>From your query, it looks as if you are letting the user select the table that they
>can query with $pop-up1. Is that correct?
If so, do all the available choices of tables have the same column names? Other than
that I can't see why this query would fail.
Martin
>>> eat pasta type fasta <[EMAIL PROTECTED]> 07/03/02 04:27PM >>>
I already got some replies for this but it wasn't it, here is the problem:
I have a simple form which queries the database based on 2 pop-up menu
choices and keywords entered by the user,
it passes on the variables as follows:
pop-up1=choice1, pop-up2=choice2, textfield=user specified data
the query is as follows
$result = mysql_query("SELECT some_id, some_description, some_feature
FROM '$pop-up1' WHERE some_description LIKE '%$texfield%' AND
some_feature LIKE '%$pop-up2'");
it tells me that column which equals to the $texfield does not exists,
meaning that the query seems to be fed wrong? No idea here, $texfield and
$pop-up2 are not requests for tables but matches in those tables
a bit buffled I am, I've these queries before on larger web servers, this
is the first time on my workstation and it does just that, simpler
queries work fine...
thanks in advance,
R>
ps. the form is guarded against empty entries
ps2. RedHat 7.3, Mysql 3.23.51, Apache 1.3+ wik PHP 4.2.1 configured for
MySQL
--
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