许多答案建议使用“itms”或“itms-apps”，但苹果并没有特别推荐这种做法。他们只提供以下方式打开App Store:

清单1从iOS应用程序启动App Store

NSString *iTunesLink = @"https://itunes.apple.com/us/app/apple-store/id375380948?mt=8";
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:iTunesLink]];

参见https://developer.apple.com/library/ios/qa/qa1629/_index.html, 2014年3月更新。

对于支持iOS 6及以上版本的应用程序，苹果提供了一种应用内机制来呈现应用程序商店:SKStoreProductViewController

- (void)loadProductWithParameters:(NSDictionary *)parameters completionBlock:(void (^)(BOOL result, NSError *error))block;

// Example:
SKStoreProductViewController* spvc = [[SKStoreProductViewController alloc] init];
spvc.delegate = self;
[spvc loadProductWithParameters:@{ SKStoreProductParameterITunesItemIdentifier : @(364709193) } completionBlock:^(BOOL result, NSError *error){ 
    if (error)
        // Show sorry
    else
        // Present spvc
}];

注意，在iOS6上，如果有错误，completion块可能不会被调用。这似乎是一个在iOS 7中解决的错误。

如果你有应用商店id，你最好使用它。特别是当您将来可能更改应用程序的名称时。

http://itunes.apple.com/app/id378458261

如果你没有应用商店id，你可以根据这个文档创建一个url https://developer.apple.com/library/ios/qa/qa1633/_index.html

+ (NSURL *)appStoreURL
{
    static NSURL *appStoreURL;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        appStoreURL = [self appStoreURLFromBundleName:[[NSBundle mainBundle] objectForInfoDictionaryKey:@"CFBundleName"]];
    });
    return appStoreURL;
}

+ (NSURL *)appStoreURLFromBundleName:(NSString *)bundleName
{
    NSURL *appStoreURL = [NSURL URLWithString:[NSString stringWithFormat:@"itms-apps://itunes.com/app/%@", [self sanitizeAppStoreResourceSpecifier:bundleName]]];
    return appStoreURL;
}

+ (NSString *)sanitizeAppStoreResourceSpecifier:(NSString *)resourceSpecifier
{
    /*
     https://developer.apple.com/library/ios/qa/qa1633/_index.html
     To create an App Store Short Link, apply the following rules to your company or app name:

     Remove all whitespace
     Convert all characters to lower-case
     Remove all copyright (©), trademark (™) and registered mark (®) symbols
     Replace ampersands ("&") with "and"
     Remove most punctuation (See Listing 2 for the set)
     Replace accented and other "decorated" characters (ü, å, etc.) with their elemental character (u, a, etc.)
     Leave all other characters as-is.
     */
    resourceSpecifier = [resourceSpecifier stringByReplacingOccurrencesOfString:@"&" withString:@"and"];
    resourceSpecifier = [[NSString alloc] initWithData:[resourceSpecifier dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES] encoding:NSASCIIStringEncoding];
    resourceSpecifier = [resourceSpecifier stringByReplacingOccurrencesOfString:@"[!¡\"#$%'()*+,-./:;<=>¿?@\\[\\]\\^_`{|}~\\s\\t\\n]" withString:@"" options:NSRegularExpressionSearch range:NSMakeRange(0, resourceSpecifier.length)];
    resourceSpecifier = [resourceSpecifier lowercaseString];
    return resourceSpecifier;
}

通过测试

- (void)testAppStoreURLFromBundleName
{
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Nuclear™"].absoluteString, @"itms-apps://itunes.com/app/nuclear", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Magazine+"].absoluteString, @"itms-apps://itunes.com/app/magazine", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Karl & CO"].absoluteString, @"itms-apps://itunes.com/app/karlandco", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"[Fluppy fuck]"].absoluteString, @"itms-apps://itunes.com/app/fluppyfuck", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Pollos Hérmanos"].absoluteString, @"itms-apps://itunes.com/app/polloshermanos", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Niños and niñas"].absoluteString, @"itms-apps://itunes.com/app/ninosandninas", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"Trond, MobizMag"].absoluteString, @"itms-apps://itunes.com/app/trondmobizmag", nil);
    STAssertEqualObjects([AGApplicationHelper appStoreURLFromBundleName:@"!__SPECIAL-PLIZES__!"].absoluteString, @"itms-apps://itunes.com/app/specialplizes", nil);
}

在SwiftUI - IOS 15.0

//
//  Test.swift
//  PharmaCodex
//

import SwiftUI

struct Test: View {
    
    var body: some View {
        
        VStack {
            
            Button(action: {
                guard let writeReviewURL = URL(string: "https://apps.apple.com/app/id1629135515?action=write-review") else {
                    fatalError("Expected a valid URL")
                }
                UIApplication.shared.open(writeReviewURL, options: [:], completionHandler: nil)
                
            }) {
                Text("Rate/Review")
            }
            
            
        }
    }
}

struct Test_Previews: PreviewProvider {
    static var previews: some View {
        Test()
    }
}

我认为苹果已经为应用程序链接提供了一个新的链接: https://apps.apple.com/app/id1445530088或https://apps.apple.com/app/ {your_app_id}

这里的大多数解决方案都过时了

为那些寻找更新的工作解决方案

// let appStoreLink = "https://apps.apple.com/app/{app-name}/{app-id}"

guard let url = URL(string: Constants.App.appStoreLink) else { return }
UIApplication.shared.open(url)

尽管这里有很多答案，但没有一个链接到开发者应用程序的建议似乎已经不管用了。

当我上次访问时，我能够使用以下格式让它工作:

itms-apps://itunes.apple.com/developer/developer-name/id123456789

这不再有效，但删除开发人员名称可以:

itms-apps://itunes.apple.com/developer/id123456789