Shell Programming and Scripting

Hello All,

I have file names in a directory that ends this way

File Names in the directory

Code:

abc.log.1.txt
abc.log.2.txt
abc.log.3.txt

I want to print the 1, 2 and 3 in the file names as the first column in my output, and some contents of the files as second and fourth columns. I wrote this for loop, but for some reasons it is not working. Can someone help me?

Code:

for i in `ls /path/`; do awk '{f='ls $i | awk -F. '{print $3}'; if($1 ~ /^SOME/){print f"\t"$5"\t""0""\t"$6}}' OFS="\t" $i; done | head

The error I am saying is

Code:

-bash: syntax error near unexpected token `{print'

Contents of abc.log.1.txt

Code:

SOME HN 098.12 HN kill 09876

Contents of abc.log.2.txt

Code:

SOME HJ 98.76 HJ mill 12345

My final output would be

Code:

#chromosome	#header2 header3 header4
1 kill 0 09876
2 mill 0 12345

The zero in the 3rd column is common for all the output records and the header is same.

Thanks in advance

That's a dangerous use of backticks -- spaces in filenames, too many files, and other such conditions may cause your code to break.

That's also a useless use of ls. There's much better ways to get a variable into awk, and you certainly don't need to use ls twice to list one file.

Your awk problem is caused by a misplaced quote in the middle, causing the awk program to end early and feed the rest into BASH.

I'd try this:

Code:

awk '{split(FILENAME, A, ".");  print A[3], $5, 0, $6 }' /path/*

If that complains there are too many files:

Code:

ls /path/ | xargs awk '{split(FILENAME, A, ".");  print A[3], $5, 0, $6 }'

FILENAME is a special variable inside awk meaning 'the current file being processed'. I split it apart on the "." and print the third item found in it along with the columns you want.

Corona,

Thanks for your detailed information. How do I add the header?

Code:

echo "header" > filename
awk '...' >> filename