我试图使用AJAX技术从MySQL数据库动态检索数据到网页,但重复div内容.我想刷新div而不重复信息.
这是代码
api.php
<?php
require 'db.php';
$query = "SELECT * FROM variables";
$result = mysqli_query($con,$query);
$data = array();
while ( $row = mysqli_fetch_row($result) )
{
$data[] = $row;
}
echo json_encode( $data );
?>
client.php
<html>
<head>
<script language="javascript" type="text/javascript" src="jquery.js">
</script>
</head>
<body>
<h2> Client example </h2>
<h3>Output: </h3>
<div id="output"></div>
<script id="source" language="javascript" type="text/javascript">
function ajaxcall()
{
$.ajax({
url: 'api.php', data: "", dataType: 'json', timeout:2000,
success: function(rows)
{
for (var i in rows)
{
var row = rows[i]; …