将 cron 表达式转换为漂亮的描述字符串？有 JAVA 和 Objective-C 的库吗？
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Convert cron expression into nice description strings? Is there a library for JAVA and Objective-C?
提问by OneWorld
I am looking for a parser that converts a cron expressionlike 45 17 7 6 * *into Every year, on June 7th at 17:45The parser should be adjustable to other languages. German for the first step.
我要寻找的是一个转换解析器cron表达式像45 17 7 6 * *成在17:45，每年6月7日的解析器应调整为其他语言。第一步是德语。
Is there a library for a
有图书馆吗
- JAVA based Android project
- Objective-C based Iphone project.
- 基于JAVA的Android项目
- 基于 Objective-C 的 Iphone 项目。
See herefor the usecase.
有关用例，请参见此处。
回答by Jigar Joshi
cronTrigger.getExpressionSummary()
Example:
例子：
CronTrigger t = new CronTrigger();
t.setCronExpression("0 30 10-13 ? * WED,FRI");
System.out.println(""+t.getExpressionSummary());
Output:
输出：
seconds: 0
minutes: 30
hours: 10,11,12,13
daysOfMonth: ?
months: *
daysOfWeek: 4,6
lastdayOfWeek: false
nearestWeekday: false
NthDayOfWeek: 0
lastdayOfMonth: false
years: *
回答by robsf
In Java, have a look into cron4j http://www.sauronsoftware.it/projects/cron4j/
在 Java 中，查看 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:
您将找到所需的解析器，但随后您必须编写代码以根据需要打印字符串。首先创建一个 SchedulingPattern 对象：
new SchedulingPattern("0 30 10-13 ? * 1,2,5")
回答by kekec
You may find cron-utilsuseful 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:
您可能会发现cron-utils对这项任务很有用，因为它以各种语言提供人类可读的描述，并且不需要完全成熟的调度程序来提供它们。支持多种 cron 格式。在文档中的代码片段下方：
//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"
