get the first 5 characters from each line in shell script

Question

Here is my sample.txt file it contains following

31113    70:54:D2 - a-31003 31114    70:54:D2 - b-31304 31111    4C:72:B9 - c-31303 31112    4C:72:B9 - d-31302 

I have to write the shell script in that I am passing first 5 characters (eg 31113) as input id to other script. For this I have tried this

#!/bin/sh filename='sample.txt' filelines=`cat $filename` while read -r line do   id= cut -c-5 $line   echo $id   #code for passing id to other script file as parameter done < "$filename" 

but it is not working this gives me error as

cut: 31113: No such file or directory cut: 70:54:D2 No such file or directory 31114 31111 31112 : No such file or directory 

How can I do this?

Answer 1

If you want to use cut this way, you need to use redirection <<< (a here string) like:

var=$(cut -c-5 <<< "$line") 

Note the use of var=$(command) expression instead of id= cut -c-5 $line. This is the way to save the command into a variable.

Also, use /bin/bash instead of /bin/sh to have it working.

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Full code that is working to me:

#!/bin/bash  filename='sample.txt' while read -r line do   id=$(cut -c-5 <<< "$line")   echo $id   #code for passing id to other script file as parameter done < "$filename" 

Answer 2

Well, its a one-liner cut -c-5 sample.txt. Example:

$ cut -c-5 sample.txt  31113 31114 31111 31112 

From there-on, you can pipe it to any other script or command:

$ cut -c-5 sample.txt | while read line; do echo Hello $line; done Hello 31113 Hello 31114 Hello 31111 Hello 31112