Assign Image SRC in php

Question

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.

Please give us an example of the output of $src which didn't work. We have no idea what your content looks like. The basic idea seems fine, but maybe something is up with the contents. — ADyson, Apr 15 '19 at 08:13
The last line with the HTML...can we assume this is part of a larger `echo` or `print` statement? Otherwise it won't output the variable's value, you'll just get the literal word $src written into the final HTML instead. Check the View Source option in your browser to see what is actually being created — ADyson, Apr 15 '19 at 08:17
"I do not want to echo it" -> I bet that yes, you want, you just don't know it, it should be something like `` — Kaddath, Apr 15 '19 at 08:18
@devpro I do not want to echo it as I mentioned in my question. — Abdulsalam Elsharif, Apr 15 '19 at 08:18
What's your objection to echo? And how do you expect to output the result of any PHP operation if you don't use it?? — ADyson, Apr 15 '19 at 08:19

score 2 · Answer 1 · answered Apr 15 '19 at 08:17

2

Try with PHP tags like below:-

<img id="img1" width="150" height="150" src="<?php echo $src;?>"/>

answered Apr 15 '19 at 08:17

Akshay Paghdar

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score 2 · Accepted Answer · answered Apr 15 '19 at 08:17

2

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; ?>"/>

answered Apr 15 '19 at 08:17

Nick

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score 2 · Answer 3 · answered Apr 15 '19 at 08:18

2

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.

answered Apr 15 '19 at 08:18

sassy_rog

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score 2 · Answer 4 · answered Apr 15 '19 at 08:38

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 ?>"/>

Assign Image SRC in php

4 Answers4