ajax调用返回php接口返回json数据的方法(必看篇)

发布时间：2020-12-12 22:07:40 所属栏目：PHP教程来源：网络整理

导读：php代码如下： header('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"); mysq

php代码如下：

header('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['nickname'] = $row['nickname'];
$user['portrait'] = $row['portrait'];
echo $_GET['callback']."(".json_encode($user).")";
}

js代码如下：

$.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); } });

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:

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。

以上这篇ajax调用返回php接口返回json数据的方法(必看篇)就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持编程之家。

（编辑：李大同）

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