mySQL Search

[billwatson]

hi

dont know much php but have been attempting to alter a downloaded piece of code to perform a search on a mysql database and return the results to flash

<?php
mysql_connect("localhost","bill","pass");
mysql_select_db("products");

$table = "produce";
$condition = $HTTP_POST_VARS['word'];
$qr = mysql_query("SELECT * FROM $table WHERE product LIKE $condition");

$r_string = 'n='.mysql_num_rows ($qr);
$i = 0;
while ($row = mysql_fetch_assoc ($qr)) {
while (list ($key, $val) = each ($row)) {
$r_string .= '&' . $key . $i . '=' . $val ;
}
$i++;
}
echo $r_string;

?>


the code returns the following errors

PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\phpdev\

http://www\public\watsons\search.php

on line 9
PHP Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:\phpdev\

http://www\public\watsons\search.php

on line 11


any pointers as to what I am doing wrong will be greatly appreciated

 

[MJB3K]

try adding or die(mysql_error()); after your mysql_query(); it tells you what is wrong with your query!

also try this query:

$qr = mysql_query("SELECT * FROM $table WHERE product LIKE '%$condition%'");

Regards,

Martin

Gaming Help And Info:

http://gaming.webrevolt.biz

 

[billwatson]

thanks martin,

tried your changes but still gives same 2 errors posted above

 

[MJB3K]

does it say what error it gives after you put or die(mysql_error()); after your query?

if so can you post it here please!


Regards,

Martin

Gaming Help And Info:

http://gaming.webrevolt.biz

 

[MDLU]

try assigning a pointer to your database connection:

$connection=mysql_connect("localhost","bill","pass");
mysql_select_db("products");

and then when you run your query, use the pointer:

$result=mysql_query("SELECT...",$connection);
$num_rows=mysql_num_rows($result);
etc....

 

[chessbot]

Are you sure that "product" is a field of $table? And try "WHERE product[red]=[/red]$condition" just for some variation.

I recently encountered this problem, and the cause was that an invalid SQL line caused my mysql resource to be defined as false, I believe, instead of a resource.

--Chessbot

"See the TURTLE of enormous girth!"
-- Stephen King, The Dark Tower series

 

[billwatson]

thanks all for your help

i found another script that i was able to make work just as i wanted