Get variable based on value of another variable

Question

Assume I have an object with 3 properties:

protected $validMainStatements;
protected $validPrimaryStatements;
protected $validSecondaryStatements;

And I got the following method:

public function selectType($stmt) {
    $stmtParts = MainServerInterface::parse($stmt);
    $type = $stmtParts[0] //returns either Main, Primary or Secondary
}

Depending on the value of type, I want to use the associated property. A simple implementation would be:

public function selectType($stmt) {
    $stmtParts = MainServerInterface::parse($stmt);
    $type = $stmtParts[0] //returns either Main, Primary or Secondary

    if($type === "Main") {
        $usedProp = $this->validMainStatements;
    } elseif($type === "Primary") {
        $usedProp = $this->validPrimaryStatements;
    } elseif($type === "Secondary") {
        $usedProp = $this->validSecondaryStatements;
    }
}

I think I don't have to mention that this is ugly and uncomfortable to use. Is there a way to implement this in an easier way? Something like (pseudocode):

$usedProp = $"valid".$type."Statements";

VolkerK · Accepted Answer · 2015-06-19 07:39:28Z

3

<?php
class Foo {
    protected $validMainStatements = 1;
    protected $validPrimaryStatements = 2;
    protected $validSecondaryStatements = 3;

    public function bar() {
        $type = 'Primary';

        return $this->{'valid'.$type.'Statements'};
    }
}

$foo = new Foo;
echo $foo->bar();

see Variable variables - Example #1 Variable property example

-- edit and btw: I'd rather do it this way:

<?php
class Foo {
    protected $validStatements = [
        'Main' => 1,
        'Primary' => 2,
        'Secondary' => 3
    ];

    public function bar() {
        $type = 'Primary';

        return $this->validStatements[$type];
    }
}

$foo = new Foo;
echo $foo->bar();

edited Jun 19, 2015 at 7:39

answered Jun 19, 2015 at 7:29

VolkerK

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1 Comment

Realitätsverlust Over a year ago

Alright, i like your 2nd solution, that's really elegant. Sold. Thank you.

Akshay Hegde · Accepted Answer · 2015-06-19 08:09:40Z

1

Demo

Try like below

$usedProp = $this->{"valid".$type."Statements"};

Test

<?php

   class test {
         public $validMainStatements = 'hello';
   }

  $instance = new test;

  $type = 'Main'; 
  echo $instance->{"valid".$type."Statements"};

?>

Result

hello

edited Jun 19, 2015 at 8:09

answered Jun 19, 2015 at 7:31

Akshay Hegde

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6 Comments

someOne Over a year ago

The valid and Statements (the ones inside the {} in the echo statement) need to be enclosed in quotes!

someOne Over a year ago

It will either generate errors or notices, and your welcome :)

Akshay Hegde Over a year ago

@someOne : thanks, edited now as you suggested, but didn't understand why I didn't get error in eval.in/, just look at demo

Eborbob Over a year ago

@someOne Absolutely, it's very dangerous relying on assumed strings - always enclose in " or '

someOne Over a year ago

Most probably because the eval.in has turned the warnings off and uses the most recent versions of PHP (which have the feature of making assumptions)!!

|

oshell · Accepted Answer · 2015-06-19 07:43:28Z

0

Just use variable variables.

$variableName = 'valid'.$type.'Statements';
$this->{$variableName};

answered Jun 19, 2015 at 7:43

oshell

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