Creating Object from variable name

Question

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

You cannot do string concatenation while object instantiation. — Shameer, Apr 01 '11 at 09:55
possible duplicate of [Instantiate new object from variable](http://stackoverflow.com/questions/5964113/instantiate-new-object-from-variable) — Tomas Gonzalez, Jul 09 '15 at 18:38

Gaurav · Accepted Answer · 2011-04-01T09:42:41.927

14

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();

edited Apr 01 '11 at 09:42

answered Apr 01 '11 at 09:31

Gaurav

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score 1 · Answer 2 · edited May 02 '12 at 03:23

1

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();

edited May 02 '12 at 03:23

htoip

427
5
19

answered Apr 01 '11 at 09:41

Jochen G.

21
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Creating Object from variable name

2 Answers2