Question

我被要求执行ajax post请求以从最后一行获取student_name和student_gender ..我还想从数据库中获取整个数据。最后一行的学生姓名和性别是＆＃34; yyyyty＆＃34;和＆＃34; F＆＃34;分别。我希望将整个数据和提取的数据（student_name和性别）分别放在控制台页面中。当我向php脚本执行ajax post请求时出现错误，该脚本声明

   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"}

我的代码位于下面

在html文件中

<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>

php脚本中的

        <?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);    
}
?>

我的问题是如何解决错误以及如何使用ajax post参数从最后一行获取学生姓名和性别，并从整个数据中显示那些数据。我想回显2次显示整个默认情况下数据并显示提取的数据......是否可能

Answer 1

您正在从服务器发送无效的json响应。

echo json_encode($json_array[count($json_array)-1]['student_name']);
echo json_encode($json_array[count($json_array)-1]['student_gender']);
echo json_encode($json_array);

这将产生无效的json字符串。不是回应它们，而是首先将它们存储在一个数组中，然后回显整个数组。

$response = [];
$response['student_name'] = $json_array[count($json_array)-1]['student_name'];
$response['student_gender'] = $json_array[count($json_array)-1]['student_gender'];
$response['whole_result'] = $json_array;

echo json_encode($response);

希望得到这个帮助。

执行ajax post参数时出错

1 个答案: