php include is not working in echo

Question

I want to add include into echo with variable and css and image, but its not working. Please help.

<?php
    if (isset($_POST['sharebutton'])) {
    $share=$_POST['sharetext'];
        $share=$share;
        echo '<div id="content">'.'<img src="img/sharepropic.png"/>'.$share.include('likecomment.php').'</div>';
    }
    ?>

$share=$share? Really? You can't concatenate an include. You'll need to do that as multiple lines. — Jessica
– Jessica, Commented Jun 11, 2013 at 18:03
Because they don't know how to use the code formatting here? — Jessica
– Jessica, Commented Jun 11, 2013 at 18:04
Yes, I don't know because i just start to learn php, and that's the reason I am here. — Krishna Torque
– Krishna Torque, Commented Jun 11, 2013 at 18:10

Hobbes · Accepted Answer · 2013-06-11 18:06:52Z

3

If you want to assign output from an include to a variable you can do this:

ob_start();
include('likecomment.php');
$out = ob_get_contents();
ob_end_clean();

echo '<div id="content">'.'<img src="img/sharepropic.png"/>'.$share.$out.'</div>';

answered Jun 11, 2013 at 18:06

Hobbes

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2 Comments

Krishna Torque Over a year ago

thank. i got it. Just i need to take another variable for include then only i can add this, right?

tftd Over a year ago

@krishnaTORQUE welcome to stack overflow. When somebody answers your question be nice and upvote & accept his answer so others know not to post answers :)

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