ajax调用返回php接口返回json数据的方法(必看篇)

开发技术作者： 2024-08-17 12:00:01

下面小编就为大家带来一篇ajax调用返回php接口返回json数据的方法(必看篇)。小编觉得挺不错的，现在就分享给大家，也给大家做个参考。一起跟随小编过来看看吧

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：

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.