[SOLVED] Query help, no results founds

[Mr Chris]

Hi All,

 

Can someone please help me, I think i've gone mad.  On my site I run a simple query:

 

$result = mysql_query("SELECT* from tble where id = 1");

if(mysql_num_rows($result) > 0)
{
	$row = mysql_fetch_array($result);
        echo '<h1 class="fr">'.$row['name'].'</h1>
        <div style="clear:right"></div>';	
} else {
	echo '';
}

 

Now if the criteria is matches in the database I get the details returned, but if there are no reulys found I get the message:

 

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

 

But surely my } else { should stop this happening?

 

Thanks

[JonnoTheDev]

You have an error in your query hence no database resource being returned. I can see it straight away:

Bad

$result = mysql_query("SELECT* from tble where id = 1");

Good

$result = mysql_query("SELECT * FROM tble WHERE id = '1'");

 

You should also catch query errors i.e.

if(!$result = mysql_query("SELECT * FROM tble WHERE id = '1'")) {
die(mysql_error());
}

[Adam]

Actually the space between the star and SELECT shouldn't cause a problem - or at least it doesn't on my version of MySQL (5.0.45) and being as 1 is an integer that shouldn't case a problem either - although  it could be possible with different data types.

 

Try the mysql_error() method shown and see what problems there are for you. Also is that all the code or is it possible there's a query later causing the error?

 

 

[Mr Chris]

Hi All,

 

I've tidied up my query, but am still having problems:

 

	

        $query_one = "SELECT strapline from cfm_seo WHERE site_id = ".SITEID." AND section_id = ".$subSecID.""; 
        $result = mysql_query($query_one)  OR die(mysql_error()); 

        while ($row = mysql_fetch_array($result)) { 
        $strapline = $row['strapline'];

        echo '<h1 class="fr">'.$strapline.'</h1>
        <div style="clear:right"></div>';	
}

 

I get this message

 

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

 

When I simply want nothing to show if it does not return any results?

 

Thanks

[Dathremar]

$query_one = "SELECT strapline from cfm_seo WHERE site_id = '".SITEID."' AND section_id = '".$subSecID."'";

 

Try this. I put single quotes around the variables

[Mr Chris]

That's worked, thanks.  Wierd one though as on certain pages a result was returned!

 

Thanks