[PHP-DB] not a valid MySQL result
I'm trying to do a full text search to a MySQL database and have a problem
being able to read the results:
here is my code:
---
$query = SELECT name,title,branch,section_one,section_two,MATCH
(section_header,section_one,section_two,section_footer) AGAINST
('.$searchCriteria.') AS score FROM pages WHERE MATCH
(section_header,section_one,section_two,section_footer) AGAINST
('.$searchCriteria.' IN BOOLEAN MODE) HAVING score 0.2 ORDER BY score
DESC;;
$results = mysql_query($query, $database) or die(mysql_error());
$num_rows = mysql_num_rows($results);
for($i=0;$i$num_rows;$i++)
{
$row = mysql_fetch_object( $results);
$body = substr($row-section_one, 0, 150);
$results .= bra
href=\index.php?page=.$row-name.\.$row-title./a
(.$row-branch.)br.$body.brbr;
}
Here is the results:
---
Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result
resource in C:\Inetpub\wwwroot\ymca\index.php on line 19
Notices:
--
Line 19 = $row = mysql_fetch_object( $results);
$num_rows returns a value of 8;
The first row retrieved from the results list works and returns the row with
Resource id #3 prefixed to the prinouts. After the first row all futher
retrievals fail.
Configuration:
--
I am using MySQL 4.0.12 and PHP 4.3.2
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Re: [PHP-DB] not a valid MySQL result
On Saturday 02 August 2003 11:48, Lee Templeton wrote: The first row retrieved from the results list works and returns the row with Resource id #3 prefixed to the prinouts. After the first row all futher retrievals fail. The variable $results is stressed by being forced to do two different jobs. Try using another variable to help it out. -- 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-db -- /* There are two kinds of pedestrians... the quick and the dead. -- Lord Thomas Rober Dewar */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] not a valid MySQL result
Hello guys This is my first posting to the list, even though I've had some experience with C++, these are my first steps with php and mysql. Well, the thing is that I'm getting the following message: Warning: Supplied argument is not a valid MySQL result resource in /var/www/bva/new/main/colecciones.php on line 35 The code is shown below: $query_cat = select author_names || ' ' || author_surnames from author, author-cat where author_cat.cat_code = $code and author.author_code = author-cat.author_code; $result_cat = mysql_query($query_cat); $num_results_cat = mysql_num_rows($result_cat); //This is line 35 I did the same with a simpler select sentence and it worked alright, so I guess there's a problem with the select sentence but haven't been able to find it. Does anyone have an idea? Thanks a lot for your help. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] not a valid MySQL result
Hello, It's always good practice to put some error detection in your code, i.e. $result = mysql_query($query) or die(mysql_error()); I think your problem is the mysql does not like your query, but you are not catching it until PHP tries to figure out the number of rows returned by the query. HTH -Brad Hello guys This is my first posting to the list, even though I've had some experienc e with C++, these are my first steps with php and mysql. Well, the thing is that I'm getting the following message: Warning: Supplied argument is not a valid MySQL result resource in /var/www/bva/new/main/colecciones.php on line 35 The code is shown below: $query_cat = select author_names || ' ' || author_surnames from author, author-cat where author_cat.cat_code = $code and author.author_code = author-cat.author_code; $result_cat = mysql_query($query_cat); $num_results_cat = mysql_num_rows($result_cat); //This is line 35 I did the same with a simpler select sentence and it worked alright, so I guess there's a problem with the select sentence but haven't been able to find it. Does anyone have an idea? Thanks a lot for your help. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] not a valid MySQL result
Hi Couple of points, you probably have an SQL error, I use echo $query; echo 'br'.mysql_error(); which shows the query and the error you use 'author-cat' in the table name and 'author_cat' and 'author-cat' in the where clause which will throw an error. HTH Peter --- Excellence in internet and open source software --- Sunmaia www.sunmaia.net tel. 0121-242-1473 --- -Original Message- From: Wilmar Perez [mailto:[EMAIL PROTECTED]] Sent: 06 September 2002 16:51 To: php-db mailing list Subject: [PHP-DB] not a valid MySQL result Hello guys This is my first posting to the list, even though I've had some experience with C++, these are my first steps with php and mysql. Well, the thing is that I'm getting the following message: Warning: Supplied argument is not a valid MySQL result resource in /var/www/bva/new/main/colecciones.php on line 35 The code is shown below: $query_cat = select author_names || ' ' || author_surnames from author, author-cat where author_cat.cat_code = $code and author.author_code = author-cat.author_code; $result_cat = mysql_query($query_cat); $num_results_cat = mysql_num_rows($result_cat); //This is line 35 I did the same with a simpler select sentence and it worked alright, so I guess there's a problem with the select sentence but haven't been able to find it. Does anyone have an idea? Thanks a lot for your help. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] not a valid MySQL result
