ID: 27727
User updated by: glmlc at hotmail dot com
Reported By: glmlc at hotmail dot com
Status: Open
Bug Type: MySQL related
Operating System: Windows 2000
PHP Version: 4.3.4
New Comment:
>Keep in mind that the info stored in the $_POST is
>selected from a table in the db called "books".
correction: it is a table in the db called "books"..
Previous Comments:
------------------------------------------------------------------------
[2004-03-27 01:23:22] glmlc at hotmail dot com
Description:
------------
I used the MySql_Query with an argument put together as a string, part
of it included the $_POST varible, and when passing the string to the
MySql_Query I got no error, yet no resource id to use for the rest of
my function. I have tried to force the stringtype to the varible and
that still didn't work. Keep in mind that the info stored in the $_POST
is selected from a table in the db called "books". Yet when I hard
code the same format in the string it returns it. Only get the problem
when I use the varible.
Reproduce code:
---------------
<?php
if ($_POST != NULL) {
echo "results for " . $_POST["Subject"];
$subject = $_POST["Subject"];
$qry = ("select * from " . $subject);
$sqlQry = (string) $qry;
GetVals();
}
function GetVals() {
$db = mysql_connect("localhost", "root");
mysql_select_db("books",$db);
$result = mysql_query($sqlQry);
echo $result;
}
?>
Expected result:
----------------
I expect it to echo the resource ID. This was a test code I developed
while debugging the rest of the project. I leveled it down to this
problem. It's not allowing me to use any of the rest of the mysql
functions because the resource id is non-existant. I don't get errors
until I try to use functions like MySql_fetch_row($result) and whatnot,
and those errors say that the argument it invalid.
Actual result:
--------------
null is stored in $result, leaving me high and dry for the rest of the
project.
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Edit this bug report at http://bugs.php.net/?id=27727&edit=1
