How to get first n characters of each line in unix data file
With cut
:
$ cut -c-22 file000000000001199998000100000000000219999800010000000000031999980001000000000004199998000100000000000519999800010000000000061999980001
If I understand the second requirement you want to split the first 22 characters into two columns of length 10 and 12. sed
is the best choice for this:
$ sed -r 's/(.{10})(.{12}).*/\1 \2/' file0000000000 0119999800010000000000 0219999800010000000000 0319999800010000000000 0419999800010000000000 0519999800010000000000 061999980001
sudo_O has provided nice cut and sed solution, I just added an awk one-liner:
awk 'BEGIN{FIELDWIDTHS="22"} {print $1}' fileecho "000000000001199998000180000 DUMMY RAG"|awk 'BEGIN{FIELDWIDTHS="22"} {print $1}'0000000000011999980001
with empty char (it depends on your requirement, you want to skip the spaces or you want to include and count them in your output)
if blank spaces should be counted and displayed in output as well: (you don't have to change the cmd above)
echo "0 0 0 0 00000001199998000180000"|awk 'BEGIN{FIELDWIDTHS="22"} {print $1}' 0 0 0 0 00000001199998
if you want to skip those spaces: (quick and dirty)
echo "0 0 0 0 00000001199998000180000"|sed 's/ //g'|awk 'BEGIN{FIELDWIDTHS="22"} {print $1}' 0000000000011999980001
This can actually be done in Bash without using any external programs (scripts using this must start with #!/bin/bash
instead of #!/bin/sh
and will not be POSIX shell compliant) using the expression ${VARIABLE:offset:length}
(where :length
is optional):
#!/bin/bashSTR="123456789"echo ${STR:0:1}echo ${STR:0:5}echo ${STR:0:10}echo ${STR:5:10}echo ${STR:8:10}
will have this output:
11234512345678967899
Note that the start offset begins at zero and the length must be at least one. You can also offset from the right side of the string using a negative offset in parentheses:
echo ${STR:(-5):4}5678
An extremely useful resource for all you'll need to know about Bash string manipulation is here: https://tldp.org/LDP/abs/html/string-manipulation.html