How to grab a part of existing URL and use it in html code?

Question

From the URL, let's say google.com/web1.php I would like to grab the part between the last slash and the extension, so: "web1" and then use it to point the specific image. Here is the example:

for url: google.com/web1.php

<img src="web1.jpg">

for url: google.com/web2.php

<img src="web2.jpg">

I know how to insert the whole URL, but don't know how to get a part of it. Here is what I use for the whole URL:

<img src="<?php echo "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]"; ?>.jpg"/>

How to get only the part I want instead of the whole URL? Thanks

Ezequiel Fernandez · Accepted Answer · 2019-03-31 21:53:56Z

1

If you print just:

echo  $_SERVER[REQUEST_URI];

It will print only the part you need.

Usually, the URI of a URL does not have the extension. In your case need to parse it.

echo preg_replace('/\\.[^.\\s]{3,4}$/', '', $_SERVER[REQUEST_URI]);

If you want to add the code on each php file, you could do this:

 echo pathinfo(__FILE__, PATHINFO_FILENAME);

edited Mar 31, 2019 at 21:53

answered Mar 31, 2019 at 21:41

Ezequiel Fernandez

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4 Comments

Piotr Ciszewski Over a year ago

will it contain only the filename without the extension or domain name too? I need just a filename, as I will name .jpg's the same way as websites and will need to have a different jpg on each of pages.

Ezequiel Fernandez Over a year ago

Yes, that code only will have that. also if you have many PHP files, on each file you could make this: (see updated post)

Piotr Ciszewski Over a year ago

Almost perfect, the only thing that needs tweaking is the last slash at the end. Currently it looks like this: <img src="web1/.jpg"> Can you help me get rid of the slash after the grabbed part? Thank you

Ezequiel Fernandez Over a year ago

echo pathinfo($_SERVER['REQUEST_URI'], PATHINFO_FILENAME);

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