problem with mysql_num_rows

Question

Here is my code :

$sql1 = 'SELECT * FROM login WHERE age= "$age", town = "$town" and ID != "$id"';
$result1 = mysql_query($sql1);
$numResults1 = mysql_num_rows($result1);

My variables are fine, they have data in them. The error is this :

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

There is a possibility that numResults could equal 0 but it still should not cause this.

Could it be the != in the first line that is causing it??

You can see MySQL errors if you change the second line to $result1 = mysql_query($sql1) or die( mysql_error() ); — JJJ
– JJJ, Commented Jul 24, 2011 at 19:41

Dirk · Accepted Answer · 2011-07-24 19:59:39Z

5

The problem is you have a bad SQL query:

$sql1 = 'SELECT * FROM login WHERE age= "$age", town = "$town" and ID != "$id"';

Note the improper , after age = "$age" in the WHERE clause. It should be something like:

$sql1 = "SELECT * FROM login WHERE age= '$age' AND town = '$town' and ID != '$id'";

edited Jul 24, 2011 at 19:59

answered Jul 24, 2011 at 19:29

Dirk

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9 Comments

Lightness Races in Orbit Over a year ago

Why should you use single quotes?

Inca Over a year ago

Actually, the quotes don't matter to mysql - it is php that doesn't replace $var with the value when enclosed in single quoted-strings.

Dirk Over a year ago

I thought you had to, but I see you don't.

Mike Over a year ago

The problem is not that he's using double quotes in his MySQL (see dev.mysql.com/doc/refman/5.1/en/string-syntax.html); the problem is that variables within single quotes are not evaluated by PHP.

Mike Over a year ago

Your previous answer was correct and you have now made it incorrect. What was wrong was your statement saying that MySQL doesn't accept double quotes. Now you are misusing quotes and the variables $age, $town and $id will not be evaluated by PHP.

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Lightness Races in Orbit · Accepted Answer · 2011-07-24 19:38:05Z

3

Your query has a syntax error. Check $result1 before attempting to use it; print mysql_error() when it's FALSE. You'll be told what the problem is.

Also, your use of single quotes for the SQL query string means that variable interpolation will not occur... so you are unlikely to ever get rows returned.

answered Jul 24, 2011 at 19:38

Lightness Races in Orbit

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Comments

Mike · Accepted Answer · 2011-07-24 19:37:28Z

1

You're getting your single and double quotes mixed up. Make all single quotes double quotes and vice versa.

Try echoing $sql1 to see what I mean

There is a possibility that numResults could equal 0 but it still should not cause this.

No. It means you have a problem with your SQL.

edited Jul 24, 2011 at 19:37

answered Jul 24, 2011 at 19:29

Mike

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2 Comments

Mike Over a year ago

Seriously, you are. $var = 3; echo '$var'; will echo literally $var instead of 3.

Lightness Races in Orbit Over a year ago

@Backslap: Why do you say that? (a) Yes, you do; (b) Your code doesn't even check.

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