linux - 删除文件中一行的开头

Question

我有一个具有这种格式的文件:

21:26:54.21 PT  disconnected
21:26:56.18 CC1 received TCMTM
21:26:56.18 CC2 received TCMTM
21:26:56.18 TFT received TCMTM
21:26:56.18 FEC received TCMTM

我想删除每行的第一部分 (21:26:54.21)，我试过这个:

sed "s/^[:digit:]*[:space:]//"  file.log

但它不起作用，我得到与条目相同的输出。

Answer 1

使用cut:

$ cut -d' ' -f2- file
PT  disconnected
CC1 received TCMTM
CC2 received TCMTM
TFT received TCMTM
FEC received TCMTM

将空格设置为分隔符并从字段 2 打印到最后一个 (f2-)。

使用awk:

$ awk '{$1=""}1' file  ## leading space :(
 PT disconnected
 CC1 received TCMTM
 CC2 received TCMTM
 TFT received TCMTM
 FEC received TCMTM

将第一个字段设置为空。

$ awk '{for (i=1;i<=NF; i++) $i=$(i+1); NF=NF-1}1' file
PT disconnected
CC1 received TCMTM
CC2 received TCMTM
TFT received TCMTM
FEC received TCMTM

它将字段编号减一，将每个字段移到前一个(因此，第一个被删除)。

使用sed:

$ sed 's/^[^ ]* //g' file
PT  disconnected
CC1 received TCMTM
CC2 received TCMTM
TFT received TCMTM
FEC received TCMTM

删除从行首到第一个空格的所有内容。

或基于您的sed:

$ sed "s/^[0-9:.]* //g" file
PT  disconnected
CC1 received TCMTM
CC2 received TCMTM
TFT received TCMTM
FEC received TCMTM

删除所有字符 0-9 或 : 或 . 从行首到第一个空格。