Question

我试图通过AJAX将数据发送到php文件。我可以使用GET方法使其正常工作，但它不能使用POST方法（我的最终文件可能变大，所以我想使用POST）。

的Javascript

function update_table（）{ var mydrop = document.getElementById（＆＃39; dropdown＆＃39;）。value;

Request1 = new XMLHttpRequest();
if (Request1) {
    var RequestObj1 = document.getElementById('Target1');
    Request1.onreadystatechange = function () {
      if (Request1.readyState == 4 && Request1.status == 200) {
       // document.getElementById('Target1').innerHTML = "test"; 
        RequestObj1.innerHTML = Request1.responseText;
      }
    }

    Request1.open("POST", "table.php"); 
    Request1.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
//  var url = "table.php?brand="+mydrop;
//  Request1.open('GET', url);  // this works fine
    Request1.send("brand" + mydrop);
} // end Request1 function

}

我的PHP

<?php
// header('Content-Type: text/xml')
ini_set('display_errors',1); 
 error_reporting(E_ALL);

 if(isset($_POST['brand'])) {
    $brandAccess = $_POST['brand'];

 }

 if(isset($_GET['brand'])) {
    $brandAccess = $_GET['brand'];

 }
print_r($_POST);
print_r($_GET);


include('./includes/connection.inc.php');
$conn = dbConnect();
$sql = "SELECT * from finished_goods WHERE BrandDesc LIKE '{$brandAccess}%' "; 

$result = $conn->query($sql);

?>

<table>
    <?php foreach ($result as $row) { ?>
    <tr>
        <td><?php echo $row['ProductNo'] ?></td>
        <td><?php echo $row['ProductName'] ?></td>
        <td><?php echo $row['BrandDesc'] ?></td>
        <td><?php echo $row['QtyOnHand'] ?></td>
    </tr>
    <?php } ?>
</table>

我把print_r函数放在那里只是为了仔细检查，POST是空的，不管我尝试的是什么。

注意我想也许我需要标题（＆＃39;内容类型：text / xml＆＃39;）但是当我把它放进去时，页面根本不起作用。

谢谢，

Answer 1

尝试

Request1.setRequestHeader("Content-type","application/x-www-form-urlencoded");
Request1.send("brand=" + mydrop);

打印$ _POST后设置模具。在chrome dev-tools中查看您对php进行的查询。

Answer 2

you must forget the famous header ajax json for jquery or another
(only necessary for php but essenssial for a good response serveur)
<?PHP
$data = /** whatever you're serializing **/;
header('Content-Type: application/json');
echo json_encode($data);

**https://stackoverflow.com/questions/4064444/returning-json-from-a-php-script**

( if errors, you must write this first, before one echo html.. (and no echo html is good for great json response)

php没有收到ajax发布数据

2 个答案: