Re: [PHP] Simple Problem

2005-04-12 Thread Joseph Connolly
I think what you mean/need is:
$sql="SELECT products.productID, products.title, 
products.number_per_box, products.stock_level, products.image, 
users.username, users.email, users.userID FROM users, products  WHERE 
products.userID = users.userID  AND userID = $userID";


John Nichel wrote:
PartyPosters wrote:
I can't figure out what I am doing wrong,
this sql string seems to filter out the information I want but it 
duplicates the all info, as 'num_rows' is total of rows in the table 
and not the correct value of the filtered information?

$sql="SELECT products.productID, products.title, 
products.number_per_box, products.stock_level, products.image, 
users.username, users.email, users.userID FROM users, products  WHERE 
products.userID = $userID";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);

this is the old sql statement which works fine - $sql="SELECT 
productID, title, number_per_box, stock_level, image, userID FROM 
products WHERE userID = '$userID'";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);

http://lists.mysql.com
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Re: [PHP] Simple Problem

2005-04-12 Thread Duncan Hill
On Tuesday 12 April 2005 16:24, PartyPosters typed:
> $sql="SELECT products.productID, products.title, products.number_per_box,
> products.stock_level, products.image, users.username, users.email,
> users.userID FROM users, products  WHERE products.userID = $userID";
> $mysql_result=mysql_query($sql,$connection);
> $num_rows=mysql_num_rows($mysql_result);

You don't specify how the query should join the tables users and products.

WHERE products.userID = users.userID and users.userID = $userID

I'll assume you do validation on the userID ahead of time to ensure that it's 
the right format.

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Re: [PHP] Simple Problem

2005-04-12 Thread Richard Davey
Hello PartyPosters,

Tuesday, April 12, 2005, 4:24:34 PM, you wrote:

P> I can't figure out what I am doing wrong, this sql string seems to
P> filter out the information I want but it duplicates the all info,
P> as 'num_rows' is total of rows in the table and not the correct
P> value of the filtered information?

P> $sql="SELECT products.productID, products.title,
P> products.number_per_box, products.stock_level, products.image,
P> users.username, users.email, users.userID FROM users, products 
P> WHERE products.userID = $userID";
P> $mysql_result=mysql_query($sql,$connection);
P> $num_rows=mysql_num_rows($mysql_result);

Your query should probably be performing a join. As it stands you're
selecting all of the products where userID = $userID, but you're also
selecting all of your users regardless (SELECT blah FROM users).

What relationship does the Users table have to the Products table? You
may need something more along the lines of:

SELECT blah FROM products LEFT JOIN users ON users.UserID =
products.UserID WHERE users.UserID = $userID

But that's just a best-guess based on what you've given so far.

Best regards,

Richard Davey
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RE: [PHP] Simple Problem

2005-04-12 Thread Pablo Gosse

I can't figure out what I am doing wrong,
this sql string seems to filter out the information I want but it
duplicates the all info, as 'num_rows' is total of rows in the table and
not the correct value of the filtered information?

$sql="SELECT products.productID, products.title,
products.number_per_box, products.stock_level, products.image,
users.username, users.email, users.userID FROM users, products  WHERE
products.userID = $userID"; $mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);


this is the old sql statement which works fine - 
$sql="SELECT productID, title, number_per_box, stock_level, image,
userID FROM products WHERE userID = '$userID'";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);

what I am I doing wrong.


In the old query you're jut pulling records from one table.  In the
second, you're joining two tables, but you're still using the same WHERE
clause, products.userID = $userID.

This is causing multiple records to be returned, since there is nothing
specifying how you are joining the users and products table so it
correctly returns a match for every row in the user table.

Change the where clause to:

WHERE products.userID = $userID
AND users.userID = $userID

and this will force the query to limit results to those records in
products that have a matching record in the users table.

HTH.

Pablo

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Re: [PHP] Simple Problem

2005-04-12 Thread John Nichel
PartyPosters wrote:
I can't figure out what I am doing wrong,
this sql string seems to filter out the information I want but it duplicates 
the all info, as 'num_rows' is total of rows in the table and not the correct 
value of the filtered information?
$sql="SELECT products.productID, products.title, products.number_per_box, 
products.stock_level, products.image, users.username, users.email, users.userID FROM 
users, products  WHERE products.userID = $userID";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);
this is the old sql statement which works fine - 
$sql="SELECT productID, title, number_per_box, stock_level, image, userID FROM products WHERE userID = '$userID'";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);
http://lists.mysql.com
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716.856.9675
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[PHP] Simple Problem

2005-04-12 Thread PartyPosters
I can't figure out what I am doing wrong,
this sql string seems to filter out the information I want but it duplicates 
the all info, as 'num_rows' is total of rows in the table and not the correct 
value of the filtered information?

