如何发布ajax并获得响应？ [关闭]

I have a problem with a form, the post and the response. In my form I call a function (javascript) with the ajax post:

var vars = "test="+test;
$.ajax({
    type: "POST",
    url: "index.php",
    data: vars
}).done(function(data) {
    alert(data);
}).fail(function(data) {
    alert(data);
});

In index.php I receive all the data:

<?php
    $test = $_POST['test'];
    //do something
?>

After I have to give back a value to the previous php. How can I do?? Thanks

var vars = "test="+test;
$.ajax({
    type: "POST",
    url: "index.php",
    data: vars
    success: function(html){
         alert(html)
    }
});

in index.php, enter follow code and check

<?php
    $test = $_POST['test'];
    //do something
    echo $test
?>

The same way you send data back for any other HTTP request.

header("Content-Type: text/plain");  # Avoid introducing XSS vulnerabilities
echo $test;

If it's a simple value you need to return, you can just echo it, and it will come back as the response.

If you need to return a more complex structure, you store it in a PHP array, say $response, and use echo json_encode($response); to output it back to javascript.

This may help

Javascript

var vars = { test : "test" };
$.ajax({
    type: "POST",
    url: "index.php",
    dataType : 'json',
    data: vars,
    success : function(data) {
        console.log(data);
    },
    error : function(resp) {
        console.log(resp.responseText);
    }
});

PHP

<?php
    $test = $_POST['test'];
    echo json_encode($test);
?>

Try it and check console log

If you have to return data in JSON format then use

echo json_encode($test);

Else you can simply echo the variable which you need in the ajax response. i.e.

echo $test;