Hi guys Thanks for your help. Peter was right I had an error in the sentence, I worked it out but still the same error came up. The mysql_error() function doesn't return any thing. Any idea? Thanks a lot. Hi Couple of points, you probably have an SQL error, I use echo $query; echo 'br'.mysql_error(); which shows the query and the error you use 'author-cat' in the table name and 'author_cat' and 'author-cat' in the where clause which will throw an error. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] not a valid MySQL result
Change your query to be: $query_cat = select concat(author_names, ' ', author_surnames) as somenewcolname from author, author-cat where author_cat.cat_code = $code and author.author_code = author-cat.author_code; Change somenewcolname to whatever you want the column to be called or leave the as somenewcolname part off if you don't care what the column name is. This should help. Jim ---Original Message--- From: Wilmar Perez Date: Friday, September 06, 2002 12:16:28 To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] not a valid MySQL result Hi guys Thanks for your help. Peter was right I had an error in the sentence, I worked it out but still the same error came up. The mysql_error() function doesn't return any thing. Any idea? Thanks a lot. Hi Couple of points, you probably have an SQL error, I use echo $query; echo 'br'.mysql_error(); which shows the query and the error you use 'author-cat' in the table name and 'author_cat' and 'author-cat' in the where clause which will throw an error. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php .
Re: [PHP-DB] not a valid MySQL result
Oops, you have one more error that I see, one one line you reference author-cat as a table name and on another you refer to it as author_cat. Make the appropriate change and try it again. One of the table references is wrong and will generate a missing table error. Jim ---Original Message--- From: Wilmar Perez Date: Friday, September 06, 2002 09:57:48 To: php-db mailing list Subject: [PHP-DB] not a valid MySQL result Hello guys This is my first posting to the list, even though I've had some experience with C++, these are my first steps with php and mysql. Well, the thing is that I'm getting the following message: Warning: Supplied argument is not a valid MySQL result resource in /var/www/bva/new/main/colecciones.php on line 35 The code is shown below: $query_cat = select author_names || ' ' || author_surnames from author, author-cat where author_cat.cat_code = $code and author.author_code = author-cat.author_code; $result_cat = mysql_query($query_cat); $num_results_cat = mysql_num_rows($result_cat); //This is line 35 I did the same with a simpler select sentence and it worked alright, so I guess there's a problem with the select sentence but haven't been able to find it. Does anyone have an idea? Thanks a lot for your help. *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php .
RE: [PHP-DB] Not a valid MySQL result resource
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 ?php //.../... 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] --- Incoming mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.295 / Virus Database: 159 - Release Date: 01/11/01 --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.295 / Virus Database: 159 - Release Date: 01/11/01 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] Not a valid MySQL result resource
?php //.../... 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 !! ? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Not a valid MySQL result resource
At 2:04 AM +0100 11/9/01, MPropre wrote:
?php
//.../... 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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Re: [PHP-DB] Not a valid MySQL result resource
?php //.../... 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Not a valid MySQL result resource
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:
?php
//.../... 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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