无法使用php从jQuery AJAX中的json_encode获取数据

Question

I have an AJAX call from jQuery to PHP where the PHP responds with a json_encode array, but the values of the array are not accessible in jQuery.
The status is OK, but the responseText is undefined .

$(document).ready(function () {
    $("#comments_form").on("submit", function(e) {
        e.preventDefault();
        e.stopPropagation();

        $.ajax({
            type: 'POST',
            url: 'process_in.php',
            data: {
                first: $("#firstname").val(), 
                second: $("#lastname").val(), 
                third: $("#mail").val(), 
                fourth: $("#phone").val(), 
                fifth: $("#message").val()
            },
            success: function(result) {                    
                var x = jQuery.parseJSON(result);
                alert(x.f);
            },
        });
    });  
})

<?php
    include ('connection.php');
    if (isset($_REQUEST['first']) && isset($_REQUEST['second']) &&   isset($_REQUEST['third']) && isset($_REQUEST['fourth']) && isset($_REQUEST['fifth'])) 
    {
        $firstname = $_REQUEST['first'];
        $lastname = $_REQUEST['second'];
        $email = $_REQUEST['third'];
        $contact = $_REQUEST['fourth'];
        $message = $_REQUEST['fifth'];

        $data = array();
        $data["f"] = xssafe($firstname);
        $data["l"] = xssafe($lastname);
        $data["e"] = xssafe($email);
        $data["c"] = xssafe($contact);
        $data["m"] = xssafe($message);
        echo json_encode($data);
    }

    function xssafe($d)
    {
        $x = filter_var($d, FILTER_SANITIZE_STRING);
        return $x;
    }  

Answer 1

A good practice is to always catch the errors too. In your ajax request there is no error callback to handle the exception. Use dataType: "JSON" instead of jQuery.parseJSON(); so that if json in unparsable you get the callback in the error block.

$.ajax({
    type: 'POST',
    url: 'process_in.php',
    dataType: 'JSON',
    data: {
        first: $("#firstname").val(), 
        second: $("#lastname").val(), 
        third: $("#mail").val(), 
        fourth: $("#phone").val(), 
        fifth: $("#message").val()
    },
    success: function(result) {                    
        console.log(result.f);
    },
    error: function (jqXHR, exception) {
        var msg = '';
        if (jqXHR.status === 0) {
            msg = 'Not connect.\n Verify Network.';
        } else if (jqXHR.status == 404) {
            msg = 'Requested page not found. [404]';
        } else if (jqXHR.status == 500) {
            msg = 'Internal Server Error [500].';
        } else if (exception === 'parsererror') {
            msg = 'Requested JSON parse failed.';
        } else if (exception === 'timeout') {
            msg = 'Time out error.';
        } else if (exception === 'abort') {
            msg = 'Ajax request aborted.';
        } else {
            msg = 'Uncaught Error.\n' + jqXHR.responseText;
        }
        console.log(msg);
    }
});

You can learn how to debug the code and check your error logs

Now lets get to your code, there are many possible cases that you are not getting the value.

It could be your php code or it could be your jquery.

In php to check whether its returning a valid json hit the url in browser like this

http://.../process_in.php?first=foo&second=foo&third=foo&fourth=foo&fifth=foo

As in your php code you haven't return any value so add an else part for the

    if (isset($_REQUEST['first']) && isset($_REQUEST['second']) &&   isset($_REQUEST['third']) && isset($_REQUEST['fourth']) && isset($_REQUEST['fifth'])) 
    {
        $firstname = $_REQUEST['first'];
        $lastname = $_REQUEST['second'];
        $email = $_REQUEST['third'];
        $contact = $_REQUEST['fourth'];
        $message = $_REQUEST['fifth'];

        $data = array();
        $data["f"] = xssafe($firstname);
        $data["l"] = xssafe($lastname);
        $data["e"] = xssafe($email);
        $data["c"] = xssafe($contact);
        $data["m"] = xssafe($message);
        echo json_encode($data);
    } else {
        echo json_encode(['error'=>'Invalid request']);
    }