Re: [PHP-DB] Problem with php and mysql

2003-03-05 Thread 1LT John W. Holmes
 One of our customers has written a php application that we are hosting.

 Basically, there is a province/state table which populates a combo box.
 When the form containing this combo box is submitted, the information is
 added into the database.

 The problem is that when nothing is selected from the combo box, it adds a
 number which seems to be random in the database, rather than the
 province/state ID which it references from the province table. Obviously
 there are many methods of avoiding this such as javascript to make sure
 something is selected, or even giving the combo box a default value, but I
 was wondering if anyone knew why the database itself isn't catching the
 error (foreign key violation) and accepting the data.

 Any ideas?

Are you using InnoDb tables? Otherwise there are no foreign key
restrictions.

How about some validation to make sure something is selected? Or, some
validation that'll give the value a default if nothing is selected.

Validate everything from the user!

---John Holmes...


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Re: [PHP-DB] Problem with php and mysql

2003-03-05 Thread Kevin Demers DSL
Hi, thanks for the help, I just had a look at the db and they are just the
default tables, so that must be it. I'll let them know to either change
them over or add validation.

Again, thanks for your help.

On Wed, 5 Mar 2003, 1LT John W. Holmes wrote:

 Date: Wed, 5 Mar 2003 10:07:04 -0500
 From: 1LT John W. Holmes [EMAIL PROTECTED]
 To: Kevin Demers DSL [EMAIL PROTECTED], [EMAIL PROTECTED]
 Subject: Re: [PHP-DB] Problem with php and mysql

  One of our customers has written a php application that we are hosting.
 
  Basically, there is a province/state table which populates a combo box.
  When the form containing this combo box is submitted, the information is
  added into the database.
 
  The problem is that when nothing is selected from the combo box, it adds a
  number which seems to be random in the database, rather than the
  province/state ID which it references from the province table. Obviously
  there are many methods of avoiding this such as javascript to make sure
  something is selected, or even giving the combo box a default value, but I
  was wondering if anyone knew why the database itself isn't catching the
  error (foreign key violation) and accepting the data.
 
  Any ideas?

 Are you using InnoDb tables? Otherwise there are no foreign key
 restrictions.

 How about some validation to make sure something is selected? Or, some
 validation that'll give the value a default if nothing is selected.

 Validate everything from the user!

 ---John Holmes...



Thanks,
Kevin

---
   Kevin Demers
 Technical/Customer Support
Internet Access - WORLDWIDE
3 East Main Street
Welland, ON  Canada L3B 3W4
(905) 714-1400   (800) 560-4560
[EMAIL PROTECTED] www.iaw.on.ca


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RE: [PHP-DB] php and local printing

2003-03-05 Thread Matthew Moldvan
Have you looked into the exec() function?  Maybe exec(lpr something or
other) may work, though I haven't tried it myself ...

Regards,
Matthew Moldvan

---
 System Administrator
 Trilogy International, Inc
  http://www.trilogyintl.com/ecommerce/
--- 

-Original Message-
From: Rajesh Fowkar [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 05, 2003 12:21 AM
To: PHP-DB Mailing List
Subject: Re: [PHP-DB] php and local printing


On Tue, Mar 04, 2003 at 06:45:54AM -0800, Mark wrote:

It dopesn't seem like you got an answer to this. 

Spot on. Yours is the first reply.

To print locally,
you'll need to use Jscript of something similar on the local end.
I've only seen this work in IE, but I haven't looked into it too
deeply. To print to the server, you'll probably need to shell out to
the OS.

Using javascript window.print() I get the print dialogue box. But what
I needed was I have got a text file which on server I can print using lp
filename. Similar thing but without any print dialogue box at the
client end.


HTH or at least starts a discussion...

Hope so.

Thank you for the reply.

