As you can see, there is no image. What's wrong with this?
$displayProdCat .= '<div class="product">
<img src="Customer/images/product'.$ItemNo.'.jpg" width="170" height="150" />
<h3>'.$ItemName.'</h3>
<p class="product_price">Php '.$Price.'</p>
Add to Cart</div>';
Probably a wrong image url is specified. Look at the source.
Related
I'm using wordpress and I want to change the logo on some pages. Since the theme I'm using (flatsome) doesn't support this I thought using php would be a good idea. I'm not sure how to do it though.
I tried this:
<?php
function change_logo_on_single($html) {
if(is_single( array(1441, 1425, 1501, 1494, 1498, 1503))){
$html = preg_replace('<a(.*?)><img(.*?)><img(.*?)></a>',
'<a href="https://example.com/" title="" rel="home">
<img width="146" height="150" src="https://example.com/wp-content/uploads/2021/02/Logo.png" class="header_logo header-logo" alt="">
<img width="146" height="150" src="https://example.com/wp-content/uploads/2021/02/Logo.png" class="header-logo-dark" alt="">
</a>', $html);
}
return $html;
}
add_filter('get_custom_logo','change_logo_on_single');
?>
I think I used the wrong pattern in preg_replace. Can someone suggest a way to do it?
After retrieve image path from database I want to pass it to <img> tag to show up,
I read this solution, but I do not want to echo it, I want to assign it as src for image tag
I tried this:
$image = mysqli_fetch_array($user_images);
$img1 = $image['file_path'];
$imageData = base64_encode(file_get_contents($img1));
$src = 'data: '.mime_content_type($img1).';base64,'.$imageData;
How can I assign $src to image, I tried this but no luck :(
<img id="img1" width="150" height="150" src="' . $src . '"/>
Thanks in advance.
Inside the HTML you still need to echo the value of $src. This should work:
<img id="img1" width="150" height="150" src="<?php echo $src; ?>"/>
Try with PHP tags like below:-
<img id="img1" width="150" height="150" src="<?php echo $src;?>"/>
Reasonable solution I can think of is
<img id="img1" width="150" height="150" src="<?php echo $src; ?>"/>
but the code where you get the image source has to be above the image tag, so you have the $src.
You are using PHP language and in Php, you can't pass variable directly in the Tag you have to set a variable in the PHP environment.
To create Php environment you can right <?php ?> or you can use to echo variable <?= ?> between these PHP tags you can pas your variables to echo and assign or anything else what you want to do.
<img id="img1" width="150" height="150" src="<?= $src ?>"/>
$images['result'][0]['image_path'] - here I have path of image. It is in public folder.
<img width="150" height="150" alt="150x150" src="<?php BASEPATH."../public/" echo $images['result'][0]['image_path'];?>" />
Can anyone help me to display this image?
Try This,
<img width="150" height="150" alt="150x150" src="<?php echo base_url('your/Folder/Structure/').$images['result'][0]['image_path'];?>" />
First concate your file name with path which causing the issue.
base_url() will also help you to fetch/show your image
I create a block from configuration. Inside this block, I write this code...
<div class="provider-bg" id="provider1">
<img alt="" class="img-responsive" src="<?php echo base_path().path_to_theme() ?>/images/provider-1.jpg" />
</div>
Then, I save with PHP in text format.
My virtual host is ... http://localhost:8888/drupal
So, the image path will be like this ...
<img alt="" src="/drupal/sites/all/themes/myancast/images/ios.png">
This image appears in the last few days ago. Today, I run the site and that image disappear immediately and got 403 error.
Failed to load resource: the server responded with a status of 403 (Forbidden).
I'm trying to find the solution the whole day. But, I still cannot solve.
Can anyone help me please ?
Add global $base_url in your code and use like below
<img alt="" class="img-responsive" src="<?php echo $base_url.'/'.path_to_theme(); ?>/images/provider-1.jpg" />
did you try this ?
<img alt="" class="img-responsive" src="<?php echo '/'.path_to_theme(); ?>/images/provider-1.jpg" />
try this,
<?php
$imgurl = file_create_url(path_to_theme().'/images/provider-1.jpg');
?>
<img alt="" class="img-responsive" src="<?php echo $imgurl ?>"/>
Pretty straight forward simple question, can you open a php code block to call image information in html? I don't think I phrased that right. Here is my code:
<img src="../inventory_images/' . <?php echo $item_number; ?> . '.jpg" width="150" height="150" border="2" />
This code is within the tags
I'm just trying to post a photo using the $item_number variable (which is also the name of the image file i.e. $item_number = T3144 and the image file is name T3144.jpg ). Also if there is a better way to accomplish this suggestions are happily accepted. Sorry to take up bandwidth with such a remedial question but for some reason I can't seem to answer this question in research. Thanks for taking the time everyone.
Your code is wrong, try:
<img src="../inventory_images/<?php echo $item_number;?>.jpg" width="150" height="150" border="2" />
with what you have it looks like the code you had would print
src="../inventory_images/' . whateveritem_numberis . '.jpg"
Yes that is perfectly fine, but make sure that this code is in a file that ends with .php or it will not get parsed by PHP. Also, you need to take out the single quotes and periods:
<img src="../inventory_images/<?php echo $item_number; ?>.jpg" width="150" height="150" border="2" />
Unless the above HTML is in an echo statement, you need to change it to this:
<img src="../inventory_images/<?php echo $item_number; ?>.jpg" width="150" height="150" border="2" />
That will in-turn look like this:
<img src="../inventory_images/T3144.jpg" width="150" height="150" border="2" />
Of course, that is going off of your example where $item_number = 'T3144';.
The single quotes and periods are used for concatenating variables inside of strings.