No, sorry
The value of VAR should be the input of my question
---------- Post updated at 09:17 PM ---------- Previous update was at 09:06 PM ----------
After this function, let there are 10 files with .rpm extension has been parsed and saved with $VAR.list .
Now the other function should take the all the 10 values of VAR i.e the function can be called inside the test() function.
that VAR value has to be written inside a file name test.txt
ex: $cat test.txt
NAME=abc
NAME=xyz......
To repeat: Do not name a function test -- ever. Name it test2 if you have to, never test. There's a real command named test which will pop up when you're not expecting it.
This is redundant:
Just do:
I have no idea what that esac is doing as you're not using case, that's got to be a syntax error.
I think you mean popd >/dev/null 2>/dev/null not popd &>/dev/null.
I think you mean [ -e ../info/$VAR.list ] not [ -e /info/$VAR.list} ]
Thanks for the quick reply, The corrections given by you are correct, I will make changes. But the last which you said.
I can't understand it., where I have to implement it and how it will collect all the values of VAR
Shell script logic
Hi
I have 2 input files like with file 1 content as (file1)
"BRGTEST-242" a.txt "BRGTEST-240" a.txt "BRGTEST-219" e.txt
File 2 contents as fle(2)
"BRGTEST-244" a.txt "BRGTEST-244" b.txt "BRGTEST-231" c.txt "BRGTEST-231" d.txt "BRGTEST-221" e.txt
I want to get... (22 Replies)