Peace

--
Rajesh
:
[ GNU/Linux One Stanza Tip (LOST) ]###

Sub : Linux Commands (#4)LOST #056

For a newbie introduction to Linux commands and shotcuts visit
URL: http://sunsite.dk/linux-newbie/Linux_commands.htm

[EMAIL PROTECTED]##
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Re: [PHP-DB] date functions (generates parse error)

2003-03-05 Thread David Rice
Sorry, I forgot to add the part at the bottom where I am calling the 
function

=

?
function tips($weekstart){
	$start = date('Ymd',strtotime($weekstart));

 	$query = SELECT * FROM Rota WHERE date = $start and date = ($start + 
INTERVAL 6 DAY) ORDER BY staffid;
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){

		if ( isset ( $tips ) ){

			if (isset ( $tips[$row[staffid]] ) ){

$hours = $row[finish] - $row[start];
$tips[$row[staffid]] = $tips[$row[staffid]] + $hours;
			}

			else{

$tips[$row[staffid]] = $row[finish] - $row[start];

}
}
		else{

			$tips = array('$row[staffid]' =( $row[finish] - $row[start] ) );

		}

	}

	return $tips;

}

function dbconnect(){
mysql_connect(localhost, filterseveuk, godisadj);
mysql_select_db(filterseveuk);
}
dbconnect();
$date = 2003-03-02;
var_dump(tips($date));
?
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[PHP-DB] ODBC/Paradox

2003-03-05 Thread Fabio Hur
Hi,

I am trying access DB Paradox via php/odbc.
Only get in W98/PWS or WME/PWS. And without success in Linux/php/odbc and
Windows 2K/php/odbc.
In W2K's Case, it appears following message:

Warning: SQL error: [Microsoft][ODBC DRIVER for Paradox] The table
external is not in the waited format., SQL state S1000 in
SQLExecDirect in D:\www\chen\quadro.inc on line 4
Unable have runs query.

Would anybody have a solution?

Thanks

Fábio Hur



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RE: [PHP-DB] date functions (generates parse error)

2003-03-05 Thread John W. Holmes
This is very weird. What are you using to create this code?

If you remove all of the spaces (?) before $query, the code doesn't give
a parse error. It shouldn't give one either way, though, if those are
spaces or a tab before your $query = ... line. 

Even if you remove all of the text and only leave ?, ?, and those
spaces, PHP will spit something like:

Notice: Use of undefined constant   - assumed ' ' in
c:\inetpub\wwwroot\test.php on line 3

So, the fix is to remove those characters and replace them with spaces
or tabs. 

---John W. Holmes...

PHP Architect - A monthly magazine for PHP Professionals. Get your copy
today. http://www.phparch.com/

 -Original Message-
 From: David Rice [mailto:[EMAIL PROTECTED]
 Sent: Wednesday, March 05, 2003 12:34 PM
 To: [EMAIL PROTECTED]; [EMAIL PROTECTED]
 Subject: Re: [PHP-DB] date functions (generates parse error)
 
 
 Sorry, I forgot to add the part at the bottom where I am calling the
 function
 
 =
 
 ?
 function tips($weekstart){
 
   $start = date('Ymd',strtotime($weekstart));
 
   $query = SELECT * FROM Rota WHERE date = $start and date =
 ($start +
 INTERVAL 6 DAY) ORDER BY staffid;
   $result = mysql_query($query);
   while ($row = mysql_fetch_array($result)){
 
   if ( isset ( $tips ) ){
 
   if (isset ( $tips[$row[staffid]] ) ){
 
   $hours = $row[finish] - $row[start];
   $tips[$row[staffid]] =
$tips[$row[staffid]] +
 $hours;
 
   }
 
   else{
 
   $tips[$row[staffid]] = $row[finish] -
$row[start];
 
   }
   }
 
   else{
 
   $tips = array('$row[staffid]' =( $row[finish] -
 $row[start] ) );
 
   }
 
   }
 
   return $tips;
 
 }
 
 function dbconnect(){
   mysql_connect(localhost, filterseveuk, godisadj);
   mysql_select_db(filterseveuk);
 }
 
 dbconnect();
 $date = 2003-03-02;
 
 var_dump(tips($date));
 ?
 
