PHP How right replace element in page?

Question

Good day.

We have test.php with next code:

<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST1; ?>" />
<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST2; ?>" />
<div class="container">
<p class="test">{Text}</p>
</div>
<!-- STYLE_TEST1 and STYLE_TEST2 - set in define as "/css/styletest1.css" and "/css/styletest2.css" -->

and index.php with code:

$content = file_get_contents("test.php");
$content = ereg_replace("{Text}", 'hello', $content);
echo $content;

in result we get code:

<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST1; ?>" />
<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST2; ?>" />
<div class="container">
<p class="test">{Text}</p>
</div>

we have problem with styles, becose echo $content; print <?php echo STYLE_TEST1; ?> and <?php echo STYLE_TEST2; ?> and not replace on values which set in define for this elements(not change on "/css/styletest1.css" and "/css/styletest2.css").

Tell me please how right make replace?

P.S.: we need that styles was include in file test.php, ie file test.php should have lines:

<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST1; ?>" />
<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST2; ?>" />

you are basically echoing php script... you can't do that... that's like this echo '<?php echo 'hi';?>'; — Arian Faurtosh
– Arian Faurtosh, Commented Mar 25, 2014 at 22:13
possible duplicate of PHP file_get_contents with php intact? — Anonymous
– Anonymous, Commented Mar 25, 2014 at 22:14
@Anonymous tell mep please how replace my code if use ob ? — user2881809
– user2881809, Commented Mar 25, 2014 at 22:23
You take the returned value and replace the values that you desire. — Anonymous
– Anonymous, Commented Mar 25, 2014 at 22:25

user2881809 · Accepted Answer · 2014-03-25 22:37:26Z

1

Thanks all for help!

This is the best decision for me on my question

function getTemplate($file) {

    ob_start(); // start output buffer

    include $file;
    $template = ob_get_contents(); // get contents of buffer
    ob_end_clean();
    return $template;

}

$content = getTemplate("test.php");
$content = ereg_replace("{Text}", 'hello', $content);
echo $content;

answered Mar 25, 2014 at 22:37

user2881809

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Arian Faurtosh · Accepted Answer · 2014-03-25 22:24:09Z

0

A way to fix this is to make it a variable....

just make sure you have

define("STYLE_TEST1", "/css/styletest1.css");
define("STYLE_TEST2", "/css/styletest2.css");

test.php

<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST1; ?>" />
<link rel="stylesheet" type="text/css" href="<?php echo STYLE_TEST2; ?>" />
<div class="container">
<p class="test">{<?=$variable?>}</p>
</div>
<!-- STYLE_TEST1 and STYLE_TEST2 - set in define as "/css/styletest1.css" and "/css/styletest2.css" -->

index.php

$variable = 'hello';

include 'test.php';

The result will be:

<link rel="stylesheet" type="text/css" href="/css/styletest1.css" />
<link rel="stylesheet" type="text/css" href="/css/styletest2.css" />
<div class="container">
<p class="test">{hello}</p>
</div>

edited Mar 25, 2014 at 22:24

answered Mar 25, 2014 at 22:16

Arian Faurtosh

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5 Comments

user2881809 Over a year ago

see p.s. in question please

Arian Faurtosh Over a year ago

the styles are included in test.php... can you elaborate what you mean?

Arian Faurtosh Over a year ago

@learner I changed it a bit... it should still work if you defined the variables correctly...

user2881809 Over a year ago

nice, why <?= (what it is)?

Arian Faurtosh Over a year ago

@learner it's called a short open tag and it basically replaces <?php echo $this; ?> to <?=$this?>

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