Linux 用grep解析字符串
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Parsing string with grep
提问by Roman Iuvshin
I need some help with parsing a string in Linux.
我需要一些帮助来解析 Linux 中的字符串。
I have a string:
我有一个字符串：
[INFO] Total time: 2 minutes 8 seconds
and want to get only
并且只想得到
2 minutes 8 seconds
采纳答案by knittl
If the line prefix is always the same, simply use sed and replace the prefix with an empty string:
如果行前缀始终相同，只需使用 sed 并将前缀替换为空字符串：
sed 's/\[INFO\] Total Time: //'
Assuming that the time is always the last thing in a line after a colon, use the following regex (replace each line with everything after the colon):
假设时间总是冒号后一行中的最后一个，使用以下正则表达式（用冒号后的所有内容替换每一行）：
sed 's/^.*: \(.*\)$//'
回答by yko
Use sedor perl:
使用sed或perl：
echo "[INFO] Total time: 2 minutes 8 seconds" | sed -e 's/^\[INFO\] Total time:\s*//'
echo "[INFO] Total time: 2 minutes 8 seconds" | perl -pe "s/^\[INFO\] Total time:\s*//;"
回答by sehe
The sed and perl options do work, but in this trivial case, I'd prefer
sed 和 perl 选项确实有效，但在这种微不足道的情况下，我更喜欢
echo "[INFO] Total time: 2 minutes 8 seconds" | cut -d: -f2
If you have something against spaces, you can just use
如果你有反对空格的东西，你可以使用
echo "[INFO] Total time: 2 minutes 8 seconds" | cut -d: -f2 | xargs
or even...
甚至...
echo "[INFO] Total time: 2 minutes 8 seconds" | cut -d: -f2 | cut -c2-
PS. Trivia: you could do this with greponly if grep implemented positive lookbehind like this egrep -o '(?<=: ).*'; Unfortunately neither POSIX extended regex nor GNU extended regex implement lookbehind (http://www.regular-expressions.info/refflavors.html)
附注。琐事：grep只有当 grep 像这样实现了正向后视，你才能做到这一点egrep -o '(?<=: ).*'；不幸的是，POSIX 扩展正则表达式和 GNU 扩展正则表达式都没有实现后视（http://www.regular-expressions.info/refflavors.html）
回答by Johnsyweb
Using grep:
使用grep：
$ echo '[INFO] Total time: 2 minutes 8 seconds' | grep -o '[[:digit:]].*$'
2 minutes 8 seconds
Or sed:
或者sed：
$ echo '[INFO] Total time: 2 minutes 8 seconds' | sed 's/.*: //'
2 minutes 8 seconds
Or awk:
或者awk：
$ echo '[INFO] Total time: 2 minutes 8 seconds' | awk -F': ' '{print }'
2 minutes 8 seconds
Or cut:
或者cut：
$ echo '[INFO] Total time: 2 minutes 8 seconds' | cut -d: -f2
2 minutes 8 seconds
And then read sed & awk, Second Edition.
回答by jaypal singh
If you prefer AWK then it is quite simple
如果你更喜欢 AWK 那么它很简单
echo "[INFO] Total time: 2 minutes 8 seconds" | awk -F": " '{ print }'
回答by Shifu
If you are getting the info from the terminal then you can grep out the info and use cut with the delimiter to remove everything before the info you want. grep INFO | cut -f2 -d:
如果您从终端获取信息，那么您可以 grep 出信息并使用带有分隔符的 cut 删除您想要的信息之前的所有内容。grep 信息 | 剪切 -f2 -d：
If you want the info out of a file then you can grep the file grep INFO somefilename | cut -f2 -d:
如果你想从文件中获取信息，那么你可以 grep 文件 grep INFO somefilename | 剪切 -f2 -d：
