我假设您正在尝试通过访问您的网站来获取信息。至少据我所知,您无法通过网络脚本知道设备的确切型号。但是您可以确定用户使用的是iPhone、iPad 还是iPod。以下是两个选项:
Javascript:
var Apple = {};
Apple.UA = navigator.userAgent;
Apple.Device = false;
Apple.Types = ["iPhone", "iPod", "iPad"];
for (var d = 0; d < Apple.Types.length; d++) {
var t = Apple.Types[d];
Apple[t] = !!Apple.UA.match(new RegExp(t, "i"));
Apple.Device = Apple.Device || Apple[t];
}
// is this an Apple device?
alert(
"Apple device? " + Apple.Device +
"niPhone? " + Apple.iPhone +
"niPod? " + Apple.iPod +
"niPad? " + Apple.iPad
);
PHP 脚本:
// Apple detection array
$Apple = array();
$Apple['UA'] = $_SERVER['HTTP_USER_AGENT'];
$Apple['Device'] = false;
$Apple['Types'] = array('iPhone', 'iPod', 'iPad');
foreach ($Apple['Types'] as $d => $t) {
$Apple[$t] = (strpos($Apple['UA'], $t) !== false);
$Apple['Device'] |= $Apple[$t];
}
// is this an Apple device?
echo
"<p>Apple device? ", ($Apple['Device'] ? 'true' : 'false'),
"</p>n<p>iPhone? ", ($Apple['iPhone'] ? 'true' : 'false'),
"</p>n<p>iPod? ", ($Apple['iPod'] ? 'true' : 'false'),
"</p>n<p>iPad? ", ($Apple['iPad'] ? 'true' : 'false'),
'</p>';
欲了解更多信息,请阅读here。
由于您的标签和声明有点混乱,如果您使用 iOS sdk 并构建 iOS 应用程序,您可以简单地从一行获取模型:
[[UIDevice currentDevice] platformString]