Re: [PHP] can't figure out this mysql error
On Tuesday 27 January 2004 02:23, Chris W. Parker wrote: If I put the two functions mysql functions within the query method I created I get an error on those two lines BUT the program continues as normal WITH the data being printed to the page just as I want it to. Clearly the mysql resource being complained about is actually valid and not invalid considering that the data I'm trying to retrieve is printed to the page. Any other ideas? Post some *concise* code which illustrates the problem. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* Your supervisor is thinking about you. */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] can't figure out this mysql error
Jason Wong mailto:[EMAIL PROTECTED] on Friday, January 23, 2004 10:03 PM said: Time to put the php debugger in action -- echo all your variables to see that contain what you expect them to contain. 'echo $this-Result;' prints Resource id #2 'print_r($this-Result);' prints Resource id #2 If I put the two functions mysql functions within the query method I created I get an error on those two lines BUT the program continues as normal WITH the data being printed to the page just as I want it to. Clearly the mysql resource being complained about is actually valid and not invalid considering that the data I'm trying to retrieve is printed to the page. Any other ideas? Thanks, Chris. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] can't figure out this mysql error
Jason Wong mailto:[EMAIL PROTECTED] on Thursday, January 22, 2004 8:49 PM said: The if-clause will never be evaluated because if there had been an error your program would have dieded on the previous line. Heh.. yeah I thought about that as I was examining the code for this post. But thanks for confirming. I'm sure if php tells you it's invalid you can bet your *** it's invalid! Check for errors and report with mysql_error() after _each_ and _every_ call to the mysql_* functions. All of what you say makes sense. Hopefully your suggestion will help me find the problem! Thanks, Chris. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] can't figure out this mysql error
Chris W. Parker on Friday, January 23, 2004 8:54 AM said: All of what you say makes sense. Hopefully your suggestion will help me find the problem! In looking through my code I see that I already do this. EVERY call to a mysql function has or die($this-stop($current_line:.__LINE__)); along with the call. $this-stop() is method that prints out mysql_errno(); and mysql_error();. In this case the or die is not even being tripped meaning there isn't an error. I'm beginning to think it's a bug, but it's more likely that it's not. Here's another bit of crazy to throw in. The function I am trying to perform is that of adding a record to my db. Guess what? The record insertion is successul each time. And no I'm not closing the resource/db link before those two functions are called. They happen immediately after the the mysql_query() function. Sounds pretty fishy Chris. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] can't figure out this mysql error
On Saturday 24 January 2004 01:07, Chris W. Parker wrote: on Friday, January 23, 2004 8:54 AM said: All of what you say makes sense. Hopefully your suggestion will help me find the problem! In looking through my code I see that I already do this. EVERY call to a mysql function has or die($this-stop($current_line:.__LINE__)); along with the call. $this-stop() is method that prints out mysql_errno(); and mysql_error();. In this case the or die is not even being tripped meaning there isn't an error. I'm beginning to think it's a bug, but it's more likely that it's not. Here's another bit of crazy to throw in. The function I am trying to perform is that of adding a record to my db. Guess what? The record insertion is successul each time. And no I'm not closing the resource/db link before those two functions are called. They happen immediately after the the mysql_query() function. Sounds pretty fishy Time to put the php debugger in action -- echo all your variables to see that contain what you expect them to contain. -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* An expert is the person who avoids the small errors while sweeping on to the grand fallacy -- Thoreau's Theories of Adaption n1 */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] can't figure out this mysql error
Hi.
I have a db class that accesses and executes querys as well as returns
the result (if you so choose to have it returned).
One method within this class is called query() (how genius!). It's
defined like so:
?php
function query($sql, $current_line)
{
$this-Result = mysql_query($sql) or die($this-stop($current_line));
if(!$this-Result)
{
echo mysql_error();
}
$this-result_fields = mysql_num_fields($this-Result); // line 127
$this-result_rows = mysql_num_rows($this-Result); // line 130
}
?
How is it possible that I'm getting the following errors?
Warning: mysql_num_fields(): supplied argument is not a valid MySQL
result resource in /home/cparker/www/schedulevark/lib/classes/db.php on
line 127
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/cparker/www/schedulevark/lib/classes/db.php on line
130
success!
I can't figure it out! It sure looks like a valid MySQL result to me! Oh
and I should mention that the query's I'm running are perfect and I can
run these two functions without error in another method (the one that
returns the results). That method is defined like so:
?php
function get_query_results()
{
$Result_Arr = array();
// THESE LINES WORK FINE BUT I DON'T WANT THEM HERE!
// // store the number of fields
// $this-result_fields = mysql_num_fields($this-Result);
//
// // store the number of rows
// $this-result_rows = mysql_num_rows($this-Result);
if($this-result_rows 0)
{
while($line = mysql_fetch_array($this-Result, MYSQL_BOTH))
{
$Result_Arr[] = $line;
}
}
mysql_free_result($this-Result);
return $Result_Arr;
}
?
The reason I don't want the two mysql counting functions in the second
method is because it forces me to execute the second method within my
page before I can access those values. Sometimes I need to access those
values but don't need to return a result so I'd like to be able to leave
that second step out.
Here is the original way they are used in my page (the way I don't
want):
?php
$sql =
SELECT field1
, field2
, field3
FROM thetable;
$object-query($sql, __LINE__);
$the_result = $object-get_query_results();
$the_result_Rows = $object-Result_rows;
$the_result_Fields = $object-Result_fields;
?
Here is the way that I WANT it to work:
?php
$sql =
SELECT field1
, field2
, field3
FROM thetable;
$object-query($sql, __LINE__);
$the_result_Rows = $object-Result_rows;
$the_result_Fields = $object-Result_fields;
?
Anyone have any ideas?
If I have not been verbose enough in this email let me know and I will
try to clarify.
Thanks,
Chris.
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Re: [PHP] can't figure out this mysql error
On Friday 23 January 2004 09:38, Chris W. Parker wrote:
?php
function query($sql, $current_line)
{
$this-Result = mysql_query($sql) or die($this-stop($current_line));
if(!$this-Result)
{
echo mysql_error();
}
The if-clause will never be evaluated because if there had been an error your
program would have dieded on the previous line.
How is it possible that I'm getting the following errors?
Warning: mysql_num_fields(): supplied argument is not a valid MySQL
result resource in /home/cparker/www/schedulevark/lib/classes/db.php on
line 127
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result
resource in /home/cparker/www/schedulevark/lib/classes/db.php on line
130
success!
I can't figure it out! It sure looks like a valid MySQL result to me! Oh
I'm sure if php tells you it's invalid you can bet your *** it's invalid!
Check for errors and report with mysql_error() after _each_ and _every_ call
to the mysql_* functions.
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
--
Search the list archives before you post
http://marc.theaimsgroup.com/?l=php-general
--
/*
If you sow your wild oats, hope for a crop failure.
*/
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