将 Objective-C 中的 NSRange 转换为 Swift

Question

如何将其转换为 Swift:

NSString *searchString = [NSString stringWithFormat:@"%@", @"Apple_HFS "];
NSRange range = [tempString rangeOfString:searchString];
NSUInteger *idx = range.location + range.length;

谢谢

Answer 1

这就是您要找的吗？

var str : NSString = "A string that include the searched string Apple_HFS inside"

let searchString : NSString = "Apple_HFS "
let range : NSRange = str.rangeOfString(searchString) // (42, 10)
let idx : Int = range.location + range.length // 52

演示：http://swiftstub.com/831969865/

Answer 2

如果你使用String，你可以直接引用endIndex:

let searchString: String = "Apple_HFS "
if let range: Range<String.Index> = tempString.rangeOfString(searchString) {
    let index = range.endIndex
    let stringAfter = tempString.substringFromIndex(index)
    // do something with `stringAfter`
} else {
    // not found
}

我包含了类型，这样您就可以看到发生了什么，但通常我会写：

let searchString = "Apple_HFS "
if let range = tempString.rangeOfString(searchString) {
    let stringAfter = tempString.substringFromIndex(range.endIndex)
    // do something with `stringAfter`
} else {
    // not found
}

Answer 3

不使用任何对象方法或类型的简单解决方案：

假设你有：

let nsrange = NSRange(location: 3, length: 5)
let string = "My string"

现在您需要将 NSRange 转换为 Range:

let range = Range(start: advance(string.startIndex, nsrange.location), end: advance(string.startIndex, nsrange.location + nsrange.length))

将 Objective-C 中的 NSRange 转换为 Swift

Convert NSRange in Objective-C to Swift

nsrange

swift