为什么getJSON传过来的数据显示不出来？

Question

test-json.php读取数据库，准备JSON格式

<?php
$conn = new mysqli("localhost", "root", "xxxx", "guestbook"); 
$result=$conn->query("select * From lyb limit 2"); 
echo '[';
$i=0;
while($row=$result->fetch_assoc()){  ?>
 {title:"<?= $row['title'] ?>",
        content:"<?= $row['content'] ?>",
        author:"<?= $row['author'] ?>",
        email:"<?= $row['email'] ?>",
        ip:"<?= $row['ip'] ?>"}
<?php 
if(
$result->num_rows!=++$i) echo ',';   
}
echo ']'    
?>

对于我的数据库，select * From lib limit 2 获取记录。

title    | content   | author   | email            |ip
welcome1 | welcome1  | welcome1 | welcome1@tom.com |59.51.24.37
welcome2 | welcome2  | welcome2 | welcome2@tom.com |59.51.24.38

php -f /var/www/html/test-json.php

[ {title:"welcome1",
         content:"welcome1",
        author:"welcome1",
         email:"welcome1@tom.com",
        ip:"59.51.24.37"},
{title:"welcome2",
         content:"welcome2",
        author:"welcome2",
         email:"welcome2@tom.com",
        ip:"59.51.24.38"}]

test-json.php 获取一些 JSON 格式的数据。

现在回调数据并显示在table.

<script src="http://127.0.0.1/jquery-3.3.1.min.js"></script>
<h2 align="center">Ajax show data in table</h2>
<table>
    <tbody id="disp">
        <th>title</th>
        <th>content</th>
        <th>author</th>
        <th>email</th>
        <th>ip</th>
    </tbody>
</table>

<script> 
$(function(){
    $.getJSON("test-json.php", function(data) {
        $.each(data,function(i,item){
            var tr = "<tr><td>" + item.title + "</td>"    +
                        "<td>"  + item.content  + "</td>" +
                        "<td>"  + item.author  + "</td>"  +
                        "<td>"  + item.email  + "</td>"   +
                        "<td>"  + item.ip  + "</td></tr>"
            $("#disp").append(tr);
        });
    });
});
</script>

键入127.0.0.1/test-json.html，为什么网页上没有test-json.php创建的数据？

得到的结果如下：

Ajax show data in table
title   content author  email   ip

我的期望如下：

Ajax show data in table
title   content author  email   ip
welcome1  welcome1  welcome1  welcome1@tom.com  59.51.24.37
welcome2  welcome2  welcome2  welcome2@tom.com  59.51.24.38

Answer 1

问题是您的 PHP 脚本的响应无效 JSON。

在 JSON 中，必须引用对象键。

与其尝试自己做出 JSON 响应，不如使用 json_encode() 为您完成。例如

<?php
$conn = new mysqli("localhost", "root", "xxxx", "guestbook"); 
$stmt = $conn->prepare('SELECT title, content, author, email, ip FROM lyb limit 2');
$stmt->execute();
$stmt->bind_result($title, $content, $author, $email, $ip);
$result = [];
while ($stmt->fetch()) {
    $result[] = [
        'title'   => $title,
        'content' => $content,
        'author'  => $author,
        'email'   => $email,
        'ip'      => $ip
    ];
}
header('Content-type: application/json; charset=utf-8');
echo json_encode($result);
exit;

_{您不必使用 prepare() 和 bind_result()，这只是我在使用 MySQLi 时的偏好。}

这会产生类似

的东西

[
  {
    "title": "welcome1",
    "content": "welcome1",
    "author": "welcome1",
    "email": "welcome1@tom.com",
    "ip": "59.51.24.37"
  },
  {
    "title": "welcome2",
    "content": "welcome2",
    "author": "welcome2",
    "email": "welcome2@tom.com",
    "ip": "59.51.24.38"
  }
]

Answer 2

您的 PHP 代码中有很多错误。

服务器端的处理方式如下(PHP):

文件名：test-json.php

从数据库中获取记录。
用已经从数据库中获取的记录填充一个数组（我在下面的代码中将该数组命名为 $data）。
将该数组编码为 JSON 格式并回显结果。

客户端的处理方式如下(JavaScript):

向 test-json.php 文件发出 AJAX 请求。
如果该请求成功，则遍历返回的 JSON 并填充一个变量（我将其命名为 'html'），该变量将保存所有 HTML将附加到 table.
将该变量（我命名为 'html'）附加到 table，这样我们就可以提高性能，因为每个 [= 只访问一次 DOM 18=]请求.

综上所述，这是解决方案：

PHP 代码-文件名：test-json.php:

<?php
// use the column names in the 'SELECT' query to gain performance against the wildcard('*').
$conn = new MySQLi("localhost", "root", "xxxx", "guestbook"); 

$result = $conn->query("SELECT `title`, `content`, `author`, `email`, `ip` FROM `lyb` limit 2"); 

// $data variable will hold the returned records from the database.
$data = [];

// populate $data variable.
// the '[]' notation(empty brackets) means that the index of the array is automatically incremented on each iteration.
while($row = $result->fetch_assoc()) {
  $data[] = [
    'title'   => $row['title'],
    'content' => $row['content'],
    'author'  => $row['author'],
    'email'   => $row['email'],
    'ip'      => $row['ip']
  ];
}

// convert the $data variable to JSON and echo it to the browser.
header('Content-type: application/json; charset=utf-8');
echo json_encode($data);

JavaScript代码

$(function(){
    $.getJSON("test-json.php", function(data) {
        var html = '';
        $.each(data,function(key, value){
            html += "<tr><td>" + value.title + "</td>"    +
                        "<td>"  + value.content  + "</td>" +
                        "<td>"  + value.author  + "</td>"  +
                        "<td>"  + value.email  + "</td>"   +
                        "<td>"  + value.ip  + "</td></tr>";

        });
        $("#disp").append(html);
    });
});

Learn more about json_encode function.

希望我能把你推得更远。

为什么getJSON传过来的数据显示不出来？

Why can't show data passed by getJSON?

javascript

ajax

json

getjson