我还没有弄清楚如何在Swift中获得字符串的子字符串:
var str = “Hello, playground”
func test(str: String) -> String {
return str.substringWithRange( /* What goes here? */ )
}
test (str)
我不能在Swift中创建一个范围。自动完成在游乐场不是超级有用的-这是它的建议:
return str.substringWithRange(aRange: Range<String.Index>)
我在Swift标准参考库中没有找到任何有用的东西。下面是另一个大胆的猜测:
return str.substringWithRange(Range(0, 1))
这:
let r:Range<String.Index> = Range<String.Index>(start: 0, end: 2)
return str.substringWithRange(r)
我看到了其他答案(在Swift字符串中查找字符索引)，似乎表明，由于字符串是NSString的桥类型，“旧”方法应该工作，但不清楚如何-例如，这也不工作(似乎不是有效的语法):
let x = str.substringWithRange(NSMakeRange(0, 3))
想法吗?
您可以使用这些扩展来改进substringWithRange
斯威夫特2.3
extension String
{
func substringWithRange(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(end))
return self.substringWithRange(range)
}
func substringWithRange(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let range = Range(start: self.startIndex.advancedBy(start), end: self.startIndex.advancedBy(start + location))
return self.substringWithRange(range)
}
}
斯威夫特3
extension String
{
func substring(start: Int, end: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if end < 0 || end > self.characters.count
{
print("end index \(end) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: end)
let range = startIndex..<endIndex
return self.substring(with: range)
}
func substring(start: Int, location: Int) -> String
{
if (start < 0 || start > self.characters.count)
{
print("start index \(start) out of bounds")
return ""
}
else if location < 0 || start + location > self.characters.count
{
print("end index \(start + location) out of bounds")
return ""
}
let startIndex = self.characters.index(self.startIndex, offsetBy: start)
let endIndex = self.characters.index(self.startIndex, offsetBy: start + location)
let range = startIndex..<endIndex
return self.substring(with: range)
}
}
用法:
let str = "Hello, playground"
let substring1 = str.substringWithRange(0, end: 5) //Hello
let substring2 = str.substringWithRange(7, location: 10) //playground
由于String是NSString的桥接类型，“旧的”方法应该可以工作，但不清楚如何工作-例如，这也不起作用(似乎不是有效的语法):
let x = str.substringWithRange(NSMakeRange(0,3))
对我来说，这是你问题中真正有趣的部分。String被桥接到NSString，所以大多数NSString方法直接作用于String。你可以自由地使用它们，不需要思考。例如，这正如你所期望的那样:
// delete all spaces from Swift String stateName
stateName = stateName.stringByReplacingOccurrencesOfString(" ", withString:"")
But, as so often happens, "I got my mojo workin' but it just don't work on you." You just happened to pick one of the rare cases where a parallel identically named Swift method exists, and in a case like that, the Swift method overshadows the Objective-C method. Thus, when you say str.substringWithRange, Swift thinks you mean the Swift method rather than the NSString method — and then you are hosed, because the Swift method expects a Range<String.Index>, and you don't know how to make one of those.
最简单的办法就是阻止霉霉这样做，明确地说:
let x = (str as NSString).substringWithRange(NSMakeRange(0, 3))
注意，这里没有涉及到重大的额外工作。“转换”并不意味着“转换”;String实际上是一个NSString。我们只是告诉Swift为了这一行代码的目的如何看待这个变量。
整个事情中真正奇怪的部分是导致所有这些麻烦的Swift方法是没有记录的。我不知道它的定义是什么;它不在NSString头文件中也不在Swift头文件中。
