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

我的解决方案是向您的实用程序类(提示提示)添加一个实用程序方法来解析系统版本并手动补偿浮点数排序。

此外,这段代码相当简单,所以我希望它能帮助一些新手。简单地传入一个目标浮点数,然后返回BOOL类型。

在你的共享类中这样声明它:

(+) (BOOL) iOSMeetsOrExceedsVersion:(float)targetVersion;

这样叫它:

BOOL shouldBranch = [SharedClass iOSMeetsOrExceedsVersion:5.0101];

(+) (BOOL) iOSMeetsOrExceedsVersion:(float)targetVersion {

/*
 Note: the incoming targetVersion should use 2 digits for each subVersion --

 example 5.01 for v5.1, 5.11 for v5.11 (aka subversions above 9), 5.0101 for v5.1.1, etc.
*/

// Logic: as a string, system version may have more than 2 segments (example: 5.1.1)
// so, a direct conversion to a float may return an invalid number
// instead, parse each part directly

NSArray *sysVersion = [[UIDevice currentDevice].systemVersion componentsSeparatedByString:@"."];
float floatVersion = [[sysVersion objectAtIndex:0] floatValue];
if (sysVersion.count > 1) {
    NSString* subVersion = [sysVersion objectAtIndex:1];
    if (subVersion.length == 1)
        floatVersion += ([[sysVersion objectAtIndex:1] floatValue] *0.01);
    else
        floatVersion += ([[sysVersion objectAtIndex:1] floatValue] *0.10);
}
if (sysVersion.count > 2) {
    NSString* subVersion = [sysVersion objectAtIndex:2];
    if (subVersion.length == 1)
        floatVersion += ([[sysVersion objectAtIndex:2] floatValue] *0.0001);
    else
        floatVersion += ([[sysVersion objectAtIndex:2] floatValue] *0.0010);
}

if (floatVersion  >= targetVersion) 
    return TRUE;

// else
return FALSE;
 }

其他回答

使用nv-ios-version项目(Apache License, Version 2.0)中包含的Version类,很容易获取和比较iOS版本。下面的示例代码转储iOS版本,并检查版本是否大于或等于6.0。

// Get the system version of iOS at runtime.
NSString *versionString = [[UIDevice currentDevice] systemVersion];

// Convert the version string to a Version instance.
Version *version = [Version versionWithString:versionString];

// Dump the major, minor and micro version numbers.
NSLog(@"version = [%d, %d, %d]",
    version.major, version.minor, version.micro);

// Check whether the version is greater than or equal to 6.0.
if ([version isGreaterThanOrEqualToMajor:6 minor:0])
{
    // The iOS version is greater than or equal to 6.0.
}

// Another way to check whether iOS version is
// greater than or equal to 6.0.
if (6 <= version.major)
{
    // The iOS version is greater than or equal to 6.0.
}

项目页面:nv-ios-version TakahikoKawasaki / nv-ios-version

博客:获取并比较iOS版本在运行时与版本类 获取并比较iOS版本在运行时与版本类

/*
 *  System Versioning Preprocessor Macros
 */ 

#define SYSTEM_VERSION_EQUAL_TO(v)                  ([[[UIDevice currentDevice] systemVersion] compare:v options:NSNumericSearch] == NSOrderedSame)
#define SYSTEM_VERSION_GREATER_THAN(v)              ([[[UIDevice currentDevice] systemVersion] compare:v options:NSNumericSearch] == NSOrderedDescending)
#define SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO(v)  ([[[UIDevice currentDevice] systemVersion] compare:v options:NSNumericSearch] != NSOrderedAscending)
#define SYSTEM_VERSION_LESS_THAN(v)                 ([[[UIDevice currentDevice] systemVersion] compare:v options:NSNumericSearch] == NSOrderedAscending)
#define SYSTEM_VERSION_LESS_THAN_OR_EQUAL_TO(v)     ([[[UIDevice currentDevice] systemVersion] compare:v options:NSNumericSearch] != NSOrderedDescending)

/*
 *  Usage
 */ 

if (SYSTEM_VERSION_LESS_THAN(@"4.0")) {
    ...
}

if (SYSTEM_VERSION_GREATER_THAN_OR_EQUAL_TO(@"3.1.1")) {
    ...
}

试试下面的代码:

NSString *versionString = [[UIDevice currentDevice] systemVersion];

快速的例子,实际工作:

switch UIDevice.currentDevice().systemVersion.compare("8.0.0", options: NSStringCompareOptions.NumericSearch) {
case .OrderedSame, .OrderedDescending:
    println("iOS >= 8.0")
case .OrderedAscending:
    println("iOS < 8.0")
}

不要使用NSProcessInfo,因为它在8.0下不能工作,所以它在2016年之前几乎没用

我知道这是一个老问题,但是应该有人在available .h中提到编译时宏。这里的所有其他方法都是运行时解决方案,不能在头文件、类类别或ivar定义中工作。

对于这些情况,使用

#if __IPHONE_OS_VERSION_MAX_ALLOWED >= __IPHONE_14_0 && defined(__IPHONE_14_0)
  // iOS 14+ code here
#else
  // Pre iOS 14 code here
#endif

不要这样回答