英文:
Parsing time in Golang
问题
我正在尝试根据名为dateAdded的输入创建一个Time结构体。我的代码如下：
dateAdded := "November 25, 2016"
layout := "September 9, 2016"
t, err := time.Parse(layout, dateAdded)
if err != nil {
fmt.Println(err)
} else {
fmt.Println(t)
}
我得到了以下错误：将时间“November 25, 2016”解析为“September 9, 2016”时出错：无法将“November 25, 2016”解析为“September 9,”。
我猜测Parse函数无法解析每种日期格式，但我想知道通常读取日期并将其解析为时间对象的方法是什么。
英文:
I'm trying to create a Time struct based on some input called dateAdded. My code is like this:
dateAdded := "November 25, 2016"
layout := "September 9, 2016"
t, err := time.Parse(layout, dateAdded)
if err != nil {
fmt.Println(err)
} else {
fmt.Println(t)
}
And I get the following error: parsing time "November 25, 2016" as "September 9, 2016": cannot parse "November 25, 2016" as "September 9, "
I assume the Parse function cannot parse every layout, but I'm curios what's the usual way of reading dates and parse them into time objects.
答案1
得分: 5
如果你没有使用时间模块预定义的常量布局之一，那么自定义的布局必须由精确的时间戳Mon Jan 2 15:04:05 -0700 MST 2006组成。请注意，每个元素都是唯一的，因此每个数字标识符都可以自动解析。基本上，它是1（月份），2（日期），3（小时），4（分钟），5（秒），6（年份），7（时区）等。
最好使用库中预定义的标准布局之一：
const (
ANSIC = "Mon Jan _2 15:04:05 2006"
UnixDate = "Mon Jan _2 15:04:05 MST 2006"
RubyDate = "Mon Jan 02 15:04:05 -0700 2006"
RFC822 = "02 Jan 06 15:04 MST"
RFC822Z = "02 Jan 06 15:04 -0700" // 带有数字时区的RFC822
RFC850 = "Monday, 02-Jan-06 15:04:05 MST"
RFC1123 = "Mon, 02 Jan 2006 15:04:05 MST"
RFC1123Z = "Mon, 02 Jan 2006 15:04:05 -0700" // 带有数字时区的RFC1123
RFC3339 = "2006-01-02T15:04:05Z07:00"
RFC3339Nano = "2006-01-02T15:04:05.999999999Z07:00"
Kitchen = "3:04PM"
// 方便的时间戳。
Stamp = "Jan _2 15:04:05"
StampMilli = "Jan _2 15:04:05.000"
StampMicro = "Jan _2 15:04:05.000000"
StampNano = "Jan _2 15:04:05.000000000"
)
英文:
The layout, if you're not using one of the pre-included constant layout that comes with the time module, must be formed from the exact timestamp Mon Jan 2 15:04:05 -0700 MST 2006. Notice that each element of it is unique, so each numeric identifier can be automatically parsed. it's basically 1 (month), 2 (day), 3 (hour), 4 (minute), 5 (second), 6 (year), 7 (time zone) etc.
It's better to use one of the pre-defined standard layouts that are included with the library:
const (
ANSIC = "Mon Jan _2 15:04:05 2006"
UnixDate = "Mon Jan _2 15:04:05 MST 2006"
RubyDate = "Mon Jan 02 15:04:05 -0700 2006"
RFC822 = "02 Jan 06 15:04 MST"
RFC822Z = "02 Jan 06 15:04 -0700" // RFC822 with numeric zone
RFC850 = "Monday, 02-Jan-06 15:04:05 MST"
RFC1123 = "Mon, 02 Jan 2006 15:04:05 MST"
RFC1123Z = "Mon, 02 Jan 2006 15:04:05 -0700" // RFC1123 with numeric zone
RFC3339 = "2006-01-02T15:04:05Z07:00"
RFC3339Nano = "2006-01-02T15:04:05.999999999Z07:00"
Kitchen = "3:04PM"
// Handy time stamps.
Stamp = "Jan _2 15:04:05"
StampMilli = "Jan _2 15:04:05.000"
StampMicro = "Jan _2 15:04:05.000000"
StampNano = "Jan _2 15:04:05.000000000"
)
答案2
得分: 1
你的日期格式是错误的。应该是"January 2, 2006"。正如规范所说：
该格式定义了参考时间的格式，参考时间被定义为
Mon Jan 2 15:04:05 -0700 MST 2006，如果它是值的话，将如何解释。
英文:
Your layout date is wrong. Should be "January 2, 2006". As the specs say:
> The layout defines the format by showing how the reference time,
> defined to be Mon Jan 2 15:04:05 -0700 MST 2006 would be interpreted
> if it were the value
答案3
得分: 1
你应该将其视为你提供给time.Provide的示例。它应该是文档中描述的具体值。
Parse函数解析一个格式化的字符串并返回它表示的时间值。布局定义了格式，通过展示参考时间的形式来定义，参考时间被定义为
Mon Jan 2 15:04:05 -0700 MST 2006
英文:
You should treat it as an example which you provide to time.Provide. And it should be of concrete value described in the documentation.
> Parse parses a formatted string and returns the time value it represents. The layout defines the format by showing how the reference time, defined to be
>
> Mon Jan 2 15:04:05 -0700 MST 2006
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