Query Problem!!

[obrienkev]

Hi all,

Why won't this work properly?

$query = "select * from answers, questions where name like '%$name%'" or die(mysql_error());
$result = mysql_query($query);

the result it gives is everything in the database. I want it to only display results very similar ti $name variable.

Even if $name is blank it displays all the results.

Why is this?

Thanks.

Kevin.

[Nedwin]

I think its because your checking for errors before submitting the query

$query = "select * from answers, questions where name like '%$name%'";
$result = mysql_query($query) or die(mysql_error());

Just a thought??

[obrienkev]

Nope. Also tried this:

$query = "select * from answers, questions where answers.name like '%".$name."%'";
		$result = mysql_query($query) or die(mysql_error());

[kpzani]

you need to join the tables like this

$sql = "SELECT answers., questions. FROM answers, questions WHERE answers.qno=questions.qno AND answers.name='" . $name . "'";

also modify the query checking to

$result = @($sql);
if (! $result) {
echo("SQL: " . $sql . "<br>Error: " . @mysql_error());
exit();
}
if (@mysql_num_rows($result) == 0) {
echo("No results returned");
exit();
}

// do stuff