php后台查询出数据,返回json数据,前台接收并输出
stu表:
index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<script src="http://www.jq22.com/jquery/jquery-3.3.1.js"></script>
</head>
<body>
<fieldset>表单
<form>
用户名:<input type='text' name='username' id='username' /><br/><br/>
性别:<input type="text" name="pwd" id='pwd' />
</form>
</fieldset>
<script type="text/javascript">
$('document').ready(function(){
var url="serve.php?inAjax=666&do=checknumber";
var data={};
$.get(url,data,function(res){
var jsobj=JSON.parse(res);
$('#username').val(jsobj.name);
$('#pwd').val(jsobj.sex);
})
});
</script>
</body>
</html>
serve.php
<?php
include 'config.php';
$inAjax=$_GET['inAjax'];
$do=$_GET['do'];
$do=$do?$do:"default";
if(!$inAjax)
{
return false;
}
switch ($do) {
case 'checknumber':
$sql="select * from stu where id=1";
$query=mysqli_query($conn,$sql);
$rs=mysqli_fetch_assoc($query);
$rs=json_encode($rs);
print_r($rs);
break;
case 'default':
die;
break;
}
?>
其中:
json_parse:将json字符串转化为对象