执行某些数据的ajax post请求时出错

时间：2017-08-23 05:50:15

标签： javascript php jquery ajax

我被要求对php脚本执行ajax post请求，以获取最后一行中的一些数据（学生姓名和学生性别，分别是yyyyty和F）。默认情况下，我想在执行ajax帖子时显示所有数据。然后，我想使用ajax post请求回显所选数据。我的代码在下面找到............ 当我执行ajax帖子时，我收到了错误信息

Status Code: 200

ErrorThrown: SyntaxError: Unexpected token { in JSON at position 1634

jqXHR.responseText:

[{"student_id":"1","student_name":"Ashfur","student_gender":"F","student_age":"19","student_religion":"Muslim","student_course_id":"1"},{"student_id":"2","student_name":"Irfan","student_gender":"M","student_age":"17","student_religion":"Islam","student_course_id":"4"},{"student_id":"3","student_name":"Alice","student_gender":"F","student_age":"21","student_religion":"Chinese","student_course_id":"2"},{"student_id":"4","student_name":"Mohit","student_gender":"M","student_age":"20","student_religion":"Christian","student_course_id":"6"},{"student_id":"5","student_name":"Susy","student_gender":"F","student_age":"27","student_religion":"Chirstian","student_course_id":"5"},{"student_id":"6","student_name":"Ida","student_gender":"F","student_age":"23","student_religion":"Islam","student_course_id":"3"},{"student_id":"7","student_name":"Abdul","student_gender":"M","student_age":"22","student_religion":"Islam","student_course_id":"1"},{"student_id":"8","student_name":"Ernest","student_gender":"M","student_age":"25","student_religion":"Chinese","student_course_id":"4"},{"student_id":"9","student_name":"Wei Ling","student_gender":"F","student_age":"23","student_religion":"Chinese","student_course_id":"2"},{"student_id":"10","student_name":"Ashtae","student_gender":"M","student_age":"23","student_religion":"Islam","student_course_id":"4"},{"student_id":"11","student_name":"Jasmine","student_gender":"F","student_age":"23","student_religion":"Chinese","student_course_id":"2"},{"student_id":"65656","student_name":"yyyyty","student_gender":"F","student_age":"65","student_religion":"anything","student_course_id":"009090"}]{"student_name":"yyyyty","student_gender":"F"}

我的html文件

<html>
<head>
<script type="text/javascript" src="/Cesium-1.34/ThirdParty/jquery-1.11.3.min.js"></script> 
</head>
<div id="resulte"</div>
<script type="text/javascript">
showData();
function showData()
{
    $.ajax({
        type: "post",
        url: "student.php",
        dataType: "json",
        data: {
            lastOnly: true,
        },      
        success: function(data){
            console.log(data);
        },
        error: function(jqXHR, textStatus, errorThrown) {
            alert('An error occurred... Look at the console (F12 or Ctrl+Shift+I, Console tab) for more information!');
            $('#resulte').html('<p>Status Code: '+jqXHR.status+'</p><p>ErrorThrown: ' + errorThrown + '</p><p>jqXHR.responseText:</p><div>'+jqXHR.responseText + '</div>');
            console.log('jqXHR:');
            console.log(jqXHR);
            console.log('textStatus:');
            console.log(textStatus);
            console.log('errorThrown:');
            console.log(errorThrown);
        },

    });
};

</script>
</body>
</html>

在php脚本中

<?php

$conn = mysqli_connect('localhost','root','netwitness') or die ("Could not connect database");
$db = mysqli_select_db($conn,'abdpractice') or die ('Could not select database');

$result = mysqli_query($conn,"select * from student");
$json_array = array();
while ($row = mysqli_fetch_assoc($result))
{
    $json_array[] = $row;
}
    echo json_encode($json_array);
if (!isset($_POST["lastOnly"])){
} else {
    $response = [];
    $response['student_name'] = $json_array[count($json_array)-1]['student_name'];
    $response['student_gender'] = $json_array[count($json_array)-1]['student_gender'];

    echo json_encode($response);    
}
?>

我想默认回显整个数据。当我执行ajax post请求时，我想再次回显所选数据...请帮助我。

谢谢。

3 个答案:

答案 0 :(得分：2)

