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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?
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Using the head Command The head command is used to display the first lines of a file. By default, the head command will print only the first 10 lines. The head command ships with the coreutils package, which might be already installed on our machine.
To access the first n characters of a string, we can use the (substring) parameter expansion syntax ${str:position:length} in the Bash shell. position: The starting position of a string extraction. length: The number of characters we need to extract from a string.
So as far as I can tell, %% doesn't have any special meaning in a bash function name. It would be just like using XX instead. This is despite the definition of a name in the manpage: name A word consisting only of alphanumeric characters and under- scores, and beginning with an alphabetic character or an under- score.
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>) 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.
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