jsonp跨域请求jsoncallback调用最简单的方法php返回的数据
jsonp跨域请求jsoncallback调用最简单的方法php返回的数据
前端json代码
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title></title>
</head>
<body>
<script type="text/javascript" src="js/jquery.min.js"></script>
<script type="text/javascript">
$(function(){
$.ajax({
type: 'get',
url: 'http://fzd.xxxxxxxxx.com/minimage/test.php?jsoncallback=?',
data:{"domain":"www.xxxxx.com"},
dataType: 'jsonp',
jsonp: "jsoncallback",
success: function(data) {
alert("用户name："+ data.name+" 密码："+ data.pass2);
$('#seolink').html(data.name)
}
});
})
</script>
<marquee scrollamount="2"><span id="seolink"></span></marquee>
</body>
</html>
php服务器端：test.php
-----------------------------------------------------------------------------------------
<?php
header('Content-Type:text/html;Charset=utf-8');
$domainstr = $_REQUEST['domain'];
$arr = array(
"name" => "<a href='232'>".$domainstr."</a>",
"pass" => "<a href='232'>222</a>",
"name2" => "<a href='232'>xxxx</a>",
"pass2" => "<a href='232'>yyyy</a>"
);
echo $_GET['jsoncallback']."(".json_encode($arr).")";
?>