错误告诉你会发生什么。

您的JSON无效：

[{"student_id":"1","student_name":"Ashfur","student_gender":"F","student_age":"19","student_religion":"Muslim","student_course_id":"1"},{"student_id":"2","student_name":"Irfan","student_gender":"M","student_age":"17","student_religion":"Islam","student_course_id":"4"},{"student_id":"3","student_name":"Alice","student_gender":"F","student_age":"21","student_religion":"Chinese","student_course_id":"2"},{"student_id":"4","student_name":"Mohit","student_gender":"M","student_age":"20","student_religion":"Christian","student_course_id":"6"},{"student_id":"5","student_name":"Susy","student_gender":"F","student_age":"27","student_religion":"Chirstian","student_course_id":"5"},{"student_id":"6","student_name":"Ida","student_gender":"F","student_age":"23","student_religion":"Islam","student_course_id":"3"},{"student_id":"7","student_name":"Abdul","student_gender":"M","student_age":"22","student_religion":"Islam","student_course_id":"1"},{"student_id":"8","student_name":"Ernest","student_gender":"M","student_age":"25","student_religion":"Chinese","student_course_id":"4"},{"student_id":"9","student_name":"Wei Ling","student_gender":"F","student_age":"23","student_religion":"Chinese","student_course_id":"2"},{"student_id":"10","student_name":"Ashtae","student_gender":"M","student_age":"23","student_religion":"Islam","student_course_id":"4"},{"student_id":"11","student_name":"Jasmine","student_gender":"F","student_age":"23","student_religion":"Chinese","student_course_id":"2"},{"student_id":"65656","student_name":"yyyyty","student_gender":"F","student_age":"65","student_religion":"anything","student_course_id":"009090"}]{"student_name":"yyyyty","student_gender":"F"}

JSON的最后一部分无效：{"student_name":"yyyyty","student_gender":"F"}

您可以使用https://jsonformatter.curiousconcept.com/之类的工具来查看您的JSON是否有效。

正如Lawrence Cherone所说，你可以使用json_encode。

$myFinalJSON = json_encode($myFirstJSON);

答案 1 :(得分：0)

如果您希望在单个json响应中返回结果，则必须将两个数组合并

while ($row = mysqli_fetch_assoc($result))
{
    $json_array[] = $row;
}
    //echo json_encode($json_array);
if (!isset($_POST["lastOnly"])){
} else {
    $response = [
     'student_name' = $json_array[count($json_array)-1]['student_name'],
     'student_gender' = $json_array[count($json_array)-1]['student_gender']
];
}
$json_array[] = $response;
echo json_encode($json_array);

答案 2 :(得分：0)

在窗口加载时发送AJAX请求，以便显示所有记录 当您触发按钮以获取最后一个数据时，您将从记录中获取最后一个数据。

注意：通常为了获取特定记录，我们发送一个搜索键，用于指定WHERE子句条件，并根据该搜索键执行我们的Sql / Mysql查询。

在这里，我没有发送任何搜索关键字，并且每次都不会获取所有记录（导致性能问题）。尝试避免这种情况，但只是为了澄清，在你的情况下，我只是使用一个按钮来触发if-else条件并且每次都获取所有记录只是为了根据条件用php演示ajax请求。

以下是代码

  

HTML文件

<form method='post'>
   <button id='last'>Fetch Last</button>
</form>
<div id="result"></div>
  

JS档案

$(window).load(function() {
    let val = 'all';
    showData(val);
});

$('#last').on('click', function() {
    let val = $(this).prop('id');
    showData(val);
});

function showData(data)
{
    $.ajax({
        type: "post",
        url: "student.php",
        dataType: "json",
        data: {
            lastOnly: data,
        },      
        success: function(data){
            console.log(data);
        },
        error: function(jqXHR, textStatus, errorThrown) {
            alert('An error occurred... Look at the console (F12 or Ctrl+Shift+I, Console tab) for more information!');
            $('#resulte').html('<p>Status Code: '+jqXHR.status+'</p><p>ErrorThrown: ' + errorThrown + '</p><p>jqXHR.responseText:</p><div>'+jqXHR.responseText + '</div>');
            console.log('jqXHR:');
            console.log(jqXHR);
            console.log('textStatus:');
            console.log(textStatus);
            console.log('errorThrown:');
            console.log(errorThrown);
        },

    });
};
  

PHP文件

$conn = mysqli_connect('localhost','root','netwitness') or die ("Could not connect database");
$db = mysqli_select_db($conn,'abdpractice') or die ('Could not select database');

$result = mysqli_query($conn,"select * from student");
$response = array();
while ($row = mysqli_fetch_assoc($result))
{
    $response[] = $row;
}
if (isset($_POST["lastOnly"]) && $_POST["lastOnly"] === 'last'){
    $size = count($json_array);
    $response['student_name'] = $json_array[$size-1]['student_name'];
    $response['student_gender'] = $json_array[$size-1]['student_gender'];
    echo json_encode($response); 
} else {
    echo json_encode($response);    
}

还有一点是注意，即 -

使用.done()和.fail()方法来处理ajax响应，而不是使用success和error。

希望你能从这个解释中得到一些帮助。 :)：）

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