ajax呼叫返回php介面回傳json資料 ajax jsonp ajax json實例 ajax取得後台json數-php教程-PHP中文網

ajax呼叫返回php介面回傳json資料 ajax jsonp ajax json實例 ajax取得後台json數

WBOY

發布： 2016-07-29 08:49:21

原創

1202 人瀏覽過

php程式碼如下：

<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>

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js程式碼如下:

<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>

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其中遇到了兩個問題:

1.第一個問題:

Uncaught SyntaxError: Unexpected token :

解決方案如下:

This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding use JSONP (as I needed to go cross-domain), and returning the JSON code callback=? and getting the error.{"foo":"bar"}

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, whichgrades will help someone in the future.

2.第二個問題:

解析json資料。從上面的javascript可以看到，我沒有使用jquery.parseJSON()這些方法，開始使用這些方法，但是總是會報

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

的錯誤，後來不用jquery.parseJSON()這個方法，反而一切正常。不知為何。

以上就介紹了ajax呼叫回傳php介面回傳json數據，包含了ajax,json方面的內容，希望對PHP教學有興趣的朋友有幫助。