[SOLVED] else statement

[squiblo]

ok my aim here is to view the profile image if they have uploaded one, but if they have not uploaded one i want to view another image, i have started already but got stuck trying to view another image if they havent uploaded one...

 

<?php

session_start();

include("checklogin.php");

$username = $_SESSION['myusername'];

$query = mysql_query("SELECT * FROM members WHERE username='$username'");
if (mysql_num_rows($query)==0)
   die("User not found!");
else
{

$row = mysql_fetch_assoc($query);
$location = $row['imagelocation'];
echo "<img src ='$location' width='225' height='275'>";
}


?>

 

the path to the image if they havent uploaded one is...

/profileimages/box.png

[arkansasonline]

If I understand what you want to do correctly.  After your if remove the die and echo the other image.

**Edit**

I did not understand correctly.

In your else you could do

if($location!='')
echo "<img src ='$location' width='225' height='275'>";
else
echo "<img src='/profileimages/box.png'>";

[p2grace]

Something like this should do the trick.

 

<?php

session_start();

include("checklogin.php");

$username = $_SESSION['myusername'];

$query = mysql_query("SELECT * FROM members WHERE username='$username'");
if (mysql_num_rows($query)==0){
die("User not found!");
}else{
$row = mysql_fetch_assoc($query);
$location = $row['imagelocation'];
if($location == ""){
	echo "<img src ='/profileimages/box.png' width='225' height='275'>";
}else{
	echo "<img src ='$location' width='225' height='275'>";
}
}


?>

[squiblo]

that worked perfect thankyou, i have linked the images, but it puts a purple border on, how can i stop this?

 

<div id="wb_Shape5" style="position:absolute;left:84px;top:240px;width:225px;height:275px;z-index:11" align="center"><a href="/upload.php">
<?php

session_start();

include("checklogin.php");

$username = $_SESSION['myusername'];

$query = mysql_query("SELECT * FROM members WHERE username='$username'");
if (mysql_num_rows($query)==0){
   die("User not found!");
}else{
   $row = mysql_fetch_assoc($query);
   $location = $row['imagelocation'];
   if($location == ""){
      echo "<img src ='/profileimages/box.png' width='225' height='275'></a>";
   }else{
      echo "<img src ='$location' width='225' height='275'>";
   }
}


?>
</a></div>

[arkansasonline]

give the image a style attribute of

echo "<img src ='/profileimages/box.png' width='225' height='275' style='border-style:none'></a>";