Quartz.Net - 延迟一个简单的触发器启动

发布于 2024-09-14 14:10:38 字数 1390 浏览 3 评论 0原文

我在 Quartz 中设置了一些作业，以按设定的时间间隔运行。问题是，当服务启动时，它会尝试立即启动所有作业...有没有办法使用 .xml 配置为每个作业添加延迟？

以下是 2 个作业触发器示例：

 <simple>
    <name>ProductSaleInTrigger</name>
    <group>Jobs</group>
    <description>Triggers the ProductSaleIn job</description>
    <misfire-instruction>SmartPolicy</misfire-instruction>
    <volatile>false</volatile>
    <job-name>ProductSaleIn</job-name>
    <job-group>Jobs</job-group>
    <repeat-count>RepeatIndefinitely</repeat-count>
    <repeat-interval>86400000</repeat-interval>        
  </simple>

 <simple>
    <name>CustomersOutTrigger</name>
    <group>Jobs</group>
    <description>Triggers the CustomersOut job</description>
    <misfire-instruction>SmartPolicy</misfire-instruction>
    <volatile>false</volatile>
    <job-name>CustomersOut</job-name>
    <job-group>Jobs</job-group>
    <repeat-count>RepeatIndefinitely</repeat-count>
    <repeat-interval>43200000</repeat-interval> 
  </simple>

如您所见，有 2 个触发器，第一个触发器每天重复，下一个触发器每天重复两次。

我的问题是，我希望第一个或第二个作业在另一个作业几分钟后开始...（因为它们最终都访问相同的 API，并且我不想超载请求）

是否有重复延迟或优先属性？我找不到任何文档这么说..

原文

I have a few jobs setup in Quartz to run at set intervals. The problem is though that when the service starts it tries to start all the jobs at once... is there a way to add a delay to each job using the .xml config?

Here are 2 job trigger examples:

 <simple>
    <name>ProductSaleInTrigger</name>
    <group>Jobs</group>
    <description>Triggers the ProductSaleIn job</description>
    <misfire-instruction>SmartPolicy</misfire-instruction>
    <volatile>false</volatile>
    <job-name>ProductSaleIn</job-name>
    <job-group>Jobs</job-group>
    <repeat-count>RepeatIndefinitely</repeat-count>
    <repeat-interval>86400000</repeat-interval>        
  </simple>

 <simple>
    <name>CustomersOutTrigger</name>
    <group>Jobs</group>
    <description>Triggers the CustomersOut job</description>
    <misfire-instruction>SmartPolicy</misfire-instruction>
    <volatile>false</volatile>
    <job-name>CustomersOut</job-name>
    <job-group>Jobs</job-group>
    <repeat-count>RepeatIndefinitely</repeat-count>
    <repeat-interval>43200000</repeat-interval> 
  </simple>

As you see there are 2 triggers, the first repeats every day, the next repeats twice a day.

My issue is that I want either the first or second job to start a few minutes after the other... (because they are both in the end, accessing the same API and I don't want to overload the request)

Is there a repeat-delay or priority property? I can't find any documentation saying so..

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痴意少年 2024-09-21 14:10:38

我知道您是通过 XML 执行此操作，但在代码中您可以将 StartTimeUtc 设置为延迟 30 秒，如下所示...

trigger.StartTimeUtc = DateTime.UtcNow.AddSeconds(30);

I know you are doing this via XML but in code you can set the StartTimeUtc to delay say 30 seconds like this...

trigger.StartTimeUtc = DateTime.UtcNow.AddSeconds(30);

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小ぇ时光︴ 2024-09-21 14:10:38

对于您的 XML 文件来说，这并不是一个完美的答案 - 但通过代码，您可以在构建触发器时使用 StartAt 扩展方法。

/* calculate the next time you want your job to run - in this case top of the next hour */
var hourFromNow = DateTime.UtcNow.AddHours(1);
var topOfNextHour = new DateTime(hourFromNow.Year, hourFromNow.Month, hourFromNow.Day, hourFromNow.Hour, 0, 0);

/* build your trigger and call 'StartAt' */
TriggerBuilder.Create().WithIdentity("Delayed Job").WithSimpleSchedule(x => x.WithIntervalInSeconds(60).RepeatForever()).StartAt(new DateTimeOffset(topOfNextHour))

This isn't exactly a perfect answer for your XML file - but via code you can use the StartAt extension method when building your trigger.

/* calculate the next time you want your job to run - in this case top of the next hour */
var hourFromNow = DateTime.UtcNow.AddHours(1);
var topOfNextHour = new DateTime(hourFromNow.Year, hourFromNow.Month, hourFromNow.Day, hourFromNow.Hour, 0, 0);

/* build your trigger and call 'StartAt' */
TriggerBuilder.Create().WithIdentity("Delayed Job").WithSimpleSchedule(x => x.WithIntervalInSeconds(60).RepeatForever()).StartAt(new DateTimeOffset(topOfNextHour))

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