Can't make PHP function work inside HTML inside PHP

Question

I wrote this code, it gets an image from a link that varies according to where you are:

<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>

I want to make that code run if a PHP condition proves true, but I cannot make it work. It seems that the function doesn't return a value instead it takes the link textually. I mean it goes to http://chusmix.com/Imagenes/grupos/.jpg literally. However the code works correctly by itself.

This is the PHP code:

<?php
    $search=get_search_query();
    $first=$search[0];

    if ($first=="#"){
          echo "<html>";
          echo "<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>";
    }
    ?>

codaddict · Accepted Answer · 2010-12-09 06:28:09Z

5

You are already inside the php tag. So there is no need for <?php and ?>.

Try:

echo "<img src='http://chusmix.com/Imagenes/grupos/".substr($search,1).".jpg'>";

answered Dec 9, 2010 at 6:28

codaddict

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r92 · Accepted Answer · 2010-12-09 06:33:03Z

0

Replace line

echo "<img src='http://chusmix.com/Imagenes/grupos/<?php echo substr(get_search_query(), 1); ?>.jpg'>";

with

echo "<img src='http://chusmix.com/Imagenes/grupos/" . substr(get_search_query(), 1) . ".jpg'>";

answered Dec 9, 2010 at 6:33

r92

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