Php: simple code validation failure

Question

I'm currently working on a simple php function for an entity in the framework Symfony2. That function keeps generating errors, although I cannot see what is wrong. Here's the code :

public function getForm(array $reqFields){
    var $formHelper = new FormHelper;
    var $fields = new array;

    if($reqFields == null){
        foreach ($this->getArray() as $k => $v) {
            array_push($fields, $formHelper->getTextField($v, $k));
        }
    }

    return $formHelper->getForm($fields);
}

I've imported FormHelper, and the functions in it exist, for that matter. When I execute it, I get the following error:

Parse error: syntax error, unexpected T_VAR

What's the problem?

Edit: Netbeans tells me that $reqFields should be initialized (but it's an argument :/), and that an identifier is expected on the return line.

I believe it was on the third one. But the problem is resolved now, thanks to Aron's answer — Gabriel Theron, Jun 15 '12 at 09:27

score 7 · Accepted Answer · edited May 23 '17 at 12:00

7

I notice two things at least:

Arrays are not objects. You cannot create an array instance like you do with new array.
The var keyword is deprecated as of PHP 5.3. (And you are using it wrong)

So your code:

var $formHelper = new FormHelper;
var $fields = new array;

Should become:

$formHelper = new FormHelper;
$fields = array();

edited May 23 '17 at 12:00

Community

1
1

answered Jun 15 '12 at 09:24

Aron Rotteveel

81,193
17
104
128

score 3 · Answer 2 · answered Jun 15 '12 at 09:21

3

Arrays arent objects!
So it is:

var $fields = array();

And you have to define class variables in the class header not in the function. So you should erase var in front too.

Like this:

class Foo
{
   public $fields = array(); // var equals public (you should use public, because var is deprecated)

   public function bar()
   {
      print_r($this->fields); // => Array ( )
   }
}

answered Jun 15 '12 at 09:21

dan-lee

14,365
5
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Thanks for the info (at least I'll know for sure what to use now), however, I still have the same error. – Gabriel Theron Jun 15 '12 at 09:22

score 1 · Answer 3 · edited Jun 15 '12 at 11:36

1

public function getForm(array $reqFields){

in php you don't declare the parameters type so you should transform it in

public function getForm($reqFields){

or

public function getForm($reqFields = array()){

if yout want the parameter to be optional also

$formHelper = new FormHelper();

also

$fields = array();

also

foreach ($this->getArray() as $k => $v) {

I assume that $this is the form...if not use $formHelper->getData();

edited Jun 15 '12 at 11:36

j0k

22,600
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answered Jun 15 '12 at 09:29

Anti

458
3
8

1

Typehinting an array, callable or class name is possible in function definitions, see [documentation](http://php.net/manual/en/language.oop5.typehinting.php). – jkrcma Jun 15 '12 at 09:55

Php: simple code validation failure

3 Answers3