CODEWITHSUNDEEP
php
shin
shin

Reputation: 32721

What do braces surrounding a member variable mean in PHP?

I am wondering what { } in the following. What { } is doing here? $this->{$key} = $value;

Thanks in advance.

In one file 

$config['field']['calendar'] = array('type'=>'boolean');
$config['field']['category'] = array('type'=>'boolean');
$config['field']['customers'] = array('type'=>'boolean');  
...
$this->preference_form->initalize($config);

And in Preference_form.php

function initalize($config = array())
{
    foreach($config as $key => $value)
    {
        $this->{$key} = $value;
    }
}

Upvotes: 0

Views: 89

Answers (2)

Matthew Blancarte
Matthew Blancarte

Reputation: 8301

It's escaping the variable expression so that the member can be set dynamically.

Take a look at the documentation here: http://php.net/manual/en/language.types.string.php#language.types.string.parsing.complex

Upvotes: 0

John Carter
John Carter

Reputation: 55271

They're optional in this case, but it's a way of making it clearer to the reader (and the parser) that you're referring to a variable.

http://www.php.net/manual/en/language.variables.variable.php

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

Another case where this syntax is useful is when expanding a function call in a string.

This doesn't work (or rather it'll evaluate $someObj as a string, and then append ->someFunc():

$myString = "$someObj->someFunc()";

But this does what you'd expect:

$myString = "{$someObj->someFunc()}";

Upvotes: 1

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