Using of variable in nested IF ELSE (PHP) [duplicate] - php

Question

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
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

Answer 1

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);

Answer 2

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