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 = $.parseJSON(result);
// alert(data.nickname);
alert(result.nickname);
}
});
</script>其中遇到了两个问题:
1.第一个问题:
Uncaught SyntaxError: Unexpected token :
解决方案如下:
This has just happened to me, and the reason was none of the reasons above. I was using the jQuery command getJSON and adding callback=? to
use JSONP (as I needed to go cross-domain), and returning the JSON code {"foo":"bar"} and
getting the error.
This is because I should have included the callback data, something like jQuery17209314005577471107_1335958194322({"foo":"bar"})
Here is the PHP code I used to achieve this, which degrades if JSON (without a callback) is used:
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{"answer":"从建筑PDF提取结构化数据。将规格、BOM、进度表及报告转为Excel/CSV/JSON。扫描件用OCR，原生PDF用pdfplumber。"}
<span>$ret</span><span>[</span><span>'foo'</span><span>]</span><span>=</span><span>"bar"</span><span>;</span><span>
finish</span><span>();</span><span>function</span><span> finish</span><span>()</span><span>{</span><span>
header</span><span>(</span><span>"content-type:application/json"</span><span>);</span><span>if</span><span>(</span><span>$_GET</span><span>[</span><span>'callback'</span><span>])</span><span>{</span><span>print</span><span> $_GET</span><span>[</span><span>'callback'</span><span>].</span><span>"("</span><span>;</span><span>}</span><span>print</span><span> json_encode</span><span>(</span><span>$GLOBALS</span><span>[</span><span>'ret'</span><span>]);</span><span>if</span><span>(</span><span>$_GET</span><span>[</span><span>'callback'</span><span>])</span><span>{</span><span>print</span><span>")"</span><span>;</span><span>}</span><span>exit</span><span>;</span><span>}</span>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。
