我试图将数据发布到一个只包含php代码的php站点,该代码应在ID #mR-RateableFramePicture在第一页上被单击时执行。这是由一个ajax请求完成的:
$('#mR-RateableFramePicture').dblclick(function() {
$.ajax({
type: "POST",
url: 'moduleRateable/scriptSavedStyle.php',
data: { rateableUserID: rateableUserID, rateablePictureID: rateablePictureID},
success: function() {
$('#DynamicContent').load('moduleRateable/scriptSavedStyle.php');
}
});
});
var rateableUserID = $('input[name="rateableUserID"]').val();
var rateablePictureID = $('input[name="rateablePictureID"]').val();下面是ajax发布到的url目标:
<?php
// Start the session (enable global $_SESSION variable).
session_start();
// Include database-link ($conn).
include '../../scriptMysqli.php';
// Make global variable to simple variable.
$userID = $_SESSION["ID"];
//Save the rateable style to one owns libary of saved styles.
$ratedUserID = $_POST['rateableUserID'];
$ratedPictureID = $_POST['rateablePictureID'];
$sql = $conn->query("UPDATE styles WHERE userID = '$ratedUserID;' AND
pictureID = '$ratedPictureID' SET savedByUser = '$userID'");
?>我收到以下错误消息:
注意:未定义的索引:第12行的C:\xampp\htdocs\mystyle\app\moduleRateable\scriptSavedStyle.php中的rateableUserID
注意:未定义的索引:第13行的C:\xampp\htdocs\mystyle\app\moduleRateable\scriptSavedStyle.php中的rateablePictureID
发布于 2017-08-25 18:47:19
您没有像下面这样在$.ajax({})调用中传递变量rateableUserID和rateablePictureID的值-
data: { rateableUserID: rateableUserID, rateablePictureID: rateablePictureID}。除非它们是全局定义的,否则您将在PHP end.Make中获得未定义的值。在调用之前,请确保您已将该值赋给了rateableUserID和rateablePictureID。但是,您仍然必须检查是否确实在post请求中传递了该变量,因为PHP找不到键名。
该函数应如下所示
$('#mR-RateableFramePicture').dblclick(function() {
var rateableUserID = $('input[name="rateableUserID"]').val();
var rateablePictureID = $('input[name="rateablePictureID"]').val();
$.ajax({
type: "POST",
url: 'moduleRateable/scriptSavedStyle.php',
data: { "rateableUserID": rateableUserID, "rateablePictureID": rateablePictureID},
success: function() {
$('#DynamicContent').load('moduleRateable/scriptSavedStyle.php');
}
});
});https://stackoverflow.com/questions/45879640
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