我有一个存储的PL / SQL程序,每10分钟启动一次(预定作业)。有时我的程序执行超过10分钟。在这种情况下,Oracle调度程序与第一个实例并行运行我的过程的另一个实例。如何在第一个过程完成之前阻止Oracle启动新过程?
答案 0 :(得分:5)
我没有看到您报告的行为,但您可以再次发布您的调度程序代码,因为有许多选项可用。
设置基本测试表明,如果同一作业已在运行,则调度程序将实际跳过作业的计划运行。换句话说,它会等到第一个实例完成后再启动另一个(非并行)。
这个例子触发了一个只能休眠90秒的自治块,但计划每分钟运行一次:
BEGIN
dbms_scheduler.create_schedule('SCH_MINUTELY', systimestamp,
repeat_interval=>'FREQ=MINUTELY; INTERVAL=1');
--dbms_scheduler.drop_job('JOB_TEST1', false);
--dbms_scheduler.drop_chain('CHAIN_TEST1', false);
--dbms_scheduler.drop_program('PROG_MAIN1', false);
dbms_scheduler.create_program(program_name=>'PROG_MAIN1',program_type=>'PLSQL_BLOCK',
program_action=>'BEGIN
dbms_lock.sleep(90);
END;',
number_of_arguments=>0,enabled=>TRUE,comments=>'Runs for 90 seconds');
-- create chain
dbms_scheduler.create_chain(chain_name=>'CHAIN_TEST1',comments=>'A chain to test scheduler behavior');
-- define chain steps
dbms_scheduler.define_chain_step(chain_name=>'CHAIN_TEST1',step_name=>'STEP_RUN_MAIN',program_name=>'PROG_MAIN1');
-- define chain rules
dbms_scheduler.define_chain_rule(chain_name=>'CHAIN_TEST1',condition=>'TRUE',action=>'START "STEP_RUN_MAIN"',rule_name=>'CHAIN_TEST_R01',comments=>'Run main pgm');
dbms_scheduler.define_chain_rule(chain_name=>'CHAIN_TEST1',condition=>'STEP_RUN_MAIN succeeded',action=>'END',rule_name=>'CHAIN_TEST_R02',comments=>'End of chain');
-- enable chain
dbms_scheduler.ENABLE ('CHAIN_TEST1');
-- create job
dbms_scheduler.create_job(job_name=>'JOB_TEST1',job_type=>'CHAIN',job_action=>'CHAIN_TEST1',schedule_name=>'SCH_MINUTELY',enabled=>TRUE,auto_drop=>FALSE,comments=>'Job to test scheduler');
END;
查看调度程序日志(dba_scheduler_job_run_details):
LOG_DATE JOB_NAME JOB_SUBNAME STATUS ACTUAL_START_DATE RUN_DURATION
-------------------------------------- ----------------------------------------------------------------- ----------------------------------------------------------------- ------------------------------ -------------------------------------- ------------
23-SEP-13 10.49.34.332727000 AM -04:00 JOB_TEST1 SUCCEEDED 23-SEP-13 10.48.04.206153000 AM -04:00 0 0:1:30.0
23-SEP-13 10.49.34.313332000 AM -04:00 JOB_TEST1 STEP_RUN_MAIN SUCCEEDED 23-SEP-13 10.48.04.302311000 AM -04:00 0 0:1:30.0
23-SEP-13 10.47.34.231511000 AM -04:00 JOB_TEST1 SUCCEEDED 23-SEP-13 10.46.04.105827000 AM -04:00 0 0:1:30.0
23-SEP-13 10.47.34.212905000 AM -04:00 JOB_TEST1 STEP_RUN_MAIN SUCCEEDED 23-SEP-13 10.46.04.200956000 AM -04:00 0 0:1:30.0
23-SEP-13 10.45.34.144779000 AM -04:00 JOB_TEST1 SUCCEEDED 23-SEP-13 10.44.04.011605000 AM -04:00 0 0:1:30.0
23-SEP-13 10.45.34.124745000 AM -04:00 JOB_TEST1 STEP_RUN_MAIN SUCCEEDED 23-SEP-13 10.44.04.113662000 AM -04:00 0 0:1:30.0
23-SEP-13 10.43.35.048820000 AM -04:00 JOB_TEST1 SUCCEEDED 23-SEP-13 10.42.04.906209000 AM -04:00 0 0:1:30.0
23-SEP-13 10.43.35.024348000 AM -04:00 JOB_TEST1 STEP_RUN_MAIN SUCCEEDED 23-SEP-13 10.42.05.012929000 AM -04:00 0 0:1:30.0
23-SEP-13 10.41.34.950533000 AM -04:00 JOB_TEST1 SUCCEEDED 23-SEP-13 10.40.04.803192000 AM -04:00 0 0:1:30.0
23-SEP-13 10.41.34.918242000 AM -04:00 JOB_TEST1 STEP_RUN_MAIN SUCCEEDED 23-SEP-13 10.40.04.900061000 AM -04:00 0 0:1:30.0
10 rows selected
我们可以看到,当Scheduler每分钟触发一次这个作业时,作业本身会运行1.5分钟,所以:
time0: scheduler runs job
time1: scheduler sees job is still running, does nothing
time2: scheduler sees this job isnt running, runs job
time3: scheduler sees job is still running, does nothing
time4: scheduler sees this job isnt running, runs job
等等
答案 1 :(得分:-1)
使用信号量。您可以创建临时表并在第一个过程启动时插入标志。如果在开始第二个程序之前第一个程序已启动,则应检查该表。这是一个简单的信号量逻辑。