$sql="SELECT products.productID, products.title, products.number_per_box, 
products.stock_level, products.image, users.username, users.email, users.userID 
FROM users, products  WHERE products.userID = $userID";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);


this is the old sql statement which works fine - 
$sql="SELECT productID, title, number_per_box, stock_level, image, userID FROM 
products WHERE userID = '$userID'";
$mysql_result=mysql_query($sql,$connection);
$num_rows=mysql_num_rows($mysql_result);

what I am I doing wrong.

Thanks
Kaan.




[PHP] Simple Problem about forms and sending info to db

2004-09-14 Thread Logan Moore
I am currently coding a cms. So far I have completed the security
login/password system but have now moved on to a new area of php. Uploading
info to a database.

This is the code as it stands now.


  

  Full Name:
  


  Email:
  


  UserName:
  


  Password:
  


  
  
  
  
  ";}

  if($_GET[action] == "adduser"){echo "Add User Function is still Under
Construction  as a result what you just did has had no effect"}
  ?>


I know there is something wrong as i get an error when I goto the page.
I want it to display the form, then when you click submit, goto the same
page except run the command to add the information from the form to the
database.

Also how do I put the information in the database.

Regards
Logan

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Re: [PHP] simple problem about authentication

2003-02-26 Thread Oliver Witt
Ernest E Vogelsinger schrieb:

> At 14:22 26.02.2003, Oliver Witt spoke out and said:
> [snip]
> >if(isset($user) && isset($pw)){
> >$user = ucwords(strtolower($user));
> >$PHP_AUTH_USER = $user;
> >$PHP_AUTH_PW = $pw;}
> >
> >[more...]
> >And this works fine for just this page. Whenever I click on a link to a page
> >which includes the same file though, it echoes 'unauthorized'...
> [snip]
>
> Sounds like you're using sessions... perhaps register_globals is off in
> your environment, so session variables are not registered? How do you set
> the "$user" and "$pw" variables?
>

no, im not using sessions and register globals is on.
the variables user and pw are set by submitting a form (method="post"). that's
where i expect the mistake to be. because as far as i know, the browser usually
saves the credentials, at least the user name after authenticating. but it just
doesn't seem to do so
Olli



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Re: [PHP] simple problem about authentication

2003-02-26 Thread Ernest E Vogelsinger
At 14:22 26.02.2003, Oliver Witt spoke out and said:
[snip]
>if(isset($user) && isset($pw)){
>$user = ucwords(strtolower($user));
>$PHP_AUTH_USER = $user;
>$PHP_AUTH_PW = $pw;}
>
>[more...]
>And this works fine for just this page. Whenever I click on a link to a page
>which includes the same file though, it echoes 'unauthorized'...
[snip]

Sounds like you're using sessions... perhaps register_globals is off in
your environment, so session variables are not registered? How do you set
the "$user" and "$pw" variables?
 

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Re: [PHP] simple problem about authentication

2003-02-26 Thread Oliver Witt
"1lt John W. Holmes" schrieb:

> > I'm trying to set up a password section on my website. But I don't want
> > a window popping up asking for a username and password. I'd rather like
> > to have a form that submits the data. I did that and it works fine, but
> > the browser seems to not save the username and password, because if i
> > click on any link, it returns an "unauthorized message"... any ideas?
> > thnx,
>
> Show some code... the possibilities are endless for what you're doing wrong.
>
> ---John Holmes...

ok, here's the code:

the form submits the user name and password in the variables $user and $pw to
another site, which includes this:

if(isset($user) && isset($pw)){
$user = ucwords(strtolower($user));
$PHP_AUTH_USER = $user;
$PHP_AUTH_PW = $pw;}

if(isset($PHP_AUTH_USER)){
$result = mysql_query("select * from _users where user = '$PHP_AUTH_USER'");
$_user = mysql_fetch_array($result);}

if(!isset($PHP_AUTH_USER)) {
header('HTTP/1.0 401 Unauthorized');
echo('unauthorized');
exit;}
else{
$result = mysql_query("select * from _users where user = '$PHP_AUTH_USER'") or
die('Error occured while connecting to database');
$rows = mysql_num_rows($result);
if($rows == 0){
header('HTTP/1.0 401 Unauthorized');
echo"window.location.href='../index.php?action=user_wrong'";
exit;}
else{
$daten = mysql_fetch_array($result);
if($PHP_AUTH_PW == $daten['pw']){
// authorized
}
else{
header('HTTP/1.0 401 Unauthorized');
echo"window.location.href='../index.php?action=pw_wrong&user=$user'";

exit;}}}

And this works fine for just this page. Whenever I click on a link to a page
which includes the same file though, it echoes 'unauthorized'...
Olli


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Re: [PHP] simple problem about authentication

2003-02-25 Thread 1LT John W. Holmes
> I'm trying to set up a password section on my website. But I don't want
> a window popping up asking for a username and password. I'd rather like
> to have a form that submits the data. I did that and it works fine, but
> the browser seems to not save the username and password, because if i
> click on any link, it returns an "unauthorized message"... any ideas?
> thnx,

Show some code... the possibilities are endless for what you're doing wrong.

---John Holmes...


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[PHP] simple problem about authentication

2003-02-25 Thread Oliver Witt
Hi,
I'm trying to set up a password section on my website. But I don't want
a window popping up asking for a username and password. I'd rather like
to have a form that submits the data. I did that and it works fine, but
the browser seems to not save the username and password, because if i
click on any link, it returns an "unauthorized message"... any ideas?
thnx,
Olli


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Re: [PHP] simple problem

2002-09-04 Thread Adam Williams

if ($i%3 = 0)
{
echo "something";
}


On Wed, 4 Sep 2002, Cirkit Braker wrote:

> what would be the best, most efficient way to print something every three
> times a loop runs
>
>   for ($i=0; $i<$num_products; $i++)
>   {
> echo "something";
>}
>
> this happens every time it runs and has to stay that way but every three
> times i want to print additional info.
>
> Any help appreciated.
>
>
>
>


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Re: [PHP] simple problem

2002-09-04 Thread Rasmus Lerdorf

Inside the loop do:

  if(!($i%3)) { whatever }

-Rasmus

On Wed, 4 Sep 2002, Cirkit Braker wrote:

> what would be the best, most efficient way to print something every three
> times a loop runs
>
>   for ($i=0; $i<$num_products; $i++)
>   {
> echo "something";
>}
>
> this happens every time it runs and has to stay that way but every three
> times i want to print additional info.
>
> Any help appreciated.
>
>
>
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>


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[PHP] simple problem

2002-09-04 Thread Cirkit Braker

what would be the best, most efficient way to print something every three
times a loop runs

  for ($i=0; $i<$num_products; $i++)
  {
echo "something";
   }

this happens every time it runs and has to stay that way but every three
times i want to print additional info.

Any help appreciated.



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Re: [PHP] Simple problem

2002-02-19 Thread Richard KS

Or you could just write

 $fp1 = Fopen("D:\\log.txt","a+");
 fwrite($fp1,"$start\n$array[0]\n$array[1]\n");


to simplify...

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"Jeff Van Campen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > I need insert some variable ito log.txt file but every item (variable)
must
>  > be on other line
>  > how to do this ?
>
> If I am understanding your question correctly, you need to add a newline
> character (\n) at then end every line.  Like so:
>
> $fp1 = Fopen("D:\\log.txt","a+");
> fwrite($fp1,"$start\n");
> fwrite($fp1,"$array[0]\n");
> fwrite($fp1,"$array[1]\n");
> .
> fwrite($fp1,$end);
> Fclose($fp1);
>
> HTH
> -jeff
>



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Re: [PHP] Simple problem

2002-02-19 Thread Jeff Van Campen

 > I need insert some variable ito log.txt file but every item (variable) must
 > be on other line
 > how to do this ?

If I am understanding your question correctly, you need to add a newline 
character (\n) at then end every line.  Like so:

$fp1 = Fopen("D:\\log.txt","a+");
fwrite($fp1,"$start\n");
fwrite($fp1,"$array[0]\n");
fwrite($fp1,"$array[1]\n");
.
fwrite($fp1,$end);
Fclose($fp1);

HTH
-jeff


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[PHP] Simple problem

2002-02-19 Thread Roman Duriancik

I need insert some variable ito log.txt file but every item (variable) must
be on other line
how to do this ?

  $fp1 = Fopen("D:\\log.txt","a+");
  fwrite($fp1,$start);
  fwrite($fp1,"$array[0]");
  fwrite($fp1,"$array[1]");
  .
  fwrite($fp1,$end);
  Fclose($fp1);


roman


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