CODEWITHSUNDEEP
phpstringclass
KoolKabin
KoolKabin

Reputation: 17643

Creating Object from variable name

How do i create my class object in single line from a variable:

$strClassName = 'CMSUsers';
$strModelName = $strClassName.'Model';
$strModelObj = new $strModelName();

The above code successfully creates my CMSUsersModel class object but when i try:

$strClassName = 'CMSUsers';
$strModelObj = new $strClassName.'Model'();

it pops error.... saying:

Parse error: syntax error, unexpected '(' in

Upvotes: 4

Views: 17086

Answers (3)

Steve4585
Steve4585

Reputation: 71

As of PHP 8.0.0, using new with arbitrary expressions is supported. This allows more complex instantiation if the expression produces a string. (Example #4 Creating an instance using an arbitrary expression)

However, the expression must be wrapped in parentheses. The new operator is of higher precedence than string concatenation . (see Operator Precedence), so $strModelObj = new $strClassName.'Model'(); means $strModelObj = (new $strClassName).('Model'());. The correct parenthesizing would be $strModelObj = new ($strClassName.'Model')();.

A minimal working example (MWE) for PHP 8:

<?php
class CMSUsersModel {}
$strClassName = 'CMSUsers';
$strModelObj = new ($strClassName.'Model')();
var_dump($strModelObj);

Upvotes: 0

Jochen G.
Jochen G.

Reputation: 21

I'm note sure because I'm not up to date with classes and PHP, but I think $strModelName is the class definition, thus I think you have to use one two lines or write something like this:

$strModelName = 'CMSUsers'.'Model'; $strModelObj = new $strModelName();

Upvotes: 1

Gaurav
Gaurav

Reputation: 28755

You can not use string concatenation while creating objects.

if you use

class aa{}

$str = 'a';
$a = new $str.'a';   // Fatal error : class a not found



class aa{}

$str = 'a';
$a = new $str.$str; // Fatal error : class a not found

So You should use

$strModelName = $strClassName.'Model';
$strModelObj = new $strModelName();

Upvotes: 14

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