What is the error in my code that I get mysql_fetch_array warning?

Question

I am trying to display some data from mySQL, the db details are correct but I get this.

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

What is the error with my code below?

<?php

$con = mysql_connect("localhost"," "," ");
mysql_select_db(" ", $con);
mysql_query('set names utf8');

$query = "SELECT * FROM products WHERE id = '13' LIMIT 1";
$row = mysql_fetch_array(mysql_query($query));
echo $row['name'];

?>

UPDATE

I found my error it was a typo. This is what you get if you still use notepad...

Sheesh. Downvoting spree... Would the anonymous coward who hates all the correct answers please step forward and explain themselves? — Marc B, Aug 15 '11 at 15:37
@Greg: Note the timestamps on the answers. They were all done at pretty much the same time. You don't get notified by SO immediately when an answer is posted. — Marc B, Aug 15 '11 at 15:49
... and no matter what, it's not good form to downvote correct answers @Greg. If you see a pile of "me too" answers, just leave them alone (or leave an acidic comment) — Pekka, Aug 15 '11 at 16:06

score 1 · Accepted Answer · answered Aug 15 '11 at 15:32

1

$query will be FALSE because the query you have entered is not valid; the return from mysql_query is FALSE when there was some error compiling or executing the query.

Change LIMT to LIMIT and try again.

answered Aug 15 '11 at 15:32

cdhowie

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score 1 · Answer 2 · answered Aug 15 '11 at 15:32

1

Wrong spelling...

$query = "SELECT * FROM products WHERE id = '13' LIMT 1";

should be

$query = "SELECT * FROM products WHERE id = '13' LIMIT 1";

answered Aug 15 '11 at 15:32

ace

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score 1 · Answer 3 · answered Aug 15 '11 at 15:32

1

You've got an error somewhere. Try:

$con = mysql_connect(...) or die(mysql_error());

mysql_select_db(...) or die(mysql_error());

$res = mysql_query(...) or die(mysql_error());
$row = mysql_fetch_array($res);

answered Aug 15 '11 at 15:32

Marc B

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score 1 · Answer 4 · answered Aug 15 '11 at 15:32

1

You have a typo, LIMT -> LIMIT

answered Aug 15 '11 at 15:32

Ed Heal

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score 1 · Answer 5 · edited May 23 '17 at 10:34

1

Use error handling to get information about errors. The message will tell you that you have a syntax error (LIMT instead of LIMIT).

A minimal example:

$query = "SELECT * FROM products WHERE id = '13' LIMT 1";

if (!$query) trigger_error("mySQL Error: ".mysql_error(), E_USER_ERROR);

The use of trigger_error() instead of die() is so you can avoid the message getting shown on a live site. Other than that, die() is also fine.

See also Reference: What is a perfect code sample using the mysql extension?

edited May 23 '17 at 10:34

Community

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answered Aug 15 '11 at 15:33

Pekka

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Thank you for the reference, it will help me as a beginner to write better code. – EnexoOnoma Aug 15 '11 at 15:38

score 0 · Answer 6 · answered Aug 15 '11 at 15:32

0

Simple typo: LIMT should be LIMIT

answered Aug 15 '11 at 15:32

JK.

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What is the error in my code that I get mysql_fetch_array warning?

6 Answers6