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ajax调用返回php接口返回json数据的方法(必看篇)
发布日期:2017-05-05 00:00:00
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php代码如下:
<?php header('Content-Type: application/json'); header('Content-Type: text/html;charset=utf-8'); $email = $_GET['email']; $user = []; $conn = @mysql_connect("localhost","Test","123456") or die("Failed in connecting database"); mysql_select_db("Test",$conn); mysql_query("set names 'UTF-8'"); $query = "select * from UserInformation where email = '".$email."'"; $result = mysql_query($query); if (null == ($row = mysql_fetch_array($result))) { echo $_GET['callback']."(no such user)"; } else { $user['email'] = $email; $user['nickname'] = $row['nickname']; $user['portrait'] = $row['portrait']; echo $_GET['callback']."(".json_encode($user).")"; } ?>