read first 8 characters of text file with bash

Question

I would like to read only the first 8 characters of a text file and save it to a variable in bash. Is there a way to do this using just bash?

score 72 · Accepted Answer · answered Jan 16 '13 at 17:47

72

You can ask head to read a number of bytes. For your particular case:

$ head -c 8 <file>

Or in a variable:

foo=$(head -c 8 <file>)

answered Jan 16 '13 at 17:47

gpoo

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score 7 · Answer 2 · answered Jan 16 '13 at 17:47

7

in bash

help read

you'll see that you can :

read -r -n 8 variable < .the/file

If you want to read the first 8, independent of the separators,

IFS= read -r -n 8 variable < .the/file

But avoid using

.... | while IFS= read -r -n 8 variable

as, in bash, the parts after a "|" are run in a subshell: "variable" would only be changed in that subshell, and it's new value lost when returing to the present shell.

answered Jan 16 '13 at 17:47

Olivier Dulac

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the last command set `IFS` to `''` just for the duration of the following command (here, the `read`) – Olivier Dulac Jan 16 '13 at 17:48
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Version 4 of `bash` introduces the `-N` option, which acts as `-n` but also ignores delimiters. – chepner Jan 16 '13 at 18:34

score 0 · Answer 3 · answered Jan 16 '13 at 17:45

0

You could use an array in bash and select only first characters. Advanced Bash Scripting guide has good examples how to use arrays.

answered Jan 16 '13 at 17:45

LetMeSOThat4U

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This is really only half an answer. How would you select just the first 8 characters from the array? – chepner Jan 16 '13 at 18:34
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echo ${arrayZ[@]:1:2} # two three It's right there in ABS. Granted, it's somewhat tedious. Alternative is to use loop. BTW solution above using "head" is not using bash builtins only. "head" is an external program. – LetMeSOThat4U Jan 16 '13 at 18:46

read first 8 characters of text file with bash

3 Answers3