Detect iOS version less than 5 with JavaScript

Question

This is related to the "fix" for position:fixed in older versions of iOS. However, if iOS5 or greater is installed, the fix breaks the page.

I know how to detect iOS 5: navigator.userAgent.match(/OS 5_\d like Mac OS X/i) but that won't work for iOS6 when it eventually comes around, or even iOS 5.0.1, only a 2 digit version.

So this is what I have atm.

$(document).bind("scroll", function() {
    if((navigator.userAgent.match(/iPhone/i)) || (navigator.userAgent.match(/iPod/i)) || (navigator.userAgent.match(/iPad/i))) {
        if (navigator.userAgent.match(/OS 5_\d like Mac OS X/i)) {
    }
    else {
        changeFooterPosition();
    }
});

Answer 1

Just simply run this code in device/browser

if(window.navigator.userAgent.match( /iP(hone|od|ad)/)){

    var iphone_version= parseFloat(String(window.navigator.userAgent.match(/[0-9]_[0-9]/)).split('_')[0]+'.'+String(window.navigator.userAgent.match(/[0-9]_[0-9]/)).split('_')[1]);

    // iPhone CPU iPhone OS 8_4 like Mac OS X

    alert(iphone_version); // its alert 8.4

    if(iphone_version >= 8){
       alert('iPhone device, iOS 8 version or greater!');
    }
}

This iphone_version variable will give you correct OS version for any iPhone device.

Answer 2

Pumbaa80's Answer was almost 100%, he just left out one part. Some iOS release have a third digit on them.

Example

Mozilla/5.0 (iPhone; U; CPU iPhone OS 4_3_3 like Mac OS X; en_us) AppleWebKit/525.18.1 (KHTML, like Gecko)

The follow allows for that

if(/(iPhone|iPod|iPad)/i.test(navigator.userAgent)) { 
    if(/OS [2-4]_\d(_\d)? like Mac OS X/i.test(navigator.userAgent)) {  
        // iOS 2-4 so Do Something   
    } else if(/CPU like Mac OS X/i.test(navigator.userAgent)) {
        // iOS 1 so Do Something 
    } else {
        // iOS 5 or Newer so Do Nothing
    }
}

That extra bit (_\d)? allows for the possibility of a third digit in the Version number. Charlie S, That should answer your question too.

Note the else because the 1st check won't work on iOS 1. iOS 1 for the iPhone and iPod didn't include a version number in its UserAgent string.

iPhone v 1.0

Mozilla/5.0 (iPhone; U; CPU like Mac OS X; en) AppleWebKit/420+ (KHTML, like Gecko) Version/3.0 Mobile/1A543 Safari/419.3

iPod v1.1.3

Mozilla/5.0 (iPod; U; CPU like Mac OS X; en) AppleWebKit/420.1 (KHTML, like Gecko) Version/3.0 Mobile/4A93 Safari/419.3

All of this can be found at the following link on Apples website here.

Answer 3

I could not really find what I was looking for, so I took ideas from this page and other pages around the net and came up with this. Hopefully others will find it useful as well.

function iOS_version() { 
    if(navigator.userAgent.match(/ipad|iphone|ipod/i)){ //if the current device is an iDevice
    var ios_info ={};
    ios_info.User_Agent=navigator.userAgent;
    ios_info.As_Reported=(navigator.userAgent).match(/OS (\d)?\d_\d(_\d)?/i)[0];
    ios_info.Major_Release=(navigator.userAgent).match(/OS (\d)?\d_\d(_\d)?/i)[0].split('_')[0];
    ios_info.Full_Release=(navigator.userAgent).match(/OS (\d)?\d_\d(_\d)?/i)[0].replace(/_/g,".");
    ios_info.Major_Release_Numeric=+(navigator.userAgent).match(/OS (\d)?\d_\d(_\d)?/i)[0].split('_')[0].replace("OS ","");
    ios_info.Full_Release_Numeric=+(navigator.userAgent).match(/OS (\d)?\d_\d(_\d)?/i)[0].replace("_",".").replace("_","").replace("OS ","");   //converts versions like 4.3.3 to numeric value 4.33 for ease of numeric comparisons
    return(ios_info);
    }
}

It allows you to get the major release and full release number either as a string or as a number for iOS.

Example User Agent String:

    Mozilla/5.0 (iPhone; CPU iPhone OS 6_1_3 like Mac OS X) AppleWebKit/536.26 (KHTML, like Gecko) Mobile/10B329

Usage:

var iOS=iOS_version();
console.log(iOS.Full_Release); //returns values in form of "OS 6.1.3"
console.log(iOS.Full_Release_Numeric); //returns values in form of 6.13
//useful for comparisons like if(iOS.Full_Release_Numeric >6.2)

console.log(iOS.Major_Release); //returns values in form of "OS 6"
console.log(iOS.Major_Release_Numeric); //returns values in form of 6
//useful for comparisons like if(iOS.Major_Release_Numeric >7)

console.log(iOS.As_Reported); //returns values in form of "OS 6_1_3"
console.log(iOS.User_Agent); //returns the full user agent string

In the case of the original question on this page, you could use this code in the following manner:

var iOS=iOS_version();
$(document).bind("scroll", function() {
    if(iOS){if(iOS.Major_Release_Numeric <5) {} 
    else {changeFooterPosition();}
    }
});   

Answer 4

This snippet of code can be used to determine any version of iOS 2.0 and later.

function iOSversion() {
  if (/iP(hone|od|ad)/.test(navigator.platform)) {
    // supports iOS 2.0 and later: <http://bit.ly/TJjs1V>
    var v = (navigator.appVersion).match(/OS (\d+)_(\d+)_?(\d+)?/);
    return [parseInt(v[1], 10), parseInt(v[2], 10), parseInt(v[3] || 0, 10)];
  }
}

ver = iOSversion();

if (ver[0] >= 5) {
  alert('This is running iOS 5 or later.');
}

Answer 5

function getIOSVersion() {
    const ua = navigator.userAgent;
    if (/(iPhone|iPod|iPad)/i.test(ua)) {
        return ua.match(/OS [\d_]+/i)[0].substr(3).split('_').map(n => parseInt(n));
    }
    return [0];
}

This will return an array with the individual version numbers like [10,0,1] for v10.0.1, or it'll default to [0] otherwise. You can check the first digit (the major version) or all of them to test for the versions you need.

Answer 6

First: Don't use match when a test is enough.

Second: You should test the other way round. Find the UAs which are known to be broken.

if(/(iPhone|iPod|iPad)/i.test(navigator.userAgent)) {
    if (/OS [1-4]_\d like Mac OS X/i.test(navigator.userAgent)) {
        changeFooterPosition();
...