Question

Why not give the correct exit code?

My full code:

<?php

$numA_m = "2";
$res = '["110","2","1"]';
$numA_s = json_decode($res);
if ($numA_m == 1) {
    $A_num_s = array("1", "2", "3", "4","110");
    $A_nam_s = array("egh", "guide", "irl", "tic", "all");
}
if ($numA_m == 2) {
    $A_num_s = array("1", "2","110");
    $A_nam_s = array("sub", "forg","all");
}
$Rsp = str_replace($A_num_s, $A_nam_s, $numA_s);
$Rsp_In = str_replace($A_nam_s, $A_num_s, $Rsp);
echo '<pre>';
print_r($Rsp);

echo '<p>';

echo '<pre>';
print_r($Rsp_In);
?>

The output should be:

Array (
[0] => all
[1] => forg
[2] => sub
)

Array (
[0] => 110
[1] => 2
[2] => 1
)

But it is like this:

Array
(
[0] => subsub0
[1] => forg
[2] => sub
)
Array
(
[0] => 110
[1] => 2
[2] => 1
)

What do i do?

DEMO: http://phpfiddle.org/main/code/srj-k3d

Answer 1

This outputs the desired result

$numA_m = "2";
$res = '["110","2","1"]';
$numA_s = json_decode($res);
if ($numA_m == 1) {
    $A_num_s = array("1", "2", "3", "4", "110");
    $A_nam_s = array("egh", "guide", "irl", "tic", "all");
}
if ($numA_m == 2) {
    $A_num_s = array("110", "2", "1"); // changed order
    $A_nam_s = array("all", "forg", "sub");
}
$count = 1;
$Rsp = str_replace($A_num_s, $A_nam_s, $numA_s, $count);
$Rsp_In = str_replace($A_nam_s, $A_num_s, $Rsp, $count);
echo '<pre>';
print_r($Rsp);

echo '<p>';

echo '<pre>';
print_r($Rsp_In);

Explanation:

str_replace searches for the strings in the order you send them. So, if in your case you are searching for 1 and then 110, it will replace all occurrences of 1 before reaching the 110.

If you replace the 110 first, it will be replaced as expected.