CODEWITHSUNDEEP
phpvariablesif-statement
Visualizer7
Visualizer7

Reputation: 325

If-statement in a Variable

I want to print the same page's name differently based on a certain variable.

Here is a corresponding code.

$metaTitle ="'if($variable=='input'){ title#1 }else {  title#2 };'";

And the produced meta title is lately used in the same file to create the page title (<title></title>)

But it keeps producing the title like

if($variable=='input'){ title#1 }else {  title#2 };

(the whole if statement as a whole. It does not recognize the if statement. It considers the statement as a plain text.)

What did I do wrong in the sentence??

Upvotes: 2

Views: 37482

Answers (4)

horsley
horsley

Reputation: 480

Because you just assign $metaTitle a STRING "'if($variable=='input'){ title#1 }else { title#2 };'" and it's not a runable statement

you should do like this

if ($variable=='input') {
    $metaTitle = "title#1";
} else {
    $metaTitle = "title#2";
}

or simply use Ternary Operator

Upvotes: 3

mbh
mbh

Reputation: 3312

Try this instead - 

if($variable=='input')
{ 
   $metaTitle = 'title#1'; 
}
else 
{ 
  $metaTitle = 'title#2'; 
}

Upvotes: -1

kazinix
kazinix

Reputation: 30103

Use ternary operator "?:":

$metaTitle = ($variable=='input')? "title#1" : "title#2";

The first part is the condition:

($variable=='input')

The second is the result when condition is true:

"title#1"

The third is the result when condition is false:

"title#2"

Source http://www.php.net/manual/en/language.operators.comparison.php#language.operators.comparison.ternary

Upvotes: 16

davoclavo
davoclavo

Reputation: 1442

The simplest and most basic solution is to set the title variable inside the if statement.

if($variable=='input'){
  $metaTitle = 'title#1';
} else {
  $metaTitle = 'title#2';
}

Upvotes: 0

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