 From: Jay Moore [mailto:[EMAIL PROTECTED] 
 Sent: Friday, August 15, 2008 7:03 AM
 To: php-general@lists.php.net
 Subject: [PHP] Re: Passing variable to a page in a frameset
 
 Jody Cleveland wrote:
  Hello,
  
  I've got a website here: http://beta.menashalibrary.org/about
  
  On every page, i've got a search box at the top. This search box 
  searches the library's web catalog. The problem is, when someone 
  searches, it takes them away from the site. What I'd like to do is 
  take what a person searches for, and load it into the 
 bottom frame of this page:
  
 http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/sa
  lamander/searchframe.html
  
  
  Is there a way, with php, to take what someone puts in the 
 search box 
  and put the results into the bottom frame of a frameset when the 
  originating page does not contain frames?
  
  - jody
 
 Frames?!
 
 As a fellow Wisconsinite and a web developer, I'm going to 
 have to ask you to leave the state.  Minnesota can have you.
 
 :P
 
 
 Jay
 
 
 PS -  No, but seriously, frames?!?!
 
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Re: [PHP] Re: Passing variable to a page in a frameset



On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:


Actually you may want to check back with basic html at the target
parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]


Target won't work for me because the originating page with the search  
box is not part of any frameset. I'm trying to get the search results  
from that page to go to a page that is part of a frameset.


- jody

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Re: [PHP] Re: Passing variable to a page in a frameset


On 15 Aug 2008, at 19:39, Jody Cleveland wrote:

On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:


Actually you may want to check back with basic html at the target
parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]


Target won't work for me because the originating page with the  
search box is not part of any frameset. I'm trying to get the search  
results from that page to go to a page that is part of a frameset.


Are you saying that the frame you want to have the search results  
shown in doesn't exist when the search form is submitted? If it does  
then it doesn't matter where the form is, just specify the target as  
the name of the frame and it will almost certainly work.


If however you want the search to create the frameset when it runs  
then you have a completely different problem which is best solved  
using some sort of session. The script that handles the POST will need  
to store the details of the search somewhere and output the frameset.  
The frame that needs to contain the results would then grab the  
details and run the search outputting the results.


If I'm completely misunderstanding you please feel free to elaborate.

-Stut

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Re: [PHP] Re: Passing variable to a page in a frameset



On Aug 15, 2008, at 1:46 PM, Stut wrote:


On 15 Aug 2008, at 19:39, Jody Cleveland wrote:

On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:


Actually you may want to check back with basic html at the target
parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]


Target won't work for me because the originating page with the  
search box is not part of any frameset. I'm trying to get the  
search results from that page to go to a page that is part of a  
frameset.


Are you saying that the frame you want to have the search results  
shown in doesn't exist when the search form is submitted? If it does  
then it doesn't matter where the form is, just specify the target as  
the name of the frame and it will almost certainly work.


If however you want the search to create the frameset when it runs  
then you have a completely different problem which is best solved  
using some sort of session. The script that handles the POST will  
need to store the details of the search somewhere and output the  
frameset. The frame that needs to contain the results would then  
grab the details and run the search outputting the results.


That is exactly what I want. I apologize for the confusion. I was  
having a hard time trying to put what I was trying to do in words.  
But, yes, your second paragraph is exactly what I want to do. My  
knowledge of PHP is very limited, and I've tried to search for  
something that will do this, but couldn't find anything.


- jody


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RE: [PHP] Re: Passing variable to a page in a frameset

2008-08-15 Thread Warren Vail

Probably something I don't understand about your implementation, but the
form parameter will allow a target parameter, and if the frameset contains a
frame that is named (even one different from the one that contains the
search form), the results should be placed in the target frame.  Without a
target, if the form is outside the frameset, it will replace the entire
frameset, exactly what you describe.

Warren 

 -Original Message-
 From: Jody Cleveland [mailto:[EMAIL PROTECTED] 
 Sent: Friday, August 15, 2008 11:40 AM
 To: php-general@lists.php.net
 Subject: Re: [PHP] Re: Passing variable to a page in a frameset
 
 
 On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:
 
  Actually you may want to check back with basic html at the target
  parameter on your search form statement.
 
  HTH,
 
  Warren Vail
  Vail Systems Technology
  [EMAIL PROTECTED]
 
 Target won't work for me because the originating page with 
 the search box is not part of any frameset. I'm trying to get 
 the search results from that page to go to a page that is 
 part of a frameset.
 
 - jody
 
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 unsubscribe, visit: http://www.php.net/unsub.php
 


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Re: [PHP] Re: Passing variable to a page in a frameset


On 15 Aug 2008, at 19:50, Jody Cleveland wrote:

On Aug 15, 2008, at 1:46 PM, Stut wrote:

On 15 Aug 2008, at 19:39, Jody Cleveland wrote:

On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:


Actually you may want to check back with basic html at the target
parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]


Target won't work for me because the originating page with the  
search box is not part of any frameset. I'm trying to get the  
search results from that page to go to a page that is part of a  
frameset.


