Using php to display an image by passing to html img tag contains variables passed from php

Question

I am passing an image link from PHP to be displayed on a HTML page. But, when the page is rendered, the image does not show, instead, a place holder is shown.

PHP

$details = getimagesize($config['thumb_dir'].$value['thumbName']);
if ($details !== false){
    $content.= "<h2>".$value['title']. "</h2>"."<p><img src= ".$config['thumb_dir'].$value['thumbName']." width =".$details[0]." height = ".$details[0] ." alt= ".$value['filename'] ." /></p>";

HTML output

<h2>Welcome to the Home page</h2><h2>Title for image 1</h2>
<p>
    <img src= /home/rraja01/public_www/w1fma/thumbs/thumb-landscape-large.jpg width = 150 height = 150 alt= landscape-large.jpg />
</p>

what is the pages extension is it .php or .html if the later then it wont work it need to be named .php etc just as you mentioned "on a HTML page" ? — Simon Davies
– Simon Davies, Commented Dec 30, 2013 at 19:18
also could you show the output of the html page as in the code onto the result your using? — Simon Davies
– Simon Davies, Commented Dec 30, 2013 at 19:19
Please, don't put spaces between your HTML attributes name and value. You should also escape characters: <img src=\"".$config['thumb_dir'].$value['thumbName']."\"/> — Jeff Noel
– Jeff Noel, Commented Dec 30, 2013 at 19:24
the src tag in your html output must have double quotes around the path. try putting escaped quotes around the php variable. — czuroski
– czuroski, Commented Dec 30, 2013 at 19:24
@JeffNoel i know that i wanted to make sure that was not the mistake in the first :-) — Simon Davies
– Simon Davies, Commented Dec 30, 2013 at 19:32

Jeff Noel · Accepted Answer · 2013-12-30 19:29:37Z

1

Your $config['thumb_dir'] variable is returning the full path on the server and including /home/rraja01/public_www/ at the front -- very few servers have the root mapped to the website. You may need to do something like this:

$website_path = substr($config['thumb_dir'], 24);
$content.= "<h2>".$value['title']. "</h2>".'<p><img src="'.$website_path.$value['thumbName'].'" width="'.$details[0].'" height="'.$details[0].'" alt="'.$value['filename'].'" /></p>';

edited Dec 30, 2013 at 19:29

Jeff Noel

7,6274 gold badges43 silver badges67 bronze badges

answered Dec 30, 2013 at 19:27

user1932079

Sign up to request clarification or add additional context in comments.

2 Comments

user3144556 Over a year ago

below is what i am putting into the $config: $config['app_dir'] = dirname(dirname(FILE)); $config['upload_dir'] = $config['app_dir'] . '/uploads/'; $config['template_dir'] = $config['app_dir'] . '/template/'; $config['view_dir'] = $config['app_dir'] . '/views/'; $config['thumb_dir'] = $config['app_dir'] . '/thumbs/';

user1932079 Over a year ago

dirname returns the path to the file, which I presume is actually __FILE__ -- there's no need to do it twice and when you echo it out to the browser, it's giving too much of a path... you only want the path relative to the domain, not the server.

Markus Kottländer · Accepted Answer · 2013-12-30 19:34:29Z

0

You have to use the relative path of your image (relative to the document root):

$details = getimagesize($config['thumb_dir'].$value['thumbName']);

$relative_path = str_replace($_SERVER['DOCUMENT_ROOT'], '', $config['thumb_dir']);

if ($details !== false) {
    $content .= '<h2>'.$value['title'].'</h2>'
             .  '<p><img src="'.$relative_path.$value['thumbName'].'" width="'.$details[0].'" height="'.$details[0].'" alt="'.$value['filename'].'" /></p>';
}

answered Dec 30, 2013 at 19:34

Markus Kottländer

8,2844 gold badges42 silver badges61 bronze badges

Collectives™ on Stack Overflow

Using php to display an image by passing to html img tag contains variables passed from php

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related