cut command top remove seconds from time. shell scripting-CodePudding

Lets say we got date and time to a log in the format of

2022-05-18-11:57:140100

I need to remove seconds from this time. It means the final output should be like

 2022-05-18-11:57

I tried the following

echo "2022-05-18-11:57:140100" | rev | cut -d/ -f6- | rev`:"

this was not successful and I have no idea how this cut command works. can anybody explain. Thanks in advance

CodePudding user response：

1st solution: With GNU awk you can simply do it like following. Simple explanation would be, set FS and OFS as : and then in main block of awk program decrease NF with 1 and print the line.

echo "2022-05-18-11:57:140100" | awk 'BEGIN{FS=OFS=":"} NF--'

2nd solution: In any awk using awk's match function please try following.

echo "2022-05-18-11:57:140100" | awk 'match($0,/.*:/){print substr($0,RSTART,RLENGTH-1)}'

3rd solution: Using sed's capability of capturing groups(keep values in temp buffer) please try following code.

echo "2022-05-18-11:57:140100" | sed -E 's/(^[^:]*):([^:]*):.*/\1:\2/'

4th solution(OP's efforts FIX): Fixing OP's efforts using rev cut rev combination here.

echo "2022-05-18-11:57:140100" | rev | cut -d':' -f2- | rev