Are you saying that the frame you want to have the search results  
shown in doesn't exist when the search form is submitted? If it  
does then it doesn't matter where the form is, just specify the  
target as the name of the frame and it will almost certainly work.


If however you want the search to create the frameset when it runs  
then you have a completely different problem which is best solved  
using some sort of session. The script that handles the POST will  
need to store the details of the search somewhere and output the  
frameset. The frame that needs to contain the results would then  
grab the details and run the search outputting the results.


That is exactly what I want. I apologize for the confusion. I was  
having a hard time trying to put what I was trying to do in words.  
But, yes, your second paragraph is exactly what I want to do. My  
knowledge of PHP is very limited, and I've tried to search for  
something that will do this, but couldn't find anything.


Ok, then I have to ask the question... why frames?

If you really need frames then you need to come up with a way to pass  
the search from the script the search form loads to the specific frame  
in the frameset it outputs. You could do this through a GET parameter,  
or via a session or in several other ways. In any case you'd be far  
better off not using frames if possible, so why frames?


-Stut

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Re: [PHP] Re: Passing variable to a page in a frameset


 That is exactly what I want. I apologize for the confusion. I was having a
 hard time trying to put what I was trying to do in words. But, yes, your
 second paragraph is exactly what I want to do. My knowledge of PHP is very
 limited, and I've tried to search for something that will do this, but
 couldn't find anything.

 - jody


In that case, I would not use frames at all.  I believe in the top frame all
you wanted to store was the search text, right?

Just have your search link do something like this:

html
head
script language=JavaScript
!--
function submitForm() {
 document.search_form.action='
http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html
';
 document.search_form.submit();
}
//--
/script
/head
body
form name=search_form method=post action=
input type=text name=search_name value=
a href=javascript:submitForm()SEARCH/a
/form
/body
/html

And then on
http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html
just
assign your posted search value and make a hidden form field.
html
head
body
?php
$searched_text = $_POST['search_name'];
?
form name=my_search method=post action=
input type=hidden name=search_text value=$searched_text
table width='800' border='0' align='center' cellpadding='2' cellspacing='2'
bordercolor='#00'
tr
td?php echo You searched for: .$searched_text; ?/td
/tr
/table
/form
/body
/html

Re: [PHP] Re: Passing variable to a page in a frameset

On Aug 15, 2008, at 2:05 PM, Stut wrote:

On 15 Aug 2008, at 19:50, Jody Cleveland wrote:

On Aug 15, 2008, at 1:46 PM, Stut wrote:

On 15 Aug 2008, at 19:39, Jody Cleveland wrote:

On Aug 15, 2008, at 1:22 PM, Warren Vail wrote:

Actually you may want to check back with basic html at the
target

parameter on your search form statement.

HTH,

Warren Vail
Vail Systems Technology
[EMAIL PROTECTED]

Target won't work for me because the originating page with the
search box is not part of any frameset. I'm trying to get the
search results from that page to go to a page that is part of a
frameset.

Are you saying that the frame you want to have the search results
shown in doesn't exist when the search form is submitted? If it
does then it doesn't matter where the form is, just specify the
target as the name of the frame and it will almost certainly work.

If however you want the search to create the frameset when it runs
then you have a completely different problem which is best solved
using some sort of session. The script that handles the POST will
need to store the details of the search somewhere and output the
frameset. The frame that needs to contain the results would then
grab the details and run the search outputting the results.

That is exactly what I want. I apologize for the confusion. I was
having a hard time trying to put what I was trying to do in words.
But, yes, your second paragraph is exactly what I want to do. My
knowledge of PHP is very limited, and I've tried to search for
something that will do this, but couldn't find anything.

Ok, then I have to ask the question... why frames?

If you really need frames then you need to come up with a way to
pass the search from the script the search form loads to the
specific frame in the frameset it outputs. You could do this through
a GET parameter, or via a session or in several other ways. In any
case you'd be far better off not using frames if possible, so why
frames?

I work for a consortium of 30 libraries. Each library has their own
website, but they all share the same web catalog. On each library's
website there is a search box to search the catalog, which is on a
completely different server from the websites. We've been finding that
once people use that search box, they get distracted with the catalog
and have no easy way to get back to the library's website. The problem
I was tasked with is, coming up with a way to search the catalog with
an easy way to return to where the user was before they initiated the
search.

The only way I thought to do this was to use a frameset for the search
results. Which, you can see here:

http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html

If anyone has any ideas, other than using frames for the results, I'd
love to hear them. The problem is, there's nothing I can do on the web
catalog end.

- jody

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Re: [PHP] Re: Passing variable to a page in a frameset

On 15 Aug 2008, at 20:21, Jody Cleveland wrote:

I work for a consortium of 30 libraries. Each library has their own
website, but they all share the same web catalog. On each library's
website there is a search box to search the catalog, which is on a
completely different server from the websites. We've been finding
that once people use that search box, they get distracted with the
catalog and have no easy way to get back to the library's website.
The problem I was tasked with is, coming up with a way to search the
catalog with an easy way to return to where the user was before they
initiated the search.

