bash lsof 应该为一组 pid 提供所有打开的文件
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lsof should give all open files for a set of pids
提问by user1071840
I have 30 instances of a process running on a server and want to log open files for each process for analysis.
我有 30 个在服务器上运行的进程实例，并希望记录每个进程的打开文件以进行分析。
I ran the following command:
我运行了以下命令：
* ps auwwx | grep PROG_NAME | awk '{print }' | xargs lsof -p | less
It complaints that, "lsof: status error on : No such file or directory"
它抱怨说，“lsof：状态错误：没有这样的文件或目录”
However, if I run lsof -p < pid >it gives me the list of open files for that process . How can I get a list of all open files for all 30 instances of the process on a FreeBSD machine.
但是，如果我运行lsof -p < pid >它，它会给我该进程的打开文件列表。如何获取 FreeBSD 机器上所有 30 个进程实例的所有打开文件的列表。
Moreover, I do not want the shared libraries to be listed. If I do -d "^txt"it isn't showing some other db files which I want to be shown. Is there any other way to grep out the .so files?
此外，我不希望列出共享库。如果我这样做，-d "^txt"它不会显示我想要显示的其他一些数据库文件。有没有其他方法可以 grep 出 .so 文件？
回答by Ben Hymanson
The lsof -poption takes a comma-separated list of PIDs. The way you're using xargswill pass the pids as separate arguments leading some to be interpreted as filenames.
该lsof -p选项采用逗号分隔的 PID 列表。您使用的方式xargs将 pids 作为单独的参数传递，导致一些被解释为文件名。
Try lsof -p $(your grep | tr '\012' ,)That's going to have a trailing comma, I'm not sure if lsofwill care but you could sedit off if necessary.
尝试lsof -p $(your grep | tr '\012' ,)这将有一个尾随逗号，我不确定是否lsof会在意，但sed如果需要，您可以将其关闭。
回答by nneonneo
You can use xargs -L1 lsof -pto run lsofonce per pid.
您可以使用每个 pidxargs -L1 lsof -p运行lsof一次。
Even better: use lsof -cto list all open files from commands matching a specific pattern:
更好的是：用于lsof -c从匹配特定模式的命令中列出所有打开的文件：
lsof -c bas # list all processes with commands starting with 'bas'
lsof -c '/ash$/x' # list all commands ending with 'ash' (regexp syntax)
