我怎样才能得到字符串的第n个字符?我尝试了括号([])访问器,没有运气。
var string = "Hello, world!"
var firstChar = string[0] // Throws error
错误:'下标'是不可用的:不能下标String与Int,请参阅文档注释讨论
我怎样才能得到字符串的第n个字符?我尝试了括号([])访问器,没有运气。
var string = "Hello, world!"
var firstChar = string[0] // Throws error
错误:'下标'是不可用的:不能下标String与Int,请参阅文档注释讨论
当前回答
你可以使用SwiftString (https://github.com/amayne/SwiftString)来做这件事。
"Hello, world!"[0] // H
"Hello, world!"[0...4] // Hello
免责声明:我写了这个扩展
其他回答
在项目中包含此扩展
extension String{
func trim() -> String
{
return self.trimmingCharacters(in: NSCharacterSet.whitespaces)
}
var length: Int {
return self.count
}
subscript (i: Int) -> String {
return self[i ..< i + 1]
}
func substring(fromIndex: Int) -> String {
return self[min(fromIndex, length) ..< length]
}
func substring(toIndex: Int) -> String {
return self[0 ..< max(0, toIndex)]
}
subscript (r: Range<Int>) -> String {
let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
upper: min(length, max(0, r.upperBound))))
let start = index(startIndex, offsetBy: range.lowerBound)
let end = index(start, offsetBy: range.upperBound - range.lowerBound)
return String(self[start ..< end])
}
func substring(fromIndex: Int, toIndex:Int)->String{
let startIndex = self.index(self.startIndex, offsetBy: fromIndex)
let endIndex = self.index(startIndex, offsetBy: toIndex-fromIndex)
return String(self[startIndex...endIndex])
}
然后像这样使用函数
let str = "Sample-String"
let substring = str.substring(fromIndex: 0, toIndex: 0) //returns S
let sampleSubstr = str.substring(fromIndex: 0, toIndex: 5) //returns Sample
Swift 4.2或更高版本
使用String的indexes属性进行范围和部分范围下标
作为@LeoDabus nice answer的变体,我们可以为DefaultIndices添加一个额外的扩展,目的是允许我们在为后者实现自定义下标(通过Int专用范围和部分范围)时使用String的indexes属性。
extension DefaultIndices {
subscript(at: Int) -> Elements.Index { index(startIndex, offsetBy: at) }
}
// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
subscript(range: Range<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start..<indices[start...][range.count]]
}
subscript(range: ClosedRange<Int>) -> SubSequence {
let start = indices[range.lowerBound]
return self[start...indices[start...][range.count]]
}
subscript(range: PartialRangeFrom<Int>) -> SubSequence {
self[indices[range.lowerBound]...]
}
subscript(range: PartialRangeThrough<Int>) -> SubSequence {
self[...indices[range.upperBound]]
}
subscript(range: PartialRangeUpTo<Int>) -> SubSequence {
self[..<indices[range.upperBound]]
}
}
let str = "foo bar baz bax"
print(str[4..<6]) // "ba"
print(str[4...6]) // "bar"
print(str[4...]) // "bar baz bax"
print(str[...6]) // "foo bar"
print(str[..<6]) // "foo ba"
感谢@LeoDabus指出我在使用索引属性作为字符串下标的(其他)替代方案的方向!
斯威夫特4.2
这个答案是理想的,因为它在一个扩展中扩展了String及其所有子序列(Substring)
public extension StringProtocol {
public subscript (i: Int) -> Element {
return self[index(startIndex, offsetBy: i)]
}
public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start...end]
}
public subscript (bounds: CountableRange<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[start..<end]
}
public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex..<end]
}
public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
let end = index(startIndex, offsetBy: bounds.upperBound)
return self[startIndex...end]
}
public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
let start = index(startIndex, offsetBy: bounds.lowerBound)
return self[start..<endIndex]
}
}
使用
var str = "Hello, playground"
print(str[5...][...5][0])
// Prints ","
这是最好最简单的方法之一
let yourString = "thisString"
print(Array(yourString)[8])
将字符串中的每个字母放入数组 然后选择第9个
看第一个字母:
first(str) // retrieve first letter
更多: http://sketchytech.blogspot.com/2014/08/swift-pure-swift-method-for-returning.html