Question

我正在从java脚本进行ajax调用，我试图从php获取json响应，如果我将dataType设置为JSON，如果没有成功块执行则会执行ajax错误块，如果我尝试不指定dataType console.log成功块中的响应我什么都没得到

JS

return (self.user + self.status + self.image_id)

PHP

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);
    }
});

请帮我弄清楚如何在PHP中使用json响应ajax调用

Answer 1

在您的JS文件中，删除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); } });
在您的php文件中，选中“include file url”和
关闭if语句阻止正确

JS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);
    }
?>

PHP文件：

let btnImage = UIImage(named: "checkmarkWhite")

categoryButtonOne.setImage(btnImage, forState: UIControlState.Selected)
categoryButtonOne.semanticContentAttribute = .ForceRightToLeft
categoryButtonOne.imageEdgeInsets = UIEdgeInsetsMake(0, (categoryButtonOne.frame.width / 2) - (btnImage?.size.width)! - 10, 0, 0)

Answer 2

从JS代码中删除contentType: "application/json"，并在PHP文件中正确关闭大括号，它将正常工作。

Answer 3

您可以使用json_encode（）函数将JSON格式的PHP数组转换为响应。发送AJAX请求时将dataType设置为“ JSON”。

来自php的JSON响应在AJAX调用上

3 个答案: