Display all lines after a matching variable

Shell Programming and Scripting
1

Hi,

I have a file called abc.txt with the following dates

2016-01-27
2016-01-28
2016-01-29
2016-01-30
2016-01-31
2016-02-01
2016-02-02
2016-02-03

I would like to print all lines below if 2016-01-31 is found, excluding that date.
I use this command -->

sed '1,/2016-01-31/d' abc.txt

If i declare the date in a variable and then try this, its not working. How do I include the variable in the sed command in Linux.

2

Hello Nagesh_1985,

Please use code tags as per forum rules for commands/inputs/codes which you are using in your posts.
Could you please try following and let me know if this helps you.

awk '($0 ~ /2016-01-31/){A=1;next} A{print}'  Input_file

Output will be as follows.

2016-02-01
2016-02-02
2016-02-03

Thanks,
R. Singh

3

Try double quoting:

DT=2016-01-31
sed "1,/$DT/d" file
2016-02-01
2016-02-02
2016-02-03
4

In short

awk 'A; /2016-01-31/{A=1}'  Input_file

All these solutions require the searched string to be present.
Here is one with a string comparison (that of course requires a sorted input)

awk '($0"" > "2016-01-31")'  Input_file
5

The command worked !!