How to add "../" into a PHP echo img src

Question

In my code I show some images in this way:

   ...
     <?php echo '<img src="' .$imageone. '" alt="" /> '?></a></div>
     <?php echo '<img src="' .$imagetwo. '" alt="" /> '?></a></div>
   ...

where $imageone, $imagetwo have path from Mysql database

e.g.
$imagesone = "images/exposition/02-05-2017-11-28-00-foto 2_b.JPG";

This code is working in the same place where I have my images folder, but now I need to put the same code in a subfolder page and I'd like to use the same image. So, I need to put dinamically "../" before my varible, something like this:

...
     <?php echo '<img src="../' .$imageone. '" alt="" /> '?></a></div>
     <?php echo '<img src="../' .$imagetwo. '" alt="" /> '?></a></div>
   ...

but it's not working.

Any suggestion?

EDIT This is my solution:

 if ($images1 != ""){

    $imageone = "../".$images1;

}

if ($images2 != ""){

    $imagetwo = "../".$images2;

}

in this way I fixed my code and it's working!

Do `ctrl+u`(on Windows) or `cmd+u`(on Mac) to see the source code, and check whether the URLs are correct or not. — Rajdeep Paul, May 08 '17 at 16:57
how it's not working? any error. seems that it need to be worked — Alive to die - Anant, May 08 '17 at 16:57
See [Absolute vs relative URLs](http://stackoverflow.com/questions/2005079/absolute-vs-relative-urls) — Masivuye Cokile, May 08 '17 at 16:57
I tried to see the source code, but there isn't "../" before the path from my database — James69, May 08 '17 at 16:58
@James69 There is a TYPO mistake in your code `$imagesone !== $imageone` (the two variable names are different) — Alive to die - Anant, May 09 '17 at 04:14
Thanks everybody! I edited my post with a solution that I found. Thanks again! — James69, May 10 '17 at 14:30

score 0 · Answer 1 · answered May 10 '17 at 16:44

0

Since you're using double quotation marks after src, you can just write down <img src="../$imageone" alt="" /> There's no need to use simple quotation marks to write the variable because the double quotation marks gets the variable number.

answered May 10 '17 at 16:44

Diéfani Favareto Piovezan

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How to add "../" into a PHP echo img src

1 Answers1