bash : 从一行中删除特定模式的字符

Question

文件a.txt的内容是：

10.39.105.16tcp(1234)   ABCDEF02  
10.49.105.13tcp(521)    ABCDEFV01  
10.19.105.12tcp(1241)   ABCDEFV01  
10.29.105.20tcp(1222)   ABCDEF03  
10.39.104.9tcp(131)     ABCDEF21  
10.49.104.34tcp(1512)   ABCDEF22  
10.89.104.120tcp(1384)  ABCDEF32

如何实现如下输出：

10.39.105.10 ABCDEF02   
10.49.105.13 ABCDEFV01   
10.19.105.13 ABCDEFV01   
10.29.105.20 ABCDEF03   
10.39.104.1 ABCDEF21   
10.49.104.2 ABCDEF22   
10.89.104.130 ABCDEF32

如何删除上述文件每一行中的模式 tcp( * )？

Answer 1

你可以使用 sed:

sed 's/tcp([0-9]\+)\s\+/ /' a.txt

这也将删除 tcp(*) 之后的尾随 space，并将其替换为输出中显示的单个 space。

要更改实际文件（不仅仅是输出），请为 inplace 添加 -i 选项：

sed -i 's/tcp([0-9]\+)\s\+/ /' a.txt

Answer 2

根据您要实现的目标，答案可能是：

    awk '{sub(/tcp(.+)/,"",)}1' file

    10.39.105.16 ABCDEF02
    10.49.105.13 ABCDEFV01
    10.19.105.12 ABCDEFV01
    10.29.105.20 ABCDEF03
    10.39.104.9 ABCDEF21
    10.49.104.34 ABCDEF22
    10.89.104.120 ABCDEF32

bash : 从一行中删除特定模式的字符

bash : Remove the specific pattern of characters from a line

bash

awk

sed

grep

tr