3

我的要求是在 Mule 中使用 DataWeave 将字符串转换为日期格式

输入字符串

8/30/2015  4:00:13 PM

输出日期需要采用 MM/DD/YYYY 格式，即

08/30/2015

我尝试使用以下逻辑

(($[1] splitBy " ")[0]) as :date{format:"[m01]/[d01]/[y0001]"}

这给出了错误

Exception while executing: 
"Order Date":(($[1] splitBy " ")[0]) as :date{format:"[m01]/[d01]/[y0001]"},
 Cannot coerce a :string to a :date                                                                                                                                                          ^

我正在使用 Mule Anypoint Studio 版本：5.2.0

4

4 回答 4

10

首先，您需要将其转换为本地日期时间（无时区），然后转换为具有所需格式的字符串。

"8/30/2015 4:00:13 PM" 
    as :localdatetime {format: "M/dd/yyyy h:mm:ss a"}
    as :string {format: "MM/dd/yyyy"}

在 Mule 4 的 DataWeave 2.0 中，它应该是

"8/30/2015 4:00:13 PM" 
    as LocalDateTime {format: "M/dd/yyyy h:mm:ss a"}
    as String {format: "MM/dd/yyyy"}

有关其他信息，请参阅。 https://docs.oracle.com/javase/8/docs/api/java/time/format/DateTimeFormatter.html

All letters 'A' to 'Z' and 'a' to 'z' are reserved as pattern letters. The following pattern letters are defined:

  Symbol  Meaning                     Presentation      Examples
  ------  -------                     ------------      -------
   G       era                         text              AD; Anno Domini; A
   u       year                        year              2004; 04
   y       year-of-era                 year              2004; 04
   D       day-of-year                 number            189
   M/L     month-of-year               number/text       7; 07; Jul; July; J
   d       day-of-month                number            10

   Q/q     quarter-of-year             number/text       3; 03; Q3; 3rd quarter
   Y       week-based-year             year              1996; 96
   w       week-of-week-based-year     number            27
   W       week-of-month               number            4
   E       day-of-week                 text              Tue; Tuesday; T
   e/c     localized day-of-week       number/text       2; 02; Tue; Tuesday; T
   F       week-of-month               number            3

   a       am-pm-of-day                text              PM
   h       clock-hour-of-am-pm (1-12)  number            12
   K       hour-of-am-pm (0-11)        number            0
   k       clock-hour-of-am-pm (1-24)  number            0

   H       hour-of-day (0-23)          number            0
   m       minute-of-hour              number            30
   s       second-of-minute            number            55
   S       fraction-of-second          fraction          978
   A       milli-of-day                number            1234
   n       nano-of-second              number            987654321
   N       nano-of-day                 number            1234000000

   V       time-zone ID                zone-id           America/Los_Angeles; Z; -08:30
   z       time-zone name              zone-name         Pacific Standard Time; PST
   O       localized zone-offset       offset-O          GMT+8; GMT+08:00; UTC-08:00;
   X       zone-offset 'Z' for zero    offset-X          Z; -08; -0830; -08:30; -083015; -08:30:15;
   x       zone-offset                 offset-x          +0000; -08; -0830; -08:30; -083015; -08:30:15;
   Z       zone-offset                 offset-Z          +0000; -0800; -08:00;

   p       pad next                    pad modifier      1

   '       escape for text             delimiter
   ''      single quote                literal           '
   [       optional section start
   ]       optional section end
   #       reserved for future use
   {       reserved for future use
   }       reserved for future use
于 2015-10-13T12:04:07.507 回答
2

在 DataWeave 中，您可以将字符串解析为各种 Java 日期时间类型，并将它们格式化回字符串。尝试解析后跟格式，如下所示：

payload.date as :localdatetime {format: "M/dd/yyyy  h:mm:ss a"} as :string {format: "MM/dd/yyyy"}

以下是有关 DataWeave 中日期格式的更多详细信息。

于 2015-10-13T12:04:16.297 回答
1

嗨，在 dataweave 中，您可以像这样编写表达式

formatedDate: |2003-10-01T23:57:59| as :string {format: "MM-dd-YYYY"}
于 2015-10-25T10:22:16.453 回答
-5

你可以用这个，

var newDate = new Date("8/30/2015  4:00:13 PM");
var dd = newDate.getDate();
var mm = newDate.getMonth()+1; //January is 0!

var yyyy = newDate.getFullYear();
if(dd<10){
    dd='0'+dd
} 
if(mm<10){
    mm='0'+mm
} 
var formattedDate = mm+'/'+dd+'/'+yyyy;
console.log(formattedDate);  //Prints "08/30/2015"

请参考这里

于 2015-10-12T07:37:51.963 回答