[英]php not receiving ajax post data
我正在嘗試通過AJAX將數據發送到php文件。 我可以使它與GET方法一起正常工作,但不能與POST方法一起工作(我的最終文件可能會變大,所以我想使用POST)。
使用Javascript
函數update_table(){var mydrop = document.getElementById('dropdown')。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 $row['QtyOnHand'] ?></td>
</tr>
<?php } ?>
</table>
我把print_r函數放在那里只是為了仔細檢查,而POST是空的不管我嘗試什么。
注意,我以為我可能需要header('Content-Type:text / xml'),但是當我把它放進去時,該頁面根本無法工作。
謝謝,
嘗試
Request1.setRequestHeader("Content-type","application/x-www-form-urlencoded");
Request1.send("brand=" + mydrop);
並在打印$ _POST之后設置模具。 在chrome dev-tools中查看您對php進行哪種查詢。
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)
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