How do I capture PHP output into a variable?

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I\'m generating a ton of XML that is to be passed to an API as a post variable when a user click on a form button. I also want to be able to show the user the XML before han

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4条回答

轻奢々

2020-11-27 04:42

You could try this:

<?php
$string = <<<XMLDoc
<?xml version='1.0'?>
<doc>
  <title>XML Document</title>
  <lotsofxml/>
  <fruits>
XMLDoc;

$fruits = array('apple', 'banana', 'orange');

foreach($fruits as $fruit) {
  $string .= "\n    <fruit>".$fruit."</fruit>";
}

$string .= "\n  </fruits>
</doc>";
?>
<html>
<!-- Show XML as HTML with entities; saves having to view source -->
<pre><?=str_replace("<", "&lt;", str_replace(">", "&gt;", $string))?></pre>
<textarea rows="8" cols="50"><?=$string?></textarea>
</html>

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南方客

2020-11-27 04:47
It sounds like you want PHP Output Buffering
```
ob_start(); 
// make your XML file

$out1 = ob_get_contents();
//$out1 now contains your XML
```
Note that output buffering stops the output from being sent, until you "flush" it. See the Documentation for more info.
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醉梦人生

2020-11-27 05:01

<?php ob_start(); ?>
<xml/>
<?php $xml = ob_get_clean(); ?>
<input value="<?php echo $xml ?>" />͏͏͏͏͏͏

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执念已碎

2020-11-27 05:07
Put this at your start:
```
ob_start();
```
And to get the buffer back:
```
$value = ob_get_contents();
ob_end_clean();
```
See http://us2.php.net/manual/en/ref.outcontrol.php and the individual functions for more information.
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