Question

我在错误语句中获得了数据库的完整结果。所以有数据通过低谷，但我不知道为什么它会出错我做错了。现在工作几个小时，找不到解决方案。希望有人可以帮助我

这是我的select_clienten.php

<?php
//database configuration
$dbHost = 'localhost';
$dbUsername = 'root';
$dbPassword = '********';
$dbName = '********';

//connect with the database
$conn = new mysqli($dbHost,$dbUsername,$dbPassword,$dbName);

$sql = "SELECT * FROM clienten";
$result = mysqli_query($conn, $sql);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
    while($row = mysqli_fetch_assoc($result)) {
        echo "<tr id=".$row["id"]."><td>".$row["roepnaam"]."</td>";
        echo "<td>".$row["achternaam"]."</td>";
        echo "<td>".$row["email"]."</td>";
        echo "<td><span class='glyphicon glyphicon-edit bewerk'></span></td>";
        echo "<td><span class='glyphicon glyphicon-trash verwijder'></span></td></tr>";
    }

} else {
    echo "0 results";
}
echo json_encode($data);
?>

这是我的index.php。

<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>title</title>
<link rel="stylesheet" href="//code.jquery.com/ui/1.12.1/themes/base/jquery- ui.css">
<link rel="stylesheet" type="text/css" href="css/bootstrap.min.css">
<link rel="stylesheet" href="/css/custom.css">
<script src="https://code.jquery.com/jquery-1.12.4.js"></script>
<script src="https://code.jquery.com/ui/1.12.1/jquery-ui.js"></script>

</head>
<body>

<script type="text/javascript">
$(document).ready(function() {
    var data;
    $.ajax(
       {type: 'get',
        url: './select_clienten.php',
        data: "",
        dataType: 'json',
        success:function(data)//we got the response of ajax :) 
        {
            $("#test").html(data);
        },
        error:function(data)
        {
            //no respons show error.
            alert  ("error!");
            alert(JSON.stringify(data)) //we give an alert of the error when there is no respons from ajax
        ;}
    }); //end of ajax function.

});
</script>
<div id="test">
<!-- show data json here -->
</div>
</body>
</html>

Answer 1

我认为问题是你在ajax函数中设置了DataSet queue = DBMgr.GetDataSet("SELECT * FROM queue"); DataTable missedQueue = queue.Tables[0].Copy();但响应是这样的html：

dataType: 'json'

您应该将所有html保存在字符串

中

echo "<tr id=".$row["id"]."><td>".$row["roepnaam"]."</td>"

并发送json响应

$data = "";
$data .= "<tr id=".$row["id"]."><td>".$row["roepnaam"]."</td>";
$data .= "<td>".$row["achternaam"]."</td>";
....

或删除echo json_encode(['data'=>$data]);并发送html

dataType: 'json'

选择查询php与Ajax Json给出错误

1 个答案: