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

php代码如下：

<span><?php
</span>header(<span>'Content-Type: application/json'</span>);
    header(<span>'Content-Type: text/html;charset=utf-8'</span>);

    <span>$email </span><span>= </span><span>$_GET[</span><span>'email'</span><span>]</span>;

    <span>$user </span><span>= </span><span>[]</span>;

    <span>$conn </span><span>= @</span>mysql_connect(<span>"localhost"</span>,<span>"Test"</span>,<span>"123456"</span>) <span>or die</span>(<span>"Failed in connecting database"</span>);
    mysql_select_db(<span>"Test"</span>,<span>$conn</span>);
    mysql_query(<span>"set names 'UTF-8'"</span>);
    <span>$query </span><span>= </span><span>"select </span><span><em>*</em></span><span> from UserInformation where email = '"</span><span>.</span><span>$email</span><span>.</span><span>"'"</span>;
    <span>$result </span><span>= </span>mysql_query(<span>$query</span>);
    <span>if </span>(<span>null </span><span>== </span>(<span>$row </span><span>= </span>mysql_fetch_array(<span>$result</span>))) <span>{
</span><span>echo </span><span>$_GET[</span><span>'callback'</span><span>]</span><span>.</span><span>"(no such user)"</span>;
    <span>} </span><span>else </span><span>{
</span><span>$user[</span><span>'email'</span><span>] </span><span>= </span><span>$email</span>;
        <span>$user[</span><span>'nickname'</span><span>] </span><span>= </span><span>$row[</span><span>'nickname'</span><span>]</span>;
        <span>$user[</span><span>'portrait'</span><span>] </span><span>= </span><span>$row[</span><span>'portrait'</span><span>]</span>;
        <span>echo </span><span>$_GET[</span><span>'callback'</span><span>]</span><span>.</span><span>"("</span><span>.</span>json_encode(<span>$user</span>)<span>.</span><span>")"</span>;
    <span>}
</span><span>?></span>

js代码如下:

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

其中遇到了两个问题:

1.第一个问题:

Uncaught SyntaxError: Unexpected token :

解决方案如下:

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This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? 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 like jQuery17209314005577471107_1335958194322({"foo":"bar"})

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

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$ret['foo']="bar";
finish();function finish(){
    header("content-type:application/json");if($_GET['callback']){print $_GET['callback']."(";}print json_encode($GLOBALS['ret']);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教程有兴趣的朋友有所帮助。