How to split a field into characters in awk? - awk

Question

I want to check if two fields in a field have common characters
first second // looking for "s"
third fourth // looking for "t" or "h" or "r"
some that
I want to print only the first two lines in which the fields have at least one common character.
I read the fields as
awk 'BEGIN{FS=" ";} CONDITION {print $0}' file.txt
In the CONDITION, I need to split one field into characters and check their existence in the other field.
// for $1
BEGIN{FS=""}
{ i=0; while(i<=NF) {
CHECK in $2; i++;
} print $0 }
but I am not sure how I should bring the fields ($1 and $2) to the second part.

Answer 1

You may use this regex way:
awk 'NF>=2 {
s = $1
p = $2
gsub(/[^[:alnum:][:blank:]_]/, "\\\\&", p)
sub("[" p "]", "", s)
}
s != $1' file
first second // looking for "s"
third fourth // looking for "t" or "h" or "r"
Here we use sub function to make a bracket expression i.e. [...] where ... is $2 and then remove there characters from $1. If $1 is not equal to s (saved value of $1) then there is still one common character between first 2 fields.

Answer 2

a straightforward implementation can be
$ awk '{for(i=1;i<=length($1);i++)
if(index($2,substr($1,i,1)))
{print; next}}' file
first second
third fourth
or,
$ awk '{n=split($1,f,"");
for(i=1;i<=n;i++)
if(index($2,f[i]))
{print; next}}' file

Answer 3

$ cat tst.awk
{
delete chars
for (i=1; i<=length($1); i++) {
chars[substr($1,i,1)]
}
for (i=1; i<=length($2); i++) {
if (substr($2,i,1) in chars) {
print
next
}
}
}
$ awk -f tst.awk file
first second // looking for "s"
third fourth // looking for "t" or "h" or "r"