Question

我需要将类似l1 = [1,1,12] l2 = [1,1,1] for elem in l1: if not elem in l2: print(elem) print(l1.index(elem))（'MMMM Do YYYY'）的字符串日期格式转换为有效日期October 18th 2019或类似的2019-10-17T23:00:00.000Z

我尝试使用将字符串解析为一刻，但我不断收到错误消息

更新：我使用了17/10/2019。收到无效日期作为错误消息，很抱歉，我需要澄清，我正在尝试将字符串moment('October 18th 2019').format()转换为有效的日期格式，

Answer 1

在构造MMMM Do YYYY对象时，只需使用以下方法之一提供输入（Moment）的格式字符串：

// this way interprets the input at the start of the day in the local time zone
moment('October 18th 2019', 'MMMM Do YYYY')

// this way interprets the input at the start of the day in UTC
moment.utc('October 18th 2019', 'MMMM Do YYYY')

// this way interprets the input at the start of the day in a specific named time zone
// (requires the moment-timezone add-on)
moment.tz('October 18th 2019', 'MMMM Do YYYY', 'Europe/London')

然后，您可以根据需要对其进行格式化和/或转换。例如：

// this way keeps the local time, includes the local time offset when formatting
moment('October 18th 2019', 'MMMM Do YYYY').format()

// this way converts from local to utc before formatting
moment('October 18th 2019', 'MMMM Do YYYY').utc().format()

// this way converts from local to utc before formatting and includes milliseconds
moment('October 18th 2019', 'MMMM Do YYYY').toISOString()

如何将字符串日期格式`October 18th 2019`转换为有效日期`2019-10-17T23：00：00.000Z`

1 个答案: