My site reads all files in a directory using glob(); statement in php. I want to use those files' names as a source for images to be displayed as
<img src="Here I want those file names from php">
I am a beginner and I hope if anyone can help me for this.
$files = glob();
<img src="$files[0]" />
<img src="$files[1]" />
You can use forloop to show all images.
<?php for($i=0;$i<count($files);$i++){ ?>
<img src="<?php echo $files[$i] ?>" />
<?php } ?>
Related
I want to keep src attribute in a variable. I am working on php and tried the following method but it doesn't display the image on html page.
Code:
<?php $path="C:/horizontal.jpg"; ?>
<image src="<?php echo $path; ?>" style="width:304px;height:228px" />
I think you can use this technique:
<?php $path="http://" . $_SERVER["SERVER_NAME"];?>
<img src="<?php echo $path.'/projectName/image.jpg'; ?>" style="width:304px;height:228px" />
Where:
$_SERVER["SERVER_NAME"]: to get the server name, for example: www.example.com.
Also you can use $_SERVER["SERVER_ADDR"], in this case you can get the address IP of your server, for example: "127.0.0.1".
I hope this information helps you.
Good Luck.
<?php $path="localhost/projectName/";?>
<img src="<?php echo $path.'image.jpg'; ?>" style="width:304px;height:228px" />
I'm trying to create a simple image gallery that displays thumbnails of uploaded images. Once a thumbnail is clicked, I would like to be directed to a page with the large version of the image, along with a comment section. So basically I'm trying to do something similar to deviantart. What I have now looks something like this:
<a href="<?php echo $image->large_image_path; ?>">
<img src="<?php echo $image->thumbnail_image_path; ?>"></a>
Clicking on a thumbnail will take to me to the large image path, which is not really what I want. Any help is greatly appreciated.
You must make the href="<?php echo $image->large_image_path; ?>" to somehing like href="show_image.php?image_path=<?php echo $image->large_image_path; ?>"
In show_image.php you can den get the path of the image by $_REQUEST['image_path'], and add it into the code like this:
<img src="<?php echo $_REQUEST['image_path']; ?> />
The you can add information or styling around the bigger image.
So, link to a PHP page instead of the image. Even better, put the image path in to a database, and use the image id to get the path and information of the image. Like this:
href="show_image.php?image_id=<?php echo $image->id; ?>"> and then in show_image.php, given that you have a method for getting the image:
<?php $image = GetImage($_REQUEST['image_id']); ?>
<img src="<?php echo $image->large_image_path; ?> />
<?php echo $image->description; ?>
<?php echo $image->date; ?>
Hope this helps you on the way.
I am trying to display image from a blob field of a MySQL table. Looks like I have some sort of error in the following line. As soon as I put "header("Content-type: image/jpeg")" things get messed up and instead of displaying webpage, all source code of the page is displayed.
Please let me know how to correct.
<div class="image" align="left">
<a href="<?php header("Content-type: image/jpeg"); echo $rec['image']; ?>">
<img src="<?php echo $rec['image']; ?>" width="150" border="0"/>
</a>
</div><!-- image -->
You normally don't put the actual image contents in the src= attribute of the image tag. Instead, you point to the URL of an image file.
(There are ways to include the image source directly in the HTML, but it doesn't work consistantly with all browsers, and you still won't have your <a> link working properly.
Instead, the best way to do this is to create a separate PHP file to serve the image.
Your HTML:
<div class="image" align="left">
<img src="myimage.php?key=<?php echo($key) ?>" width="150" border="0"/>
</div><!-- image -->
myimage.php:
<?php
header("Content-type: image/jpeg");
$key = $_GET['key'];
// todo: load image for $key from database
echo $rec['image'];
You're trying to put the image data inline inside the content. The only feasible way to do this is via a Data URI data URI. Something like:
<img src="data:image/jpeg;base64,<?= base64_encode($rec['image']) ?>" width="150" border="0" />
However, what you probably want to do is put it into a separate script. So your HTML would be:
<img src="showimage.php?id=XXX" width="150" border="0" />
And your showimage.php script would be:
<?php
// Get $rec from database based on the $_GET['id']
header('Content-Type: image/jpeg');
echo $rec['image'];
?>
I've done something like that retrieving blob from my database in another way that you may find useful, here is the code example.. see if it suits your needs and if you needed anymore help let me know.
while ($row = mysql_fetch_array($hc_query2)) {
$title = $row['title'];
$text = $row['text'];
$image = $row ['image'];
$output ='<div class="HCInstance"><img src="data:image/jpeg;base64,' . base64_encode($image) . '" alt="High Council" width="100px" height="100px"/>
<div class="HCHeader"><h2>'.$title.'</h2></div><br/><div class="HCDetails"><p>'.$text.'</p></div></div>';
echo $output;
}
I have a PHP echo function inside of a HTML link, but it isn't working. I want to have an image location, defined in img src, be in part of the clickable link of the image. The page will have multiple images doing the same thing, so I am trying to use PHP to automate this.
<a href="http://statuspics.likeoverload.com/<?php echo $image; ?>">
<img src="<?php $image=troll/GrannyTroll.jpg?>" width="100" height="94" />
</a>
Turn
<?php $image=troll/GrannyTroll.jpg?>
into
<?php echo "troll/GrannyTroll.jpg"; ?>
?
Or provide more details on what you are trying to achieve.
Also, you might consider urlencode-ing some of those URL parameters.
Edit:
So you might try setting the variable beforehand:
<?php $image = "troll/GrannyTroll.jpg"; ?>
<img src="<?php echo $picture; ?>" width="100" height="94" />
So now i understand what you are trying to do.
One error is that you didn't enclose $image=troll/GrannyTroll.jpg with quotes like this:
$image = 'troll/GrannyTroll.jpg';
The second error is that you do it in the wrong order, you have to define $image first, before you use it.
That's what I believe you want to do:
<?php
$image = "troll/GrannyTroll.jpg";
?>
<img src="<?php echo $image; ?>" width="100" height="94"/>
I'm using Drupal 6.x. This is my code on my node-product.tpl.php template. I've created a custom jquery gallery for the products. It works great, but I'm just missing tool tips from my images (both large and small thumbnails). To upload images I'm using a CCK field named field_images. There I input the image titles when I upload the images. How can I add the tool tip code snippet to make it work?
<div class="product-large">
<img src="/sites/default/files/imagecache/360x280/<?php print $node->field_images[0]['filename']; ?>" />
</div>
<div class="product-small">
<?php
// get all images
foreach ($node->field_images as $images) {
?>
<img src="/sites/default/files/imagecache/120x90/<?php print $images['filename']; ?>" rel="/sites/default/files/imagecache/360x280/<?php print $images['filename']; ?>" />
<?php
}
?>
</div>
Thanks much appreciated!
Chris
The "tooltip" is a result of the title HTML attribute. You'll want to add title="foo" to your img tag.
Perhaps:
<img src="/sites/default/files/imagecache/120x90/<?php print $images['filename']; ?>" rel="/sites/default/files/imagecache/360x280/<?php print $images['data']['filename']; ?>" title="<?php print $images['title']; ?>"/> for the second image and similarly $node->field_images[0]['data']['title'] for the first.