RE: [PHP-DB] not a valid MySQL result resource

2003-01-20 Thread John W. Holmes
When you get that error, it generally means your query failed, or you
have the wrong variable in a mysql_fetch_*() function. Use mysql_error()
to see what the error is if the query failed. 

---John W. Holmes...

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

> -Original Message-
> From: Addison Ellis [mailto:[EMAIL PROTECTED]]
> Sent: Monday, January 20, 2003 2:30 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] not a valid MySQL result resource
> 
> hello and thank you for your time.
>   can you tell me why the below: i think this whole section is
> line 22 because whatever i change in these first four lines still has
> the error message point to line 22.
> $cobj =  mysql_db_query($dbname,"select * from category where
> id=$category");  //line 22
> $crow =  mysql_fetch_object($cobj);
> $scobj = mysql_db_query($dbname,"select * from subcategory where
> category=$category");
> $scrow = mysql_fetch_object($scobj);
> 
>   name;?>
>name;?>
> 
>   gives me this:
> Warning: Supplied argument is not a valid MySQL result resource in
> /users/infoserv/web/register/ca/direct.php on line 22
> 
> 
> perhaps i should use something different. i am trying to print the
> results from two previous selections on this third page. the second
> page works with
> $cobj =  mysql_db_query($dbname,"select * from category where
> id=$category");
> $crow =  mysql_fetch_object($cobj);
> name;?>
> 
> and then they make selection two...
> thank you again. addison
> --
> Addison Ellis
> small independent publishing co.
> 114 B 29th Avenue North
> Nashville, TN 37203
> (615) 321-1791
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]
> subsidiaries of small independent publishing co.
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]



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RE: [PHP-DB] Not a valid MySQL result resource

2001-11-09 Thread matt stewart

don't you need it to be SELECT login, password...
lowercase first letter? - if you're comparing the columns user.login and
user.password.

-Original Message-
From: DL Neil [mailto:[EMAIL PROTECTED]]
Sent: 09 November 2001 01:34
To: MPropre
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Not a valid MySQL result resource


>  //.../... first part of the code is to connect to the right DB on a MySQL
> server. It works fine
> //This code to show the query. It runs well under MySQL and gives 1 result
:
>  $query="SELECT Login, Password FROM `user` WHERE user.login=''$login''
and
> user.password=''$password''";
> //This var should contain a query result ressource:
>  $result_query=mysql_query($query);
> //Here's my error message after executing:
> // Supplied argument is not a valid MySQL result resource in this line:
>  $result_table=mysql_fetch_array($result_query);
> // I thought that this var ($result_table) should contain the row of 2
cells
> //And I expected to read $result_table['login'] or $result_table[0] as the
> first cell of the row...
> // I any of you can help me, poor php newby... :o) Great thanks !!
> ?>


=Please check the archives for advice about checking the result of every
mysql_...() call. It's much better to
know/be told by PHP/MySQL than to say "I thought that...I expected to ..."
=the $query assignment statement contains multiple double-quotes. These
cannot be 'nested'. Use a mixture of
single and double-quotes or 'escape' the inner set(s).
=also (and this may be a function of our email packages not a PHP thing) are
those single quotes around user?

=dn



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Re: [PHP-DB] Not a valid MySQL result resource

2001-11-08 Thread Miles Thompson

I'd check for mysql_num_rows( $result_query) > 0 as well.
It's entirely possible the user mis-typed the login or password values.
Miles

At 07:26 PM 11/8/01 -0600, Paul DuBois wrote:
>At 2:04 AM +0100 11/9/01, MPropre wrote:
>>>//.../... first part of the code is to connect to the right DB on a MySQL
>>server. It works fine
>>
>>
>>//This code to show the query. It runs well under MySQL and gives 1 result :
>>  $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
>>user.password=''$password''";
>>
>>//This var should contain a query result ressource:
>>  $result_query=mysql_query($query);
>>
>>//Here's my error message after executing:
>
>Not so fast.  Where's your error checking to verify that the
>query actually succeeded?
>
>if (!$result_query)
>{
> die ("Gee, I guess error checking is a good thing after all!"
>  . mysql_error ());
>}
>
>>// Supplied argument is not a valid MySQL result resource in this line:
>>  $result_table=mysql_fetch_array($result_query);
>>
>>// I thought that this var ($result_table) should contain the row of 2 cells
>>//And I expected to read $result_table['login'] or $result_table[0] as the
>>first cell of the row...
>>// I any of you can help me, poor php newby... :o) Great thanks !!
>>
>>?>
>>
>>
>>
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>>To unsubscribe, e-mail: [EMAIL PROTECTED]
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>>To contact the list administrators, e-mail: [EMAIL PROTECTED]
>
>
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Re: [PHP-DB] Not a valid MySQL result resource

2001-11-08 Thread DL Neil

>  //.../... first part of the code is to connect to the right DB on a MySQL
> server. It works fine
> //This code to show the query. It runs well under MySQL and gives 1 result :
>  $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
> user.password=''$password''";
> //This var should contain a query result ressource:
>  $result_query=mysql_query($query);
> //Here's my error message after executing:
> // Supplied argument is not a valid MySQL result resource in this line:
>  $result_table=mysql_fetch_array($result_query);
> // I thought that this var ($result_table) should contain the row of 2 cells
> //And I expected to read $result_table['login'] or $result_table[0] as the
> first cell of the row...
> // I any of you can help me, poor php newby... :o) Great thanks !!
> ?>


=Please check the archives for advice about checking the result of every mysql_...() 
call. It's much better to
know/be told by PHP/MySQL than to say "I thought that...I expected to ..."
=the $query assignment statement contains multiple double-quotes. These cannot be 
'nested'. Use a mixture of
single and double-quotes or 'escape' the inner set(s).
=also (and this may be a function of our email packages not a PHP thing) are those 
single quotes around user?

=dn



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Re: [PHP-DB] Not a valid MySQL result resource

2001-11-08 Thread Paul DuBois

At 2:04 AM +0100 11/9/01, MPropre wrote:
>//.../... first part of the code is to connect to the right DB on a MySQL
>server. It works fine
>
>
>//This code to show the query. It runs well under MySQL and gives 1 result :
>  $query="SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
>user.password=''$password''";
>
>//This var should contain a query result ressource:
>  $result_query=mysql_query($query);
>
>//Here's my error message after executing:

Not so fast.  Where's your error checking to verify that the
query actually succeeded?

if (!$result_query)
{
 die ("Gee, I guess error checking is a good thing after all!"
  . mysql_error ());
}

>// Supplied argument is not a valid MySQL result resource in this line:
>  $result_table=mysql_fetch_array($result_query);
>
>// I thought that this var ($result_table) should contain the row of 2 cells
>//And I expected to read $result_table['login'] or $result_table[0] as the
>first cell of the row...
>// I any of you can help me, poor php newby... :o) Great thanks !!
>
>?>
>
>
>
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>To unsubscribe, e-mail: [EMAIL PROTECTED]
>For additional commands, e-mail: [EMAIL PROTECTED]
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