ajax调用返回php接口返回json数据 ajax jsonp ajax json实例 ajax获取后台json数

后端开发|php教程

ajax,json

后端开发-php教程

php代码如下：

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<?phpheader(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[ ickname] = $row[ ickname]; $user[portrait] = $row[portrait]; echo $_GET[callback]."(".json_encode($user).")"; }?>

js代码如下:

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$.ajax({ url: "http://test.localhost/UserInterfaceForChatroom/UserInformation.php?email=pshuyue@", type: "GET", dataType: jsonp, // crossDomain: true, success: function (result) {//data = $.parseJSON(result);//alert(data.nickname);alert(result.nickname); } });

其中遇到了两个问题:

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1.第一个问题:

Uncaught

SyntaxError: Unexpected token :

解决方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and addingcallback=?to

use JSONP (as I needed to go cross-domain), and returning the JSON code{"foo":"bar"}and

getting the error.

This is because I should have included the callback data, something likejQuery17209314005577471107_1335958194322({"foo":"bar"})

Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:

$ret[foo]="bar";finish();function finish(){ header("content-type:application/json");if($_GET[callback]){print $_GET[callback]."(";}print json_encode($GLOBALS[ et]);if($_GET[callback]){print")";}exit;}

Hopefully that will help someone in the future.

2.第二个问题:

解析json数据。从上面的javascript中可以看到，我没有使用jquery.parseJSON()这些方法，开始使用这些方法，但是总是会报

VM219:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1

的错误，后来不用jquery.parseJSON()这个方法，反而一切正常。不知为何。

以上就介绍了ajax调用返回php接口返回json数据，包括了ajax,json方面的内容，希望对PHP教学有兴趣的朋友有所帮助。