我想检查设备的iOS版本是否大于3.1.3 我尝试了以下方法:
[[UIDevice currentDevice].systemVersion floatValue]
但是不管用，我只想要一个:
if (version > 3.1.3) { }
我怎样才能做到这一点呢?
当前回答
一个在obj - c++ 11中更通用的版本(你可能会用NSString/C函数替换其中的一些东西，但这不是那么冗长。这给了你两种机制。splitSystemVersion为你提供了一个包含所有部分的数组，如果你只是想打开主版本(例如switch([self splitSystemVersion][0]) {case 4: break;案例5:断裂;})。
#include <boost/lexical_cast.hpp>
- (std::vector<int>) splitSystemVersion {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
std::vector<int> versions;
auto i = version.begin();
while (i != version.end()) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
versions.push_back(boost::lexical_cast<int>(versionPart));
}
return versions;
}
/** Losslessly parse system version into a number
* @return <0>: the version as a number,
* @return <1>: how many numeric parts went into the composed number. e.g.
* X.Y.Z = 3. You need this to know how to compare again <0>
*/
- (std::tuple<int, int>) parseSystemVersion {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
int versionAsNumber = 0;
int nParts = 0;
auto i = version.begin();
while (i != version.end()) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
int part = (boost::lexical_cast<int>(versionPart));
versionAsNumber = versionAsNumber * 100 + part;
nParts ++;
}
return {versionAsNumber, nParts};
}
/** Assume that the system version will not go beyond X.Y.Z.W format.
* @return The version string.
*/
- (int) parseSystemVersionAlt {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
int versionAsNumber = 0;
int nParts = 0;
auto i = version.begin();
while (i != version.end() && nParts < 4) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
int part = (boost::lexical_cast<int>(versionPart));
versionAsNumber = versionAsNumber * 100 + part;
nParts ++;
}
// don't forget to pad as systemVersion may have less parts (i.e. X.Y).
for (; nParts < 4; nParts++) {
versionAsNumber *= 100;
}
return versionAsNumber;
}
其他回答
一个在obj - c++ 11中更通用的版本(你可能会用NSString/C函数替换其中的一些东西，但这不是那么冗长。这给了你两种机制。splitSystemVersion为你提供了一个包含所有部分的数组，如果你只是想打开主版本(例如switch([self splitSystemVersion][0]) {case 4: break;案例5:断裂;})。
#include <boost/lexical_cast.hpp>
- (std::vector<int>) splitSystemVersion {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
std::vector<int> versions;
auto i = version.begin();
while (i != version.end()) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
versions.push_back(boost::lexical_cast<int>(versionPart));
}
return versions;
}
/** Losslessly parse system version into a number
* @return <0>: the version as a number,
* @return <1>: how many numeric parts went into the composed number. e.g.
* X.Y.Z = 3. You need this to know how to compare again <0>
*/
- (std::tuple<int, int>) parseSystemVersion {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
int versionAsNumber = 0;
int nParts = 0;
auto i = version.begin();
while (i != version.end()) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
int part = (boost::lexical_cast<int>(versionPart));
versionAsNumber = versionAsNumber * 100 + part;
nParts ++;
}
return {versionAsNumber, nParts};
}
/** Assume that the system version will not go beyond X.Y.Z.W format.
* @return The version string.
*/
- (int) parseSystemVersionAlt {
std::string version = [[[UIDevice currentDevice] systemVersion] UTF8String];
int versionAsNumber = 0;
int nParts = 0;
auto i = version.begin();
while (i != version.end() && nParts < 4) {
auto nextIllegalChar = std::find_if(i, version.end(), [] (char c) -> bool { return !isdigit(c); } );
std::string versionPart(i, nextIllegalChar);
i = std::find_if(nextIllegalChar, version.end(), isdigit);
int part = (boost::lexical_cast<int>(versionPart));
versionAsNumber = versionAsNumber * 100 + part;
nParts ++;
}
// don't forget to pad as systemVersion may have less parts (i.e. X.Y).
