I am trying to display an image on my webpage using a PHP script to determine which image is displayed. The image link is as follows:
<a href="gallery.php?image=image01">......</a>
My PHP script is thus:
<?php
$result = $_GET['image'];
echo '<img src="images/gallery/'.$result.'.jpg">';
?>
So what I am trying to achieve in terms of HTML is:
<img src="images/gallery/image01.jpg">
The result I am getting is '"; ?>' displayed on the page. Any help would be much appreciated!
You have to change your code like this
<?php
$result = $_GET['image'];
?>
<img src="images/gallery/<?php echo $result; ?>.jpg">
<?php
$result = filter_input ( INPUT_GET , 'image' );
if (isset($result) && !empty($result)) {
echo '<img src="images/gallery/'.$result.'.jpg">';
}
?>
You used echo wrong, here is how you should use it.
<?php
$result = $_GET['image'];
?>
<img src="images/gallery/<?php echo $result ?>.jpg">
I would change the gallery.php to this:
<?php $result = $_GET['image']; ?>
<img src="images/gallery/<?php echo $result; ?>.jpg">
That would simply it a little bit. You should echo out the result to see what you are getting when the variable is passed to the gallery page.
php here... <?php $result = $_GET['image']; ?> <img src="images/gallery/<?php echo $result; ?>.jpg"> as in @ExCluSiv3 answer.echo"<img src='{$image}'>";
$image = uploads/myImage.jpg
I think this is the simplest code. To use a php variable while echoing out html, use curly {} brackets to insert any php variable. For instance, a file upload...
<?php
if(isset($_POST['submit'])){
$filename=$_FILES['file']['name'];
$temp_dir=$_FILES['file']['tmp_name'];
$image = "img/".$filename;
}
?>
<?php if($row2['pack1']==1){ echo "<img src=".BASE_URL."images/1seo.png"; } ?>
<a href="gallery.php?image=image01">......</a>, it will reload to execute$result = $_GET['image']; echo '<img src="images/gallery/'.$result.'.jpg">';?