I want to write html image tag inside php but the below code is not working, Please help.
<?php
....
....
echo '<img src="../admin/upload/'<?=$display_img?>" width="120px" height="120px"/>
.....
?>
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2 Answers
You don't need the php tags you just need to break out of the string:
$width = 120;
echo '<img src="../admin/upload/' . $display_img . '" width="' . $width . 'px" height="120px"/>';
9 Comments
user3653474
thanks for your answer but it is not working there is syntax error some where in this line.....
Fluffeh
Fixed up the typo in the answer.
Cjmarkham
Fixed the other typo :P
Fluffeh
@CarlMarkham LMAO. Cheers for that. I swear my head was tilted for a second looking at it. +1
Cjmarkham
Yeh, quotes can be a pain when there is no syntax highlighting
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- you are inserting a php code block
<? ?>when you are already in another. <?=$display_img?>the equal sign doesn't belong here and the code is attached to the php tags, so the compiler may not be able to read it. It was better this way<?php echo( $display_img ) ?>, of course if you weren't already in php.Echoing is a bit messy for me, I would've exited php first like this.
<?php .... .... ?> <img src="../admin/upload/<?php echo( $display_img ) ?>" width="120px" height="120px"/> <?php ..... ?>
or here is your original code corrected:
<?php
....
....
echo '<img src="../admin/upload/' . $display_img . '" width="120px" height="120px" />';
// OR
echo '<img src="../admin/upload/{$display_img}" width="120px" height="120px" />';
.....
?>
4 Comments
Cjmarkham
echo is a statement not a function. No need for brackets.
profimedica
Prior to PHP 5.4.0, this short syntax only works with the short_open_tag configuration setting enabled for using <?=