对于iOS 5.1和6,如何让iOS应用程序在iPhone 3GS上运行?(How to make an iOS app work on iPhone 3GS and up, for both iOS 5.1 and 6?)
这是一个感兴趣的问题,因此请尽可能提供您的答案信息。
我最近将Xcode更新为4.5并且无法访问iOS SDK 5.1。 使用几个SO帖子(使用以前的Xcode安装程序中的图像)中描述的方法,我再次添加了iOS SDK 5.1。
我希望通过我的应用程序支持的环境是iPhone 3GS,iPhone 4,iPhone 4S和iPhone 5.显然,iOS SDK 5.1和6.0。 这些要求的推荐构建设置是什么?
现在我在5.1上设置
Deployment Target
,在6.0上设置Base SDK
,在armv7和armv7s上设置架构。 它是否正确?现在,我相信iOS SDK 5.1不是为armv7s(iPhone 5架构)编译的。 如何确保为iOS 6.0编译的代码也适用于iPhone <5? 现在我收到链接器错误
ld: file is universal (4 slices) but does not contain a(n) armv7s slice
指向iOS 5.1 SDKld: file is universal (4 slices) but does not contain a(n) armv7s slice
,当不检查Build For Active Architectures Only
。 如何确保在发布到App Store时,应用程序仍可在iOS SDK 5.1上运行。 它似乎是与iOS 5.1结合编译armv7s - 这是行不通的。任何帮助是极大的赞赏。
This is a question out of interest, so please be as informative as possible with your answer.
I recently updated Xcode to 4.5 and lost access to iOS SDK 5.1. Using the method described in several SO posts (use an image from previous Xcode installer) I added the iOS SDK 5.1 again.
The environments I want to support with my apps are iPhone 3GS, iPhone 4, iPhone 4S and iPhone 5. Obviously with iOS SDK 5.1 and 6.0. What are the recommended build settings for these requirements?
Right now I set the
Deployment Target
on 5.1,Base SDK
on 6.0, architectures on armv7 and armv7s. Is this correct?Now, I believe iOS SDK 5.1 is not compiled for armv7s (iPhone 5 architecture). How do I make sure that the code compiled for iOS 6.0 also works on iPhones < 5? Right now I'm getting linker errors
ld: file is universal (4 slices) but does not contain a(n) armv7s slice
pointing to the iOS 5.1 SDK, when not checkingBuild For Active Architectures Only
. How do I make sure that, when releasing to the App Store, the app will still run on iOS SDK 5.1. It appears to be compiling for armv7s in combination with iOS 5.1 -- which won't work.Any help is greatly appreciated.
原文:https://stackoverflow.com/questions/12950777
最满意答案
建议使用int ...你的价值可能是:
Sec + Min * 60 + Hour * 3600
24:30:00,您将获得88200。
从DB加载值时,您可以通过简单的数学公式反转您的值:
Hour = int(value / 3600) Min = int(value % 3600 / 60) Sec = value % 3600 % 1800
Suggest to use int for that... your value could be:
Sec + Min * 60 + Hour * 3600
For the 24:30:00, you will get 88200.
When loading your value from DB, you could reverse your value by simple math equation:
Hour = int(value / 3600) Min = int(value % 3600 / 60) Sec = value % 3600 % 1800
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