bash 如何使用管道字符分隔符分隔字段
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How to separate fields with pipe character delimiter
提问by morandg
I know this question has already been asked but no of the solution I've found worked for me! I have a program that has an output like this:
我知道已经有人问过这个问题，但我找到的解决方案都没有对我有用！我有一个程序，它有这样的输出：
COUNT|293|1|lps
I'm interested in having the second field however no one of these tries worked:
我对第二个领域很感兴趣，但是这些尝试都没有奏效：
./spawn 1 | cut -d '|' -f2
./spawn 1 | cut -d \| -f2
./spawn 1 | awk -F "|" '{print }'
./spawn 1 | awk 'BEGIN{FS="|"} {print }'
./spawn 1 | sed 's/|/;/g'
./spawn 1 | sed 's/\|/;/g'
But the output is always the same:
但输出总是相同的：
COUNT|293|1|lps
Is there a bug somewhere in bash? I would be surprised, the results is the same on my Linux host and on my embedded device using busybox's ash! Any pointer is strongly appreciated!
bash 中的某个地方是否存在错误？我会感到惊讶，结果在我的 Linux 主机和我的嵌入式设备上使用 busybox 的 ash 是一样的！强烈感谢任何指针！
EDITMy fault, the output was in stderr ... ._.
编辑我的错，输出在 stderr ... ._。
./spawn 1 2>&1 | cut -d '|' -f2
4615
Sorry for ennoying!
抱歉打扰了！
回答by Mark Longair
Just repeating what I guessed in a comment as an answer, now that the questioner has confirmed that this is the problem.
只是重复我在评论中的猜测作为答案，现在提问者已经确认这是问题所在。
The problem here is that ./spawn 1is outputting to standard error, not standard output. You can redirect the output using 2>&1, so the following should work:
这里的问题./spawn 1是输出到标准错误，而不是标准输出。您可以使用重定向输出2>&1，因此以下内容应该有效：
./spawn 1 2>&1 | cut -d '|' -f2
回答by morandg
$ echo 'COUNT|293|1|lps' | cut -d'|' -f2
293
It works here.
它在这里工作。
