Iterate through array of strings applying a Regex to each element in Korn Shell

Question

I have an array of strings called names containing names with some subsequent garbage data. Like this

Jill Shortz, City Contractor, America
Bill Torts, Family Doctor, Canada
Will Courtz, Folk DJ, Bulgaria
Phil-Lip Warts, Juggler, India

I want to iterate through names extracting only the first two words with the regex (^\w+-*( *\w+)*) and overwriting them back into names so it will contain

Jill Shortz
Bill Torts
Will Courtz
Phil-Lip Warts

this is how I attempted it but my AIX machine does not like the -P argument to execute in Perl mode

for((i=0;i<${#names[@]};++i)); do
        names[$i]=`grep -P '(^\w+-*( *\w+)*)' -o <<<"${names[i]}"`
done

That looks like a CSV; could you strip everything from the first comma onwards, instead? — Jeff Schaller
– Jeff Schaller ♦, Commented Nov 5, 2019 at 16:33
... which, at least with AT&T ksh93 on Linux, appears to be possible using parameter expansion ex. names[$i]="${names[$i]%%,*}" — steeldriver
– steeldriver, Commented Nov 5, 2019 at 16:37
Thank you for both of your comments @steeldriver this has accomplished what I was intending to do. I am however still curious as to how I would accomplish string substitution with a regex in ksh so I'm going to keep the question open for the time being. — CyberStems
– CyberStems, Commented Nov 5, 2019 at 17:42
@CyberStems you could probably use basic regular expression ^[^,]* (a - possibly empty - sequence of non-, characters, anchored to the start) — steeldriver
– steeldriver, Commented Nov 5, 2019 at 17:48
@steeldriver this regex is exactly what I need and appears more efficient than mine, my problem is actually having that regex applied to my string and then assigning that value back into the array. Sorry for poor wording earlier as English is not my first language. — CyberStems
– CyberStems, Commented Nov 5, 2019 at 17:58

Kusalananda · Accepted Answer · 2019-11-05 19:37:22Z

I don't really see a use for regular expressions at all to complete this task, assuming you just want to remove everything after the first comma.

names=( "${names[@]%%,*}" )
printf '"%s"\n' "${names[@]}"

This is removes the first comma and everything after it from each array element individually (literally "the longest suffix matching the globbing pattern ,*"). The resulting list of modified names is then re-assigned to the names array (and printed with printf).

The code, given your initial initialisation of the array to the list in the question, would generate

"Jill Shortz"
"Bill Torts"
"Will Courtz"
"Phil-Lip Warts"

(the double quotes are added by the printf format string).

The code would work with ksh93, bash, zsh, and yash.

glenn jackman · Accepted Answer · 2019-11-05 20:17:16Z

I don't see anywhere in the ksh man page that you can match a string against a regular expression, and use capturing parentheses to extract substrings (like you would do in bash with

[[ $str =~ ^([[:alnum:]]+([ -]+[[:alnum:]]+)+) ]] && echo "${BASH_REMATCH[1]}"

However, you can use extended regular expressions in glob patterns with ~(E:regex), so you can do this:

for n in "${names[@]}"; do
  # remove the pattern from the start of the string
  tmp=${n##~(E:\w+([ -]+\w+)*)}
  # and then remove what remained from the end of the string
  echo "[${n%$tmp}]"
done

[Jill Shortz]
[Bill Torts]
[Will Courtz]
[Phil-Lip Warts]

... and for maximum write-only unreadability

for n in "${names[@]}"; do
  echo "${n%${n##~(E:\w+([ -]+\w+)*)}}"
done

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