Gawk - Regexp - unable to get results - regex

Question

I have a two column file named names.csv. Field 1 has names with alphabet characters in them. I am trying to find out names where a character repeats e.g. Viijay (and not Vijay)
The command below works and returns all the rows in Field 1
gawk "$1 ~ /[a-z]/ {print $0}" names.csv
To meet the requirement stated above (viz. repeating characters), I have actually used the command below, which does not return any rows
gawk "$1 ~ /[a-z]{1,}/ {print $0}" names.csv
What is the correction needed to get what I am looking for?
To further elaborate, if the values in Column 1/Field 1 are Vijay, Viijay and Vijayini, i want only Viijay to be returned. That is, only values where a character ("i" in the example here) is repeated (not "recurring" as in Vijayini where the character "i" is recurring in the string but not clustered together.)
Requested sample data is:
Vijay 1
Viijay 2
Vijayini 3
and the expected output:
Viijay 2

Answer 1

As awk regex doesn't support backreferences in matching, you need to find the duplicated characters some other way. This one duplicates every character in $1 and adds them to a variable which is then matched against the original string in, ie. Viijay -> re="(VV|ii|ii|jj|aa|yy)"; if($1~re)... (notice, that it does not test if the entry is already in re, you might want to consider adding some checking, more checking considerations in the comments):
$ awk '
{ # you should test for empty $1
re="(" # reset re
for(i=1;i<=length($1);i++) # for each char in $1
re=re (i==1?"":"|") (b=substr($1,i,1)) b # generate dublicated re entry
re=re ")" # terminating )
if($1~re) # match
print # and print if needed
}' file
Output:
Viijay 2
Ironically or exemplarily it fails on Busybox awk—in which the backreferences can be used Ɑ:
$ busybox awk '$1~"(.)\\1" {print $0}' file
Viijay,2

Answer 2

Since awk doesn't support backreferences in a regexp you're better off using grep or sed for this:
$ grep '^[^[:space:]]*\([a-z]\)\1' file
Viijay 2
$ sed -n '/^[^[:space:]]*\([a-z]\)\1/p' file
Viijay 2
That might be GNU-only, google to check.
With awk you'd have to do something like the following to first create a regexp that matches 2 repetitions of any character in your specific character set of a-z:
$ awk '{re=$1; gsub(/[^a-z]/,"",re); gsub(/./,"&{2}|",re); sub(/\|$/,"",re)} $1 ~ re' file
Viijay 2
FYI to create a regexp from $1 that would match 2 repetitions of any character it contains, not just a-z, would be:
re=$1; gsub(/[^\\^]/,"[&]{2}|",re); gsub(/[\\^]/,"\\\\&{2}|",re); sub(/\|$/,"",re);
You have to handle ^ differently from other characters as that's the only character that has a different meaning than literal when it's the first character in a bracket expression (i.e. negation) so you have to escape it with a backslash rather than putting it inside a bracket expression to make it literal. You have to handle \ different because [\] means the same as [] which is an unterminated bracket expression because [ is the start but ] is just the first character inside the bracket expression, it's not the ] needed to terminate it.