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
Rob Napier已经用下标给出了一个很棒的答案。但我觉得有一个缺点，因为没有检查出约束条件。这可能会导致崩溃。所以我修改了扩展，在这里
extension String {
subscript (r: Range<Int>) -> String? { //Optional String as return value
get {
let stringCount = self.characters.count as Int
//Check for out of boundary condition
if (stringCount < r.endIndex) || (stringCount < r.startIndex){
return nil
}
let startIndex = self.startIndex.advancedBy(r.startIndex)
let endIndex = self.startIndex.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: startIndex, end: endIndex)]
}
}
}
下面的输出
var str2 = "Hello, World"
var str3 = str2[0...5]
//Hello,
var str4 = str2[0..<5]
//Hello
var str5 = str2[0..<15]
//nil
所以我建议总是检查if let
if let string = str[0...5]
{
//Manipulate your string safely
}
在操场上试试这个
var str:String = "Hello, playground"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,8))
它会给出"ello, p"
然而，有趣的是，如果你使最后一个索引大于playground中的字符串，它将显示你在str:o之后定义的任何字符串
Range()似乎是一个泛型函数，因此它需要知道它正在处理的类型。
你还必须给它实际的字符串，你感兴趣的游乐场，因为它似乎持有所有的刺在一个接一个的序列与他们的变量名之后。
So
var str:String = "Hello, playground"
var str2:String = "I'm the next string"
let range = Range(start:advance(str.startIndex,1), end: advance(str.startIndex,49))
我是下一条线
即使str2使用let定义也有效
:)
