以上所有答案都不适用于所有版本的iOS(包括iOS 13)上的所有主要浏览器。下面是一个适用于所有iOS版本的Safari、Chrome和Firefox的解决方案:

var isIOS = (function () {
    var iosQuirkPresent = function () {
        var audio = new Audio();

        audio.volume = 0.5;
        return audio.volume === 1;   // volume cannot be changed from "1" on iOS 12 and below
    };

    var isIOS = /iPad|iPhone|iPod/.test(navigator.userAgent);
    var isAppleDevice = navigator.userAgent.includes('Macintosh');
    var isTouchScreen = navigator.maxTouchPoints >= 1;   // true for iOS 13 (and hopefully beyond)

    return isIOS || (isAppleDevice && (isTouchScreen || iosQuirkPresent()));
})();

请注意，此代码片段的编写优先考虑的是可读性，而不是简洁性或性能。

解释:

If the user agent contains any of "iPod|iPhone|iPad" then clearly the device is iOS. Otherwise, continue... Any other user agent that does not contain "Macintosh" is not an Apple device and therefore cannot be iOS. Otherwise, it is an Apple device, so continue... If maxTouchPoints has a value of 1 or greater then the Apple device has a touch screen and therefore must be iOS since there are no Macs with touch screens (kudos to kikiwora for mentioning maxTouchPoints). Note that maxTouchPoints is undefined for iOS 12 and below, so we need a different solution for that scenario... iOS 12 and below has a quirk that does not exist in Mac OS. The quirk is that the volume property of an Audio element cannot be successfully set to any value other than 1. This is because Apple does not allow volume changes on the Audio element for iOS devices, but does for Mac OS. That quirk can be used as the final fallback method for distinguishing an iOS device from a Mac OS device.

如果你正在使用React，有一个很棒的库可以解决这类问题:React - agent。(使用ua-parser-js构建。)

https://github.com/medipass/react-ugent

可用的浏览器有:

Chrome, chromium, edge, firefox，即，lynx, safari, opera

可用的操作系统包括:

安卓，黑莓，铬操作系统，debian, ios, linux, MAC操作系统，ubuntu, unix, Windows

可选设备包括:

游戏机、电脑、手机、平板电脑、智能电视、可穿戴、嵌入式

易于使用:

<Ugent browser="safari" os="ios">
  <div>
    This text only shows on Safari on iOS.
  </div>
</Ugent>

如果你不使用React，你可以使用- ua-parser-js

https://github.com/faisalman/ua-parser-js

一个简化的、易于扩展的版本。

var iOS = ['iPad', 'iPhone', 'iPod'].indexOf(navigator.platform) >= 0;

更新:我最初的答案不包括在桌面模式下的iPad(在即将推出的iPadOS 13及更高版本中，默认更改为桌面模式)。 这对我的用例很好，如果不适合你，使用这个更新:

// iPhone and iPad including iPadOS 13+ regardless of desktop mode settings

iOSiPadOS = /^iP/.test(navigator.platform) ||
           /^Mac/.test(navigator.platform) && navigator.maxTouchPoints > 4;

这应该是安全的，只要 台式mac电脑根本不支持触摸事件 或不超过4个触摸点(当前iOS设备支持5个触摸点) 这是因为regexp ^首先检查平台字符串的起始位置，如果没有“iP”就停止(无论如何都比搜索长UA字符串直到结束快)。 它比导航仪安全。userAgent检查为导航器。平台不太可能是伪造的 检测iPhone / iPad模拟器

---

ORIGINAL ANSWER:
Wow, a lot of longish tricky code here. Keep it simple, please!

以我之见，这一个速度快，省钱，而且运行良好:

 iOS = /^iP/.test(navigator.platform);

  // or, if you prefer it verbose:
 iOS = /^(iPhone|iPad|iPod)/.test(navigator.platform);

我在几年前写过这个，但我相信它仍然有效:

if(navigator.vendor != null && navigator.vendor.match(/Apple Computer, Inc./) && navigator.userAgent.match(/iPhone/i) || (navigator.userAgent.match(/iPod/i))) 

    {

        alert("Ipod or Iphone");

    }

else if (navigator.vendor != null && navigator.vendor.match(/Apple Computer, Inc./) && navigator.userAgent.match(/iPad/i))  

    {

        alert("Ipad");

    }

else if (navigator.vendor != null && navigator.vendor.match(/Apple Computer, Inc./) && navigator.userAgent.indexOf('Safari') != -1)

    {

        alert("Safari");

    }

else if (navigator.vendor == null || navigator.vendor != null)

    {

        alert("Not Apple Based Browser");

    }

检测iOS(均<12和13+)

社区维基，因为编辑队列说它是满的，所有其他答案目前是过时或不完整的。

const iOS_1to12 = /iPad|iPhone|iPod/.test(navigator.platform);

const iOS13_iPad = (navigator.platform === 'MacIntel' && navigator.maxTouchPoints > 1));

const iOS1to12quirk = function() {
  var audio = new Audio(); // temporary Audio object
  audio.volume = 0.5; // has no effect on iOS <= 12
  return audio.volume === 1;
};

const isIOS = !window.MSStream && (iOS_1to12 || iOS13_iPad || iOS1to12quirk());