我有一个问题,为了让job1在LSF中使用变量名动态等待job2,这是我的代码
#!/bin/bash
JOB1=first
JOB2=second
bsub -I -q reg -J $JOB1 "ls /bin*" &
bsub -I -q reg -w 'done("$JOB1")' -J $JOB2 ls /usr/ &"发布于 2015-03-07 08:21:52
几个猜测。
bash不会展开单引号内的变量。你可能是这个意思bsub -I -q reg -w "done($JOB1)“-J $JOB2 "ls /usr/ &”提交第一个作业时,
strace表明这是可能的。pid 8413 18:48:30.533859 execve(".../bsub","bsub","-I","-J","first","ls /bin /* ",/* 45 vars */ <未完成...> pid 8414 18:48:30.534415 execve(".../bsub","bsub","-I","-w","done(first)","-J","second","ls /usr/ &",/* 45 vars */ <未完成的...> ...pid 8414 18:48:30.572521 connect(5,{sa_family=AF_INET,sin_port=htons(6881),sin_addr=inet_addr("a.b.c.d")},16 <未完成的...> pid 8413 18:48:30.573422 connect(5,{sa_family=AF_INET,sin_port=htons(6881),sin_addr=inet_addr(“A.B.C.D”))},16 <未完成的...>
https://stackoverflow.com/questions/28902160
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