Question

我需要返回与日期字符串相同的日期字符串，但是将其转换为日期并返回后丢失了几个字符

var dateStr = "2019-08-02T11:46:46.5117312Z"

let formatter = DateFormatter()
formatter.calendar = Calendar(identifier: .iso8601)
formatter.locale = Locale(identifier: "en_US_POSIX")
formatter.timeZone = TimeZone(secondsFromGMT: 0)
formatter.dateFormat = "yyyy-MM-dd'T'HH:mm:ss.SSSZZZZZ"

let date = formatter.date(from: dateStr)

 var str = formatter.string(from: date!) // ===>>> "2019-08-02T11:46:46.511Z"

Answer 1

您可以使用自定义格式化程序来实现。但是请注意，Double精度不足以存储日期。结果是2019-08-02T11：46：46.511731 1 Z。代码：

class CustomDateFormatter {
    private let dateFormatter: DateFormatter = {
        let formatter = DateFormatter()
        formatter.calendar = Calendar(identifier: .iso8601)
        formatter.locale = Locale(identifier: "en_US_POSIX")
        formatter.timeZone = TimeZone(secondsFromGMT: 0)
        formatter.dateFormat = "yyyy-MM-dd'T'HH:mm:ss"
        return formatter
    }()

    func date(fromString str: String) -> Date? {
        let strs = str.components(separatedBy: CharacterSet(charactersIn: ".Z"))
        if strs.count != 3 {
            return nil
        }

        let dateStr = strs[0]
        let secondsStr = strs[1]
        guard let date = dateFormatter.date(from: dateStr),
            let seconds = Double("0."+secondsStr) else {
            return nil
        }

        let timeinteval = date.timeIntervalSince1970 + seconds
        return Date(timeIntervalSince1970: timeinteval)
    }

    func string(fromDate date: Date) -> String {
        let clippedDate = Date(timeIntervalSince1970: floor(date.timeIntervalSince1970))
        let seconds = date.timeIntervalSince1970.truncatingRemainder(dividingBy: 1)
        var lastPart = String(format: "%.7f", seconds)
        lastPart = (lastPart as NSString).substring(from: 1)
        return "\(dateFormatter.string(from: clippedDate))\(lastPart)Z"
    }
}

let dateStr = "2019-08-02T11:46:46.5117312Z"
let formatter = CustomDateFormatter()
let date = formatter.date(fromString: dateStr)!

print(formatter.string(fromDate: date)) // =====>> 2019-08-02T11:46:46.5117311Z

如何从日期中以“ yyyy-MM-dd'T'HH：mm：ss.SSSZZZZZZZ”格式获取日期字符串？

1 个答案: