awk 删除以字符结尾的行

Question

I have a string:

COL1	COL2
PRE	test1/
PRE	test1/
PRE	test1/
2023-01-27	12:37:16
2023-01-27	12:37:16
2023-01-27	12:37:16
2023-01-27	12:37:16
2023-01-27	12:37:16

Want left a black space with awk the complete lines that finish with the character "/" but i cannot guess it.

I test it for example but doesn't work:

awk '{gsub("*/",""); print $1 $2}'

Thanks!

Answer 1

Use the pattern expression to run different code depending on whether the line ends with /$ or not.

awk '!/\/$/ {print $1, $2}
     /\/$/ {print ""}' filename

Answer 2

The issue with the command you tried is that it only replaces the string "*/" with an empty string, but it doesn't check for lines that end with the character "/". Here's a corrected version of the command:

awk '$2 ~ /\/$/ {gsub("\/$",""); print $1, $2}'

This command uses the ~ operator to check if the second field ( $2 ) ends with the character / , and if so, removes the / from the end of the line before printing both fields.