没有' ..<'候选人产生预期的上下文结果类型&#NS; NSRange Swift 3

时间:2016-09-28 16:11:10

标签: ios swift swift3

我已迁移到Swift 3.0,现在我在这一行上收到错误:

let lastFourDigits = (accountNumber as NSString).substringWithRange(accountNumber.endIndex.advancedBy(-4)..<accountNumber.endIndex)

No '..<' candidates produce the expected contextual result type 'NSRange' (aka '_NSRange')。我在这里做错了什么?

1 个答案:

答案 0 :(得分:2)

索引操作的语法已更改(请参阅Get nth character of a string in Swift programming language和完整的基本原理:Swift Evolution 0065):

let accountNumber = "XX0000000000000000001234"

let lastFourDigits = accountNumber[accountNumber.index(accountNumber.endIndex, offsetBy: -4)..<accountNumber.endIndex]

print("Last 4: \(lastFourDigits)")

无需在Swift中使用NSStringNSRange

更简单的语法也是:

let lastFourDigits = accountNumber.substring(from: accountNumber.index(accountNumber.endIndex, offsetBy: -4))