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How define a variable to display an Image
Posted: Mon May 12, 2003 10:01 pm
by guycrossley
Hello,
I want to display an image, who's location is stored in a string. For Example:
$temp = "Images/UserIcons/Brettsface.png";
But, when i try
<img src= $temp width="100" height="100">
I dont get anything... What have i done wrong?
--Guy
(PS- However, this works <img src= "Images/UserIcons/Brettsface.png" width="100" height="100"> )
Posted: Mon May 12, 2003 10:34 pm
by AVATAr
Code: Select all
<?php
$temp = "Images/UserIcons/Brettsface.png";
echo '<img src= "'.$temp.'" width="100" height="100">'; //check de simple quotes and the double quotes around $temp
?>
Posted: Tue May 13, 2003 1:17 am
by Czar
you can also leave the php block and do it like;
Code: Select all
<?php
$temp = "Images/UserIcons/Brettsface.png";
?>
<img src=<?php echo $temp; ?> width="100" height="100">
Posted: Tue May 13, 2003 9:04 am
by lc
So many options... I usually do it like this:
Code: Select all
<?php
print "<img src="$temp" width="100" height="100">";
?>
Posted: Tue May 13, 2003 8:14 pm
by guycrossley
ah, thanks guys. The following code worked:
Code: Select all
<?php $temp = "Images/UserIcons/Brettsface.png";
echo '<img src="'.$temp.'">';
?>
But this didn't, why?
Code: Select all
<?php $temp = $row_rsUsersї'USER_ICON'];
echo '<img src="'.$temp.'">';
?>
Is this a database issue, or is there a datatype mismatch or something?
Posted: Tue May 13, 2003 8:28 pm
by guycrossley
Ignore the last post,
$row_rsUsers['USER_ICON']
was returning an array because i didnt define the recordset correctly.
Thx for all your help guys!
