How to 'drop'/delete characters from in front of a string?

Question

I have a string that I would like to manipulate. The string is H08W2345678 how would I be able to manipulate it so the output is just W2345678?

Similarly if the I wanted to drop the last 4 characters from H08W2345678 so that I get H08W234 how would I do this?

Answer 1

Just using bash (or ksh93 where that syntax comes from or zsh):

string="H08W2345678"

echo "${string:3}"
W2345678

echo "${string:0:-4}"
H08W234

See the Wooledge wiki for more on string manipulation.

Answer 2

$ echo "H08W2345678" | sed 's/^.\{3\}//'
W2345678

sed 's/^.\{3\}//' will find the first three characters by ^.\{3\} and replace with blank. Here ^. will match any character at the start of the string (^ indicates the start of the string) and \{3\} will match the the previous pattern exactly 3 times. So, ^.\{3\} will match the first three characters.

$ echo "H08W2345678" | sed 's/.\{4\}$//'
H08W234

Similarly, sed 's/.\{4\}$//' will replace the last four characters with blank ($ indicates the end of the string).

Answer 3

How to 'drop'/delete characters from in front of a string?

I have a string that I would like to manipulate. The string is H08W2345678 how would I be able to manipulate it so the output is just W2345678?

echo "H08W2345678" | cut -c 4-

Similarly if the I wanted to drop the last 4 characters from H08W2345678 so that I get H08W234 how would I do this?

echo "H08W2345678" | cut -c -7

Answer 4

If you have a file in which every line is an eleven-character (or whatever) string that you want to chop up, sed is the tool to use.  It’s fine for manipulating a single string, but it’s overkill.  For a single string, Jason’s answer is probably the best, if you have access to bash version 4.2 or higher.  However, the ${parameter:offset} and ${parameter:offset:length} syntaxes appear to be unique to bash (well, bash, ksh93, mksh, and zsh) — I don’t see them in The Open Group Base Specifications for Shell Command Language.  If you’re stuck with a POSIX-compliant shell that doesn’t support substring expansion (extraction), you can use

$ printf "%s\n" "${string#???}"
W2345678

$ printf "%s\n" "${string%????}"
H08W234

using printf instead of echo to guard against strings like abc-e, where, when you drop the first three characters, you are left with -e (and echo -e doesn’t do what you would want).

And, if you’re not using a Bourne-family shell at all (or you’re using an ancient, pre-POSIX system), these should still work:

$ expr " $string" : ' ...\(.*\)'
W2345678

$ expr " $string" : ' \(.*\)....'
H08W234

The extra leading space is to avoid problems with values of $string that are actual expr operators (e.g., +, /, index or match) or options (e.g., --, --help or --version).

Answer 5

With:

string="H08W2345678"

Matching 3 or 4 characters seems simple (for most shells):

$ printf '%s\t%s\n' "${string#???}" "${string%????}"
W2345678      H08W234

For the older shells (like the Bourne shell), use:

$ string=H08W2345678

$ expr " ${string}" : " ...\(.*\)"
W2345678

$ expr " ${string}" : " \(.*\)...." '
H08W234

If it is needed a numeric count of characters, use:

$ expr " ${string}" : " .\{3\}\(.*\)"
W2345678

$ expr " ${string}" : " \(.*\).\{4\}" '
H08W234

Of course, those regex work also with sed, awk and bash 3.0+:

$ echo "$string" | sed 's/^.\{3\}//'
W2345678

$ echo "$string" | sed 's/.\{4\}$//'
H08W234

$ echo "$string" | awk '{sub(/^.{3}/,"")}1'
W2345678

$ echo "$string" | awk '{sub(/.{4}$/,"")}1'
H08W234

$ r='^.{3}(.*)$'; [[ $a =~ $r ]] && echo "${BASH_REMATCH[1]}"
W2345678

$ r='^(.*).{4}$'; [[ $a =~ $r ]] && echo "${BASH_REMATCH[1]}"
H08W234