PHP, new variable class in namespace

Question

I'm experimenting with PHP 5.3's namespacing functionality and I just can't figure out how to instantiate a new class with namespace prefixing.

This currently works fine:

<?php
new $className($args);
?>

But how I can prepend my namespace in front of a variable classname? The following example doesn't work.

<?php
new My\Namespace\$className($args);
?>

This example yields: Parse error: syntax error, unexpected T_VARIABLE

sMyles · Accepted Answer · 2019-10-10 16:16:02Z

20

Try this:

$class = "My\Namespace\\$className";
new $class();

There must be two backslashes \\ before the variable $className to escape it

edited Oct 10, 2019 at 16:16

sMyles

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answered Oct 29, 2011 at 2:06

JRL

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6 Comments

Stevanicus Over a year ago

wont u need to escape the back slash? "My\\Namespace\\$className";

Big McLargeHuge Over a year ago

I don't get it... why is this necessary? I had to do the same thing. Is it something wrong with the PHP install?

Dandy Over a year ago

The reason for the double backslash is that the backslash is a functional character and needs to be escaped itself.

Stichoza Over a year ago

Someone please explain this. Why the hell it's not possible to instantiate a namespaced class without using strings :| weird.

Shardj Over a year ago

@Stichoza it is possible, but in the question asked OP is mixing a namespace with a string which makes no sense, why would php expect a random string dropped into the middle of the line. secure.php.net/manual/en/language.namespaces.rules.php

|

ᴍᴇʜᴏᴠ · Accepted Answer · 2019-01-29 17:55:19Z

1

Here's how I did it:

$classPath = sprintf('My\Namespace\%s', $className);
$class = new $classPath;

Note the single quotes instead of double.

answered Jan 29, 2019 at 17:55

ᴍᴇʜᴏᴠ

5,3494 gold badges48 silver badges61 bronze badges

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