The only way I thought to do this was to use a frameset for the
search results. Which, you can see here:

http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html

Is POST the only way to get the search results, or will it work with a
GET?

If GET will work then you need to set the search form to post to a
script on your site which then outputs a frameset with a URL on your
server that shows the header, and the URL for the shared search server
with all the POSTed variables as GET parameters as the second frame.
Job done.

If not then you're going to need to play silly wotsits with a hidden
form in the top frame which reposts the search to the bottom form. Not
pretty and would require JS but it should work.

-Stut

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Re: [PHP] Re: Passing variable to a page in a frameset

On Aug 15, 2008, at 2:27 PM, Stut wrote:

On 15 Aug 2008, at 20:21, Jody Cleveland wrote:

I work for a consortium of 30 libraries. Each library has their own
website, but they all share the same web catalog. On each library's
website there is a search box to search the catalog, which is on a
completely different server from the websites. We've been finding
that once people use that search box, they get distracted with the
catalog and have no easy way to get back to the library's website.
The problem I was tasked with is, coming up with a way to search
the catalog with an easy way to return to where the user was before
they initiated the search.

The only way I thought to do this was to use a frameset for the
search results. Which, you can see here:

http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html

Is POST the only way to get the search results, or will it work with
a GET?

If GET will work then you need to set the search form to post to a
script on your site which then outputs a frameset with a URL on your
server that shows the header, and the URL for the shared search
server with all the POSTed variables as GET parameters as the second
frame. Job done.

GET should work too. Do you know of any examples anywhere online for
this? My brain shuts off at the thought of how I'd do that.

- jody

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Re: [PHP] Re: Passing variable to a page in a frameset


 I work for a consortium of 30 libraries. Each library has their own
 website, but they all share the same web catalog. On each library's website
 there is a search box to search the catalog, which is on a completely
 different server from the websites. We've been finding that once people use
 that search box, they get distracted with the catalog and have no easy way
 to get back to the library's website. The problem I was tasked with is,
 coming up with a way to search the catalog with an easy way to return to
 where the user was before they initiated the search.

 The only way I thought to do this was to use a frameset for the search
 results. Which, you can see here:

 http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html

 If anyone has any ideas, other than using frames for the results, I'd love
 to hear them. The problem is, there's nothing I can do on the web catalog
 end.

 - jody



Easiest solution - Open the search page in a new window. Then they can just
close it to get back to the previous window...

I now understand what you're trying to say in regards to the frames.

The top frame resides completely on your server so you can place a Go back
to homepage link to direct people back to YOUR libraries homepage. And the
bottom frame contains the search results page that you have no control over
and cannot alter to simply place a Home link on it. And that should remain
since the global search results page is accessed by multiple libraries.

Like I said above, the easiest thing to do is just open it in a seperate
window so they can close it at any time and still be at the same place on
your libraries website.

Re: [PHP] Re: Passing variable to a page in a frameset

On 15 Aug 2008, at 20:34, Jody Cleveland wrote:

On Aug 15, 2008, at 2:27 PM, Stut wrote:

On 15 Aug 2008, at 20:21, Jody Cleveland wrote:

I work for a consortium of 30 libraries. Each library has their
own website, but they all share the same web catalog. On each
library's website there is a search box to search the catalog,
which is on a completely different server from the websites. We've
been finding that once people use that search box, they get
distracted with the catalog and have no easy way to get back to
the library's website. The problem I was tasked with is, coming up
with a way to search the catalog with an easy way to return to
where the user was before they initiated the search.

The only way I thought to do this was to use a frameset for the
search results. Which, you can see here:

http://beta.menashalibrary.org/sites/beta.menashalibrary.org/themes/salamander/searchframe.html

Is POST the only way to get the search results, or will it work
with a GET?

If GET will work then you need to set the search form to post to a
script on your site which then outputs a frameset with a URL on
your server that shows the header, and the URL for the shared
search server with all the POSTed variables as GET parameters as
the second frame. Job done.

GET should work too. Do you know of any examples anywhere online for
this? My brain shuts off at the thought of how I'd do that.

Off the top of my head and very untested...

?php
$vars = array();
foreach ($_POST as $k = $v)
{
$vars[] = urlencode($k).'='.urlencode($v);
}
$searchurl = 'http://search.server.com/search.php?'.implode('',
$vars);

?
frameset
frame src=/header.html /
frame src=?php echo $searchurl; ? /
/frameset

Modify to your own frameset/url requirements but that's the basic idea.

-Stut

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Re: [PHP] Re: Passing variable to a page in a frameset


 GET should work too. Do you know of any examples anywhere online for this?
 My brain shuts off at the thought of how I'd do that.

 - jody


When you GET a value you are retrieving a passed value that appears in the
address bar:
Example

http://www.mysite.com?name=joe

www.mysite.com is the website

?name=joe is the value being passed by GET

To put this value into a PHP variable you would simply do:
?php
$name = $_GET['name'];
? http://www.google.com/search?hl=enq=encrypt+javascript

Re: [PHP] Re: Passing variable to a page in a frameset