CODEWITHSUNDEEP
awksedcutash
Deepali Joshi
Deepali Joshi

Reputation: 73

Extracting beginning alphabets of a string in bash/ash

How can I extract the beginning alphabetic letters from a string? I want to extract alphabets occurring in the beginning before I hit the first non-alphabetic character.

e.g. If the input string is abcd045tj56 the output should be abcd

Similarly, if the input is jkl657890 the output should be jkl

Can it be done in shell script using awk/sed/cut?

I tried

echo "XYZ123" | awk 'sub(/[[:alpha:]]*/, "")'

But it gives 123 instead of xyz

Then I tried

echo "XYZ123" | awk '{print (/[[:alpha:]]*/)}'

but it gives 1

I want the answer to be XYZ

Upvotes: 4

Views: 217

Answers (7)

blhsing
blhsing

Reputation: 106995

You can use bash's parameter expansion to remove the first non-alphabet and all characters after it:

s=XYZ123
echo ${s%%[^[:alpha:]]*}

Demo: https://onlinegdb.com/OzjGf53T-

Note that this approach has the performance benefit of avoiding the overhead of spawning a separate process.

Upvotes: 1

The fourth bird
The fourth bird

Reputation: 163477

Using gnu awk you can print the first 1 or more alphabetic letters:

echo "XYZ123" | awk 'match($0, /[[:alpha:]]+/, a) {print a[0]}'

Output

XYZ

If there should be at least a single a non alphabetic character following, you can use a capture group and print that value:

echo "XYZ123" | awk 'match($0, /([[:alpha:]]+)[^[:alpha:]]/, a) {print a[1]}'

Upvotes: 1

Daweo
Daweo

Reputation: 36680

I tried

echo "XYZ123" | awk 'sub(/[[:alpha:]]*/, "")'

But it gives 123 instead of xyz

You instructed GNU AWK to replace zero-or-more alphabetic characters using empty string, if you wish to do this task using sub select non-alpha character followed by zero-or-more any characters, namely

echo "XYZ123" | awk '{sub(/[^[:alpha:]].*/, "");print}'

gives output

XYZ

(tested in GNU Awk 5.1.0)

Upvotes: 1

blhsing
blhsing

Reputation: 106995

You can use awk with a non-alphabet as the field separator so you can get the leading alphabets by printing the first field:

awk -F'[^[:alpha:]]' '{print $1}'

Demo: https://awk.js.org/?snippet=g7eajb

Upvotes: 1

RavinderSingh13
RavinderSingh13

Reputation: 133680

Converting my comment to an answer here. Using any awk version.

awk '
match($0,/^[a-zA-Z]+/){
  print substr($0,RSTART,RLENGTH)
}
' Input_file

OR:

awk '
match($0, /[^[:alpha:]]/){
  print substr($0, 1, RSTART-1)
}
' Input_file

Upvotes: 5

anubhava
anubhava

Reputation: 785721

You may use this sed:

sed 's/[^[:alpha:]].*$//'

This sed matches a non-alpha character and everything afterwards and substitutes with an empty string.

Examples:

sed 's/[^[:alpha:]].*$//' <<< 'abcd045tj56'
abcd

sed 's/[^[:alpha:]].*$//' <<< 'XYZ123'
XYZ

sed 's/[^[:alpha:]].*$//' <<< 'jkl657890'
jkl

If you want to do this in bash then:

s='abcd045tj56'
echo "${s/[^[:alpha:]]*}"

abcd

Upvotes: 4

Paolo
Paolo

Reputation: 26211

Use grep:

$ grep -Eo '^[A-Za-z]+' <<<"XYZ123"

to only match alphabetic letters at the beginning of the string.

Upvotes: 2

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