没有'.. <'候选者产生预期的上下文结果类型'NSRange Swift 3
[英]No '..<' candidates produce the expected contextual result type 'NSRange Swift 3
问题描述
I've migrated to Swift 3.0 and I'm now getting an error on this line: 
我已经迁移到Swift 3.0，现在在这条线上出现错误：
let lastFourDigits = (accountNumber as NSString).substringWithRange(accountNumber.endIndex.advancedBy(-4)..<accountNumber.endIndex)
No '..<' candidates produce the expected contextual result type 'NSRange' (aka '_NSRange') . No '..<' candidates produce the expected contextual result type 'NSRange' (aka '_NSRange') 。 What am I doing wrong here? 我在这里做错了什么？
1 个解决方案
解决方案1
2 2016-09-28 16:15:14
The syntax for index operations changed (see Get nth character of a string in Swift programming language and the full rationale: Swift Evolution 0065 ): 索引操作的语法已更改（请参阅Swift编程语言中的获取字符串的n个字符以及完整的基本原理： Swift Evolution 0065 ）：
let accountNumber = "XX0000000000000000001234"
let lastFourDigits = accountNumber[accountNumber.index(accountNumber.endIndex, offsetBy: -4)..<accountNumber.endIndex]
print("Last 4: \(lastFourDigits)")
No need to use NSString or NSRange in Swift. 在Swift中无需使用NSString或NSRange 。
A simpler syntax would also be: 一个更简单的语法是：
let lastFourDigits = accountNumber.substring(from: accountNumber.index(accountNumber.endIndex, offsetBy: -4))
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