awk to capture element upto space or using special escape character - awk

Question

Trying to extract the 5th element in $1 after the - upto the space or \\. If a / was used then the script awk -F'[-/'] 'NR==0{print; next} {print $0"\t""\t"$5}' file works as expected. Thank you :).
file --tab-delimited--
00-0000-L-F-Male \\path\to xxx xxx
00-0001-L-F-Female \\path\to xxx xxx
desired (last field has two tabs before)
00-0000-L-F-Male \\path\to xxx xxx Male
00-0001-L-F-Female \\path\to xxx xxx Female
awk
awk -F'-[[:space:]][[:space:]]+' 'NR==0{print; next} {print $0"\t""\t"$5}' file
00-0000-L-F-Male \\path\to xxx xxx
00-0001-L-F-Female \\path\to xxx xxx
awk 2
awk -F'[-\\]' 'NR==0{print; next} {print $0"\t""\t"$5}' file
awk: fatal: Unmatched [ or [^: /[-\]/

Answer 1

Using any awk:
$ awk -F'[-\t]' -v OFS='\t\t' '{print $0, $5}' file
00-0000-L-F-Male \\path\to xxx xxx Male
00-0001-L-F-Female \\path\to xxx xxx Female
Regarding your scripts:
awk
awk -F'-[[:space:]][[:space:]]+' 'NR==0{print; next