【发布时间】：2017-08-23 04:12:16
【问题描述】：
我被要求执行 ajax 发布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全部数据。最后一行的学生姓名和性别分别为“yyyyty”和“F”。我希望将整个数据和提取的数据（学生姓名和性别）分别放在控制台页面中。当我对声明的 php 脚本执行 ajax 发布请求时出现错误
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"}
我的代码在下面找到
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>
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);
}
?>
我的问题是如何解决错误以及如何使用 ajax 发布参数从最后一行获取学生姓名和性别，并从整个数据中单独显示这些数据。我想回显 2 次，这些数据显示了整个数据默认数据并显示提取的数据...可以吗
【问题讨论】：
-
您的 json 无效（从 js 方面看）因为在 php 方面您有两个回显：第一次是来自数据库查询的数组，另一个是在您的“else”子句中。
-
但我希望默认显示所有数据.. 然后当我执行 ajax 发布请求时，我希望显示提取的数据
-
无论你想要什么，你都会得到你发送的：无效的 json。找到一种正确编码你的东西的方法，以便前端可以获取数据并执行它的操作。