 _
 Express yourself with cool emoticons http://messenger.msn.co.uk
 
 
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Re: [PHP-DB] php and local printing

2003-03-05 Thread Jason Wong
On Wednesday 05 March 2003 13:20, Rajesh Fowkar wrote:
 On Tue, Mar 04, 2003 at 06:45:54AM -0800, Mark wrote:
 It dopesn't seem like you got an answer to this.

 Spot on. Yours is the first reply.

Not surprising as this question really does not belong here.

 To print locally,
 you'll need to use Jscript of something similar on the local end.
 I've only seen this work in IE, but I haven't looked into it too
 deeply. To print to the server, you'll probably need to shell out to
 the OS.

 Using javascript window.print() I get the print dialogue box. But what
 I needed was I have got a text file which on server I can print using lp
 filename. Similar thing but without any print dialogue box at the
 client end.

Exactly, this is a client-side problem (Javascript, ActiveX/script, etc).

-- 
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design  Hosting * Internet  Intranet Applications Development *
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http://marc.theaimsgroup.com/?l=php-db
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Firmware update in the coffee machine
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[PHP-DB] Column Sort Question

2003-03-05 Thread Jerry
Hi All,

This is a newbie question. I am using PHP Version 4.2.3 with
register_globals=off. I am trying to creating a simple address book with
mysql as the backend. I have a  form that passes what the user wants to
query from the address book and it returns everything in a table with column
heading just fine. I use $myvar1=$_POST['myvar1']; to pass the info from the
form to the php file.

I am using the following in the header fields of the table:
th nowrap a href=?php echo $PHP_SELF?order_by=FIRSTNAME; ?First
Name/a /th

This doesn't work (I get a bunch of errors) and I think it's because the
variable from $_POST are not staying persistent. I tried playing around with
session with no luck.

What is the best way to resort on a column?

Thanks,

Jerry



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RE: [PHP-DB] Column Sort Question

2003-03-05 Thread Gary . Every
Shouldn't you be using $_GET or, since register_globals are set to off,
$order_by should contain FIRSTNAME

Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
Pay It Forward
mailto:[EMAIL PROTECTED]
http://accessingram.com


-Original Message-
From: Jerry [mailto:[EMAIL PROTECTED]
Sent: Wednesday, March 05, 2003 3:40 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Column Sort Question


Hi All,

This is a newbie question. I am using PHP Version 4.2.3 with
register_globals=off. I am trying to creating a simple address book with
mysql as the backend. I have a  form that passes what the user wants to
query from the address book and it returns everything in a table with column
heading just fine. I use $myvar1=$_POST['myvar1']; to pass the info from the
form to the php file.

I am using the following in the header fields of the table:
th nowrap a href=?php echo $PHP_SELF?order_by=FIRSTNAME; ?First
Name/a /th

This doesn't work (I get a bunch of errors) and I think it's because the
variable from $_POST are not staying persistent. I tried playing around with
session with no luck.

What is the best way to resort on a column?

Thanks,

Jerry



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[PHP-DB] Storing browser stats

2003-03-05 Thread Bruce Levick
Is there a method for storing each users browser stats within a database
table.

So everytime someone visits the browser info is gathered and stored as a
unique row in the 'browser' table in the database.

Any advice??

Cheers

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[PHP-DB] Converting Dates...

2003-03-05 Thread Doug Coning
Hi,

I need help.  I need to convert a Date column in MySQL where the date is
stored like 2003-01-15 to the following seconds 1042621252 .  I'm
connecting the database through PHP to third party solution that requires
this.  Here is the only information that was given:

---
 - Dates and times are passed in as seconds (not milliseconds) since
1/1/1970
   and converted into Date Objects by multiplying by 1000 and then add the
   TimeZoneOffset to get it to local time:

   eg:
  startTime = 10710812; // startTime is the number of seconds between the
req'd date and 1/1/1970
  startTimeObj = Date(startTime * 1000 + (new Date().getTimezoneOffset() *
6));

 - startDate and endDate combine the the date and time of events
-

I have no idea how to convert my date colums into this type.  Any help would
be GREATLY appreciated!

Thanks,

Doug Coning




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