为什么我没有收到从JS到PHP的变量？

Question

I am trying to send a variable from JS to php through ajax but I'm not able to get in php file.

JS

var names = ['lee','carter'] ;
$.ajax({
    type: "POST",
    url: "http://localhost/test/ajax.php",
    data: {name:names},
}).done(function() {
    location.href = 'http://localhost/test/ajax.php' ;
});

PHP

print_r($_POST);

this is showing an empty array but when I do console.log(data) it shows an array in console.log

   var names = ['lee','carter'] ;
    $.ajax({
        type: "POST",
        url: "http://localhost/test/ajax.php",
        data: {name:names},
    }).done(function(data) {
        console.log(data) ;
    });

Edit: (by mega6382) I believe OP wants to open a page in browser with post params, which cannot be done by AJAX. All others who answered got mistaken by the AJAX code in the question and started providing AJAX solutions, without realizing what OP is trying to do. If you were to read OP's comments on Jeroen's answer.

Answer 1

The problem is with what you do when the ajax request finishes:

}).done(function() {
    location.href = 'http://localhost/test/ajax.php' ;
});

Here you are re-directing to http://localhost/test/ajax.php (requesting it a second time...), using a GET request so $_POST is indeed empty.

Answer 2

只需在您的php文件中执行以下操作即可接收json格式的字符串

echo json_encode(['success' => true]);

Answer 3

Instead of ajax try sending a dynamically generated form like:

var names = ['lee','carter'] ;
var newForm = $('<form>', {
    'action': "http://localhost/test/ajax.php",
    'target': '_top',
    'method': 'POST'
});
names.forEach(function (item, index)
{
    newForm.append($('<input>', {
        'name': 'name[]',
        'value': item,
        'type': 'hidden'
    }));
});
$(document.body).append(newForm);
newForm.submit();

This will send the values over POST via a form. It will do both redirect to the new page and send post vals.

Answer 4

Since you're using an AJAX request, in this case POST , you could use the _REQUEST method in php for obtaining the JS variable. In your case try:

 $names = $_REQUEST['name']

 echo $names;

/* Or try JSON_ENCODE if echo isn't working*/

$json = json_encode($names)

echo ($json);

Answer 5

  var names = ['lee','carter'] ;
    JSON.stringify(names);
    $.ajax({
        type: "POST",
        dataType: 'json',
        url: "http://localhost/test/ajax.php",
        data: {name:names}
    }).done(function() {
            console.log('ok');
    });

This is a successful ajax call. Now in order to "check by yourself" that is working you don't have to redirect to the other page because that way you refresh and lose the data. What you have to do is:

1. Open developer tab in your browser
2. Go to network tab
3. Locate your ajax call (it has the name of your php class that you do the call)
4. go to response tab and there you have your php output

This link is to help you understand how to debug ajax call