i have grep 2 columns , suppose col1 and col2. In col2 i want to remove a pattern which occurs in every line. how to use awk/sed for this purpose?
我有grep 2列，假设col1和col2。在col2中，我想删除每行中出现的模式。如何使用awk / sed达到此目的？
suppose ps -eaf | grep b would result into following output:
假设ps -eaf | grep b将导致以下输出：
col1 col2 col3
1 a/b/rac 123
2 a/b/rac1 456
3 a/b/rac3 789
I want output to get stored in a file like this :
我希望输出存储在这样的文件中：
1 rac
2 rac1
3 rac3
2 个解决方案
#1
2
Vaguely speaking, this might do what you want:
含糊地说，这可能会做你想要的：
$ awk 'sub(/.*b\//,"",$2){print $1, $2}' file
1 rac
2 rac1
3 rac3
assuming file contains:
假设文件包含：
col1 col2 col3
1 a/b/rac 123
2 a/b/rac1 456
3 a/b/rac3 789
#2
0
You will want to pipe the output to
您将要输出管道
| awk '{sub(/^.*\//, "", $2); print $1, $2}'
The first command in the awk program changes the contents of the second field to only contain whatever was after the final / character, which appears to be what you want. The second command prints the first two fields.
awk程序中的第一个命令将第二个字段的内容更改为仅包含最终/字符之后的内容，这似乎是您想要的。第二个命令打印前两个字段。
#1
2
Vaguely speaking, this might do what you want:
含糊地说，这可能会做你想要的：
$ awk 'sub(/.*b\//,"",$2){print $1, $2}' file
1 rac
2 rac1
3 rac3
assuming file contains:
假设文件包含：
col1 col2 col3
1 a/b/rac 123
2 a/b/rac1 456
3 a/b/rac3 789
#2
0
You will want to pipe the output to
您将要输出管道
| awk '{sub(/^.*\//, "", $2); print $1, $2}'
The first command in the awk program changes the contents of the second field to only contain whatever was after the final / character, which appears to be what you want. The second command prints the first two fields.
awk程序中的第一个命令将第二个字段的内容更改为仅包含最终/字符之后的内容，这似乎是您想要的。第二个命令打印前两个字段。