Posted in PHP onAugust 07, 2014
昨天写一个功能:点击手机验证的同时获取json端的数据。
javascript代码如下:
$(".check_mobile").click(function(){ var mobile = $('.mobile').val(); $.getJSON("http://www.test.com/user.php?mobile="+mobile+"&format=json&jsoncallback=?", function(data){ if (data.succ == 1) { var html = "<input type='hidden' name='cityid' value='"+data.data.cityid+"'><input type='hidden' name='communityid' value='"+data.data.communityid+"'>"; $(".r_m").append(html); } }); });
user.php代码如下:
<?php if($_GET){ $mobile = $_GET['mobile']; if ($mobile == 'XXXX') { $user = array( 'city' =>'石家庄', 'cityid' =>'1', 'community' =>'紫晶悦城', 'communityid'=>'1' ); $sucess = 1; $return = array( 'succ' =>$sucess, 'data' => $user ); }else { $sucess = 2; $return = array( 'succ' =>$sucess ); } echo json_encode($return); } ?>
相应如下:
问题出来了:
在火狐浏览器中: SyntaxError: missing ; before statement
解决方法如下:
header("Access-Control-Allow-Origin:http:www.test.com"); $b = json_encode($return); echo "{$_GET['jsoncallback']}({$b})"; exit;
最后完整代码:
<?php header("Access-Control-Allow-Origin:http:www.test.com"); if($_GET){ $mobile = $_GET['mobile']; if ($mobile == '18831167979') { $user = array( 'city' =>'石家庄', 'cityid' =>'1', 'community' =>'紫晶悦城', 'communityid'=>'1' ); $sucess = 1; $return = array( 'succ' =>$sucess, 'data' => $user ); }else { $sucess = 2; $return = array( 'succ' =>$sucess ); } $b = json_encode($return); echo "{$_GET['jsoncallback']}({$b})"; exit; } ?>
如果在 PHP 中少了 header("Access-Control-Allow-Origin:http:www.test.com"); 代码,则会出现
XMLHttpRequest cannot load ''. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin ' ' is therefore not allowed access.
如果少了 echo "{$_GET['jsoncallback']}({$b})"; 代码
在谷歌浏览器中:Uncaught SyntaxError: Unexpected token :
在火狐浏览器中:SyntaxError: missing ; before statement
getJSON跨域SyntaxError问题分析
- Author -
hebedich声明:登载此文出于传递更多信息之目的,并不意味着赞同其观点或证实其描述。
Reply on: @reply_date@
@reply_contents@