grep date from 2nd line of a text file

I have text file named (document_determinationv1.txt). This text file contains two lines which is mentioned below. I want to get the date(20210805) from the second line and want to store in a variable using shell script commands. I am new to Linux. Thanks in advance.

[DOCUMENT_DETERMINATION]
#REVISION:v1;DATE:20210805

As always, there's more than one way to do it, depending on the exact requirements.

As one of the simplest ways, you could do:

AVARIABLE=$(sed -ne 's/.*DATE://p' document_determinationv1.txt)

This doesn't actually care which line the date is on. It will work as long as your file contains exactly one line containing the keyword DATE followed by a colon and the date you want to extract. It will also not check whether DATE appears as a word of its own, whether the part after DATE: is actually a date, or whether it is followed by any additional junk. If your file contains more than one line containing DATE followed by a colon then it will put all of the parts after each DATE: into the variable, separated by newlines, which may or may not wreak havoc with your further processing. So if your file contains, for example:

[DOCUMENT_DETERMINATION]
#REVISION:v1;DATE:20210805;WEIGHT:123kg
[REAL_DETERMINATION]
#PRIME:13;CANDIDATE:BART SIMPSON

then the command will happily put

20210805;WEIGHT:123kg
BART SIMPSON

into the variable, including the newline between the letters g and B.

But as long as you can guarantee the file has exactly the format you quoted in your question it will work fine.

You could do the following:

cat file.txt | sed "2q;d" | awk -F: '{print $3}'

• cat prints out the file content into a pipe
• sed takes this pipe and outputs the second line
• which awk takes, splits by colons and finally prints the third argument

You can try each command separately. And if you want to store the whole stuff into a variable, use the $(command) syntax:

VAR=$(cat file.txt | sed "2q;d" | awk -F: '{print $3}')

And print the result:

echo $VAR