[SOLVED] echo other picture

[squiblo]

if my database cannot find a member profile picture how can i echo a different picture?

 

<font face="arial">

<?php

//get data
$button = $_GET['submit'];
$search = $_GET['search'];

if (!$button)
header("location:unititled.php");
else
{
if (strlen($search)<3)
	header("location:unititled.php");
else
{
	echo "";

	//connect to our database
	mysql_connect("localhost","","");
	mysql_select_db("");


		//explode our search term
		$search_exploded = explode(" ",$search);

		foreach($search_exploded as $search_each)
		{

		//construct query
		$x++;
		if ($x==1)
		    $construct .= "username LIKE '%$search_each%'";
		else
		    $construct .= " OR username LIKE '%$search_each%'";			
	    
		}


	//echo out construct

	$construct = "SELECT * FROM members WHERE $construct";
	$run = mysql_query($construct);

	$foundnum = mysql_num_rows($run);

	if ($foundnum==0)
	   echo "No results found.";
	else
	{
	   echo "You searched for <b>$search</b><br>$foundnum result(s) found!<p><hr size='1'>";
	   
	   while ($runrows = mysql_fetch_assoc($run))
	   {
	 //get data
                $state = ucwords($runrows['state']);
                $url = $runrows['url'];
                $username = ucwords($runrows['username']);
                $imagelocation = $runrows['imagelocation'];

	echo "
                <img src ='$imagelocation' width='100' height='110' border='0' align='left' style='padding-right:10px'><br>
	<b>$username</b><br>
	$state<br>
	<a href='$url'>View Profile</a><br><br><br>
	<hr size='1' width='300' align='left'>
      		   
      		
      		
        	
  		";

  		
             }


           }   



        }	



}	



?>
</font>

 

the profile pictures are in ['imagelocation']

[rvdb86]

If I understand correctly you are retrieving the data from a database, therefore if they have not selected a picture the field should be blank right? therefore this should work:

 

if ($imagelocation == ""){
   $imagelocation = "path to defualt image";
}