bash - Get first line of stdin which is not in a file -

i trying write function in bash script gets lines stdin , picks out first line not contained in file.

here approach:

doubles=file.txt  firstnotdouble(){      while read input_line;              found=0;              cat $doubles |              while read double_line;                      if [ "$input_line" = "$double_line" ]                                                   found=1;                              break                      fi              done              if [ $found -eq 0 ] # no double found, echo , break!                                  echo $input_line                      break              fi      done } 

after debugging attempts realized when found set 1 in first if block, not keep value until next if block. that's why it's not working. why script act if there 2 found variables in different "scopes"?

the second question if approach whole optimized.

as indicated in comments, issue environment variables commands in pipeline (that is, series of commands separated |) run in subshells, , each subshell has own environment variables. have avoided problem avoiding uuoc (useless use of cat), writing:

while read ...; ... done < "$doubles" 

instead of pipeline.

---

a (much) faster way using while read loop repeatedly through doubles file use grep:

# specify file scanned first argument firstnotdouble() {   while ifs= read -r double_line;     if ! grep -qxf "$double_line" "$1";       echo "$double_line"       return     fi   done   return 1 } 

in grep:

• -q suppress print out, , stop on first match
• -x pattern must match entire line
• -f pattern simple string instead of regular expression.

in read:

• ifs= avoids spaces being trimmed
• -r avoids backslashes being deleted

with gnu grep, use -xf -m1 (or -xfm1 if being cryptic) instead of -qxf, , leave out echo. grep extension -m n limits number of matches found n.