Php 如何使用Json解析来自服务器的数据？

这是我的php文件。我想将android中的数据解析为JSON对象。我可以这样做吗

<?php
include("../include/connection.php");
    $old_pass="1234";
    $new_pass="123";
    $re_pass="123";
    $contact_no="8285663445";
     $chg_pwd="select * from users where contact_no='$contact_no'";
    $data=mysqli_query($conn,$chg_pwd) or die(mysqli_error($conn));
    $chg_pwd1=mysqli_fetch_array($data);
     $data_pwd=$chg_pwd1['password'];
    if($data_pwd==$old_pass){
    if($new_pass==$re_pass){
        $update_pwd="update users set password='$new_pass' where contact_no='$contact_no'";
        $updata=mysqli_query($conn,$update_pwd) or die(mysqli_error($conn));
        $json = array("status" => 1, "msg" => "password reset successfully");
    }
    else{

        $json = array("status" => 0, "msg" => "Your new and Retype Password is not match!");
    }
    }
    else
    {
    $json = array("status" => 0, "msg" => "Request method not accepted");
    }
header('Content-type: application/json');
echo json_encode($json);
?>

这就是我在日志中看到的

0-09 15:33:23.578 23721-25138/com.pickpamphlet.easydeals E/mapvalue{old_pass=12345，contact_no=8947897487，re_pass=zxcvbnm，new_pass=zxcvbnm} 10-09 15:33:24.813 23721-23721/com.pickpamphlet.easydeals E/log:{状态：0，消息：请求方法未接受}'

---

Android代码中的JSON解析已正确完成

1234不是数据库中的密码。请检查您的情况是否正确

if($data_pwd==$old_pass){

}
else
{
    $json = array("status" => 0, "msg" => "Request method not accepted");
}

---

正确检查php条件，确保传递的值正确。

谢谢。问题出在我的php脚本中。

if($data_pwd==$old_pass){

}
else
{
    $json = array("status" => 0, "msg" => "Request method not accepted");
}