Using of variable in nested IF ELSE (PHP)

Question

I have a problem in using of array as a variable in nested if..else.. conditions. Its used to create custom web-service APIs. Here is the structure of code:

if (condition) {
    // code to be executed in case of true

    if (condition) {
        // code to be executed in case of true

        $result['key1'] = $variable1;
        $result['key2'] = $variable2;
        $result['key3'] = $variable3;

    } else {
        // code to be executed in case of false
    }

} else {
    $result['error'] = 'Something went wrong!!!';
}
echo json_encode($result);    // line 121

On execution of code it displays the following error:

Notice: Undefined variable: result in C:\xampp.. on line 121

If both of your if conditions are false, you will get that notice. Simply initialize your `$result` array before your first `if` — Hanky Panky, Sep 15 '14 at 09:06
Define $result and $variable1 before checking thus outside if — Youness, Sep 15 '14 at 09:06

score 0 · Answer 1 · answered Sep 15 '14 at 09:06

Variables need to be declared before using it inside functions like json_encode()! Do it this way:

$result = array();
if (condition) {
    // code to be executed in case of true

    if (condition) {
        // code to be executed in case of true

        $result['key1'] = $variable1;
        $result['key2'] = $variable2;
        $result['key3'] = $variable3;

    } else {
        // code to be executed in case of false
    }

} else {
    $result['error'] = 'Something went wrong!!!';
}
echo json_encode($result);

score 0 · Answer 2 · answered Sep 15 '14 at 09:07

U need to declare your variable first if you are directly entering the value to an array.

$result = array();

if (condition) {
    // code to be executed in case of true

    if (condition) {
        // code to be executed in case of true

        $result['key1'] = $variable1;
        $result['key2'] = $variable2;
        $result['key3'] = $variable3;

    } else {
        // code to be executed in case of false
    }

} else {
    $result['error'] = 'Something went wrong!!!';
}
echo json_encode($result);    // line 121

use a upvote rather giving same – Rakesh Sharma Sep 15 '14 at 09:10 — Rakesh Sharma, Sep 15 '14 at 09:10

Using of variable in nested IF ELSE (PHP)

2 Answers2