SQL query fails when adding a text field to the WHERE clause.

[AaronL]

Hi guys,

I just started to learn MySql and php and am relatively new to how databases work. I setup my database and wrote the following code and I get the error:

Warning: mysql_num_rows( ): supplied argument is not a valid MySQL result resource

[code=php]
mysql_select_db ("breedknowledg e", $con);

$qheight = $_POST["height"];
$qminwgt = $_POST["minwgt"];
$qmaxwgt = $_POST["maxwgt"];
$qexercize = $_POST["exercize"];
$qgrooming = $_POST["grooming"];

if ($qheight == "") { $qheight = "99"; }
if ($qminwgt == "") { $qminwgt = "1"; }
if ($qmaxwgt == "") { $qmaxwgt = "999"; }

$result = mysql_query("SE LECT * FROM breeds WHERE weight >= $qminwgt AND weight <= $qmaxwgt AND exercize = $qexercize ORDER BY breedname");
$numrows = mysql_num_rows( $result);
[/code]
Now when I remove "AND exercize = $qexercize" I don't get the error. The only difference is that this field is a text field. I can run the query just fine in my MySQL admin program through my hosting site. The code looks fine I don't get it! Please help! Thanks!

[AaronL]

Nevermind I figured it out, I found that text variable references need to be delimited by a ' so I created a $sqlbuild variable and built the select code differently

[Atli]

Hi.

Did you try just adding the single-quotes?
[code=php]
mysql_query("SE LECT ... exercize = '$qexercize' ...");
[/code]
In any case, glad you found a solution.

P.S.
Please use &#91;code] tags when posting your code examples.
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&#91;code] ... Code goes here... &#91;/code]

P.P.S.
I changed the title of the thread.
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Thank you.
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