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I'm using the code below to choose an image based on the domain.

the only difference between environments is wamp is running PHP 7 and my host 7.1 and I've been unable to get a copy of wamp with PHP 7.1 included.

on the live site, the end result is a blank white page with the PHP error on the 'else' line

I'm not sure if this is correct but after $logo_img = "3_logo.svg" if I add the semicolon you get a white page but without works just fine.

<?php
                    $host = $_SERVER['HTTP_HOST']; 
                    $logo_img = '';

                    if($host == "1.co.uk") {
                        $logo_img = "1_logo.svg"; 
                    }
                    else if($host == "2.co.uk") {
                        $logo_img = "2_logo.svg"; 
                    }
                    else($host == "3.co.uk") {
                        $logo_img = "3_logo.svg" //with semicolon errors - unexpected ';'
                    }



                    ?>

any help appreciated!!

UPDATE:

var_dump ($_SERVER['HTTP_HOST']); returns

Localhost

string 'localhost' (length=9)

Live site

string(23) "www.domain.co.uk"
blackhill24
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1 Answers1

0

Please fix this here is syntax error

else($host == "3.co.uk") {
                        $logo_img = "3_logo.svg" //with semicolon errors - unexpected ';'
                    }

you need to write

elseif($host == "3.co.uk") {
                        $logo_img = "3_logo.svg";
                    }
else{ ..... }
Davit Huroyan
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