test-json.php读取数据库,并以JSON格式准备。
<?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 | [email protected] |59.51.24.37
welcome2 | welcome2 | welcome2 | [email protected] |59.51.24.38
php -f /var/www/html/test-json.php
[ {title:"welcome1",
content:"welcome1",
author:"welcome1",
email:"[email protected]",
ip:"59.51.24.37"},
{title:"welcome2",
content:"welcome2",
author:"welcome2",
email:"[email protected]",
ip:"59.51.24.38"}]
test-json.php
以JSON格式获取一些数据。
现在回调数据并在表格中显示它。
<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>
输入127.0.0.1/test-json.html
,为什么test-json.php
在网页上没有数据创建?
我得到的是如下:
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 [email protected] 59.51.24.37
welcome2 welcome2 welcome2 [email protected] 59.51.24.38
问题是PHP脚本的响应是无效的JSON。
在JSON中,必须引用对象键。
而不是尝试滚动自己的JSON响应,而是使用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()
,这是我在使用MySQLi时的偏好。
这将产生类似的东西
[
{
"title": "welcome1",
"content": "welcome1",
"author": "welcome1",
"email": "[email protected]",
"ip": "59.51.24.37"
},
{
"title": "welcome2",
"content": "welcome2",
"author": "welcome2",
"email": "[email protected]",
"ip": "59.51.24.38"
}
]
你的PHP
代码中有很多错误。
以下是服务器端应该处理的事情(PHP
):
文件名:test-json.php
$data
)。JSON
格式并回显结果。以下是客户端(JavaScript
)应该如何处理的事情:
AJAX
文件发出test-json.php
请求。JSON
并填充一个变量(我将其命名为'html'),该变量将保存将附加到表中的所有HTML
代码(以及接收的数据)。DOM
请求只访问AJAX
一次。尽管如此,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关于
json_encode
功能。
希望我进一步推动你。