How to extract the first two characters of a string in shell scripting?

Question

For example, given:

USCAGoleta9311734.5021-120.1287855805

I want to extract just:

US

Answer 1

perl -ple 's/^(..).*/$1/'

Answer 2

You've gotten several good answers and I'd go with the Bash builtin myself, but since you asked about sed and awk and (almost) no one else offered solutions based on them, I offer you these:

echo "USCAGoleta9311734.5021-120.1287855805" | awk '{print substr($0,0,2)}'

and

echo "USCAGoleta9311734.5021-120.1287855805" | sed 's/\(^..\).*/\1/'

The awk one ought to be fairly obvious, but here's an explanation of the sed one:

• substitute "s/"
• the group "()" of two of any characters ".." starting at the beginning of the line "^" and followed by any character "." repeated zero or more times "*" (the backslashes are needed to escape some of the special characters)
• by "/" the contents of the first (and only, in this case) group (here the backslash is a special escape referring to a matching sub-expression)
• done "/"

Answer 3

easiest way is

${string:position:length}

Where this extracts $length substring from $string at $position.

This is a bash builtin so awk or sed is not required.

Answer 4

Is this what your after?

my $string = 'USCAGoleta9311734.5021-120.1287855805';

my $first_two_chars = substr $string, 0, 2;

ref: substr

Answer 5

If your system is using a different shell (not bash), but your system has bash, then you can still use the inherent string manipulation of bash by invoking bash with a variable:

strEcho='echo ${str:0:2}' # '${str:2}' if you want to skip the first two characters and keep the rest
bash -c "str=\"$strFull\";$strEcho;"

Answer 6

Just grep:

echo 'abcdef' | grep -Po "^.."        # ab