mjhjmtu

I have a big text file I want to print only the first 4 and the first 5 and the first 8 characters of each line in one command line.

For example
I have the lines:

123456789ab
ABCdefgih55

So the output have to be:

1234
ABCd
12345
ABCde
12345678
ABCdefgh

for len in 4 5 8; do
 cut -c "1-$len" file
done

This uses cut -c repeatedly to cut out the first part of each line of the file called file. The length of the cut out bit is depending on the loop variable len.

If you're strict about that "one line" criteria:

for len in 4 5 8; do cut -c "1-$len" file; done

Or, as an easy to use shell function:

cut_to_lengths () 
 file=$1; shift
 for len do
 cut -c "1-$len" "$file"
 done

Using it:

$ cut_to_lengths file 4 5 8 1
1234
ABCd
12345
ABCde
12345678
ABCdefgi
1
A

In comments you specify that you don't want to output lines if they are shorter the cut length.

To do this, we can change the cut command into an awk command:

awk -v len="$len" 'length >= len print substr($0, 1, len) '

Replace the cut -c "1-$len" with the above awk command in the code above.

Your example is not well formated. I guess the whitespace should be a new line? If so the simplest way is to use cut

$ cut -c -4 input will print the first 4 character of each line. Repeat it with any number of characters you like.

try the following

for line in `cat input`; do echo $line | cut -c 1-4 && echo $line | cut -c 1-5 && echo $line | cut -c 1-8; done

this gives

1234
12345
12345678
ABCd
ABCde
ABCdefgi