我正在从java脚本进行ajax调用,并且尝试从php获取json响应,如果我将dataType设置为JSON,则如果未执行成功 block ,则将执行ajax错误 block ,如果我尝试执行,则不指定dataType console.log 成功 block 中的响应我什么也没得到
JS
let CurrentDate = Date();
console.log(CurrentDate);
jsonObject = {
'TrackName' : 'Material Science',
'TrackDesc' : 'Test Text Test Text Test Text Test Text Test Text Test Text Test Text ',
'Timestamp' : CurrentDate
}
console.log(jsonObject);
$.ajax({
type:'post',
url:'../../../../PHP/adminScripts/addNewTrack.php',
contentType: "application/json",
data: {trackDetails:jsonObject},
dataType: "json",
success: function(response) {
console.log('SUCCESS BLOCK');
console.log(response);
},
error: function(response) {
console.log('ERROR BLOCK');
console.log(response);
}
});
PHP
<?php
header('Content-type: application/json');
include('../connection.php');
if($_POST) {
$obj = $_POST['trackDetails'];
$TrackName = mysql_real_escape_string($obj['TrackName']);
$TrackDesc = mysql_real_escape_string($obj['TrackDesc']);
$TrackAdderID = 'Admin '; //$_SESSION["userID"];
$Timestamp = mysql_real_escape_string($obj['Timestamp']);
$response_array['status'] = 'status123';
echo json_encode($response_array);
请帮我弄清楚如何在 PHP 中获取 ajax 调用的 json 响应
请您参考如下方法:
- 在 JS 文件中,删除
contentType: "application/json",
- 在您的 php 文件中,选中“包含文件 url”并
- 正确关闭 if 语句 block
JS 文件:
let CurrentDate = Date();
jsonObject = {
'TrackName' : 'Material Science',
'TrackDesc' : 'Test Text Test Text Test Text Test Text Test Text Test Text Test Text ',
'Timestamp' : CurrentDate
}
$.ajax({
type:'post',
url:'addNewTrack.php',
data: {trackDetails:jsonObject},
dataType: "json",
success: function(response) {
console.log('SUCCESS BLOCK');
console.log(response);
},
error: function(response) {
console.log('ERROR BLOCK');
console.log(response);
}
});
PHP 文件:
<?php
header('Content-type: application/json');
include('../connection.php');
if($_POST) {
$obj = $_POST['trackDetails'];
$TrackName = mysql_real_escape_string($obj['TrackName']);
$TrackDesc = mysql_real_escape_string($obj['TrackDesc']);
$TrackAdderID = 'Admin '; //$_SESSION["userID"];
$Timestamp = mysql_real_escape_string($obj['Timestamp']);
$response_array['status'] = 'status123';
echo json_encode($response_array);
}
?>