for (; nParts < 4; nParts++) {
versionAsNumber *= 100;
}
return versionAsNumber;
}
Try:
NSComparisonResult order = [[UIDevice currentDevice].systemVersion compare: @"3.1.3" options: NSNumericSearch];
if (order == NSOrderedSame || order == NSOrderedDescending) {
// OS version >= 3.1.3
} else {
// OS version < 3.1.3
}
在主界面轻按“设置”>常规>关于。 设备的软件版本应该出现在这个屏幕上。 检查版本号是否大于3.1.3。
在Swift中查看iOS版本的解决方案
switch (UIDevice.currentDevice().systemVersion.compare("8.0.0", options: NSStringCompareOptions.NumericSearch)) {
case .OrderedAscending:
println("iOS < 8.0")
case .OrderedSame, .OrderedDescending:
println("iOS >= 8.0")
}
这个解决方案的缺点:不管你用哪种方式检查OS版本号，这都是一种糟糕的做法。永远不要以这种方式硬编码依赖关系，总是检查特性、功能或类的存在。考虑这一切;苹果可能会发布一个类的向后兼容版本，如果他们这样做了，那么你建议的代码将永远不会使用它，因为你的逻辑是寻找操作系统版本号，而不是类的存在。
(信息来源)
在Swift中检查类存在的解决方案
if (objc_getClass("UIAlertController") == nil) {
// iOS 7
} else {
// iOS 8+
}
不要使用if (NSClassFromString("UIAlertController") == nil)，因为它在使用iOS 7.1和8.2的iOS模拟器上正常工作，但如果你在使用iOS 7.1的真实设备上测试，你会不幸地注意到你永远不会通过代码片段的else部分。
这是一个Swift版本的yasirmturk宏。希望它能帮助一些人
// MARK: System versionning
func SYSTEM_VERSION_EQUAL_TO(v: String) -> Bool {
return UIDevice.currentDevice().systemVersion.compare(v, options: NSStringCompareOptions.NumericSearch) == NSComparisonResult.OrderedSame
}
func SYSTEM_VERSION_GREATER_THAN(v: String) -> Bool {
return UIDevice.currentDevice().systemVersion.compare(v, options: NSStringCompareOptions.NumericSearch) == NSComparisonResult.OrderedDescending
}
func SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO(v: String) -> Bool {
return UIDevice.currentDevice().systemVersion.compare(v, options: NSStringCompareOptions.NumericSearch) != NSComparisonResult.OrderedAscending
}
func SYSTEM_VERSION_LESS_THAN(v: String) -> Bool {
return UIDevice.currentDevice().systemVersion.compare(v, options: NSStringCompareOptions.NumericSearch) == NSComparisonResult.OrderedAscending
}
func SYSTEM_VERSION_LESS_THAN_OR_EQUAL_TO(v: String) -> Bool {
return UIDevice.currentDevice().systemVersion.compare(v, options: NSStringCompareOptions.NumericSearch) != NSComparisonResult.OrderedDescending
}
let kIsIOS7: Bool = SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO("7")
let kIsIOS7_1: Bool = SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO("7.1")
let kIsIOS8: Bool = SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO("8")
let kIsIOS9: Bool = SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO("9")
推荐文章
- 如何删除默认的导航栏空间在SwiftUI导航视图
- 如何在iOS中使用Swift编程segue
- Swift -整数转换为小时/分钟/秒
- Swift:声明一个空字典
- 为什么ARC仍然需要@autoreleasepool ?
- 在成功提交我的应用程序后，“太多符号文件”
- 首先添加一个UIView，甚至是导航栏
- 我如何改变UIButton标题颜色?
- 在Swift中如何调用GCD主线程上的参数方法?
- NSLayoutConstraints是可动画的吗?
- iOS -构建失败，CocoaPods无法找到头文件
- Xcode 4挂在“附加到(应用程序名称)”
- CFNetwork SSLHandshake iOS 9失败
- 请求失败:不可接受的内容类型:文本/html使用AFNetworking 2.0
- 缺少推荐的图标文件-该包不包含iPhone / iPod Touch的应用程序图标，像素为“120x120”，png格式
