Convert cron expression into nice description strings? Is there a library for JAVA and Objective-C? [closed]

问题

I am looking for a parser that converts a cron expression like 45 17 7 6 * * into Every year, on June 7th at 17:45 The parser should be adjustable to other languages. German for the first step.

Is there a library for a

JAVA based Android project
Objective-C based Iphone project.

See here for the usecase.

回答1:

cronTrigger.getExpressionSummary()

Example:

    CronTrigger t = new CronTrigger();
    t.setCronExpression("0 30 10-13 ? * WED,FRI");
    System.out.println(""+t.getExpressionSummary());

Output:

seconds: 0
minutes: 30
hours: 10,11,12,13
daysOfMonth: ?
months: *
daysOfWeek: 4,6
lastdayOfWeek: false
nearestWeekday: false
NthDayOfWeek: 0
lastdayOfMonth: false
years: *

Api Java Doc

回答2:

In Java, have a look into cron4j http://www.sauronsoftware.it/projects/cron4j/

You will find the parser you need but then you have to write your code to print the string as you need it. Start by creating a SchedulingPattern object:

new SchedulingPattern("0 30 10-13 ? * 1,2,5")

回答3:

You may find cron-utils useful for this task, since provides human readable descriptions in various languages and does not require a fully fledged scheduler to provide them. Supports multiple cron formats. Below a code snippet from the docs:

//create a descriptor for a specific Locale
CronDescriptor descriptor = CronDescriptor.instance(Locale.UK);

//parse some expression and ask descriptor for description
String description = descriptor.describe(parser.parse("*/45 * * * * *"));
//description will be: "every 45 seconds"

来源：https://stackoverflow.com/questions/4469276/convert-cron-expression-into-nice-description-strings-is-there-a-library-for-ja

标签

java

objective-c

datetime

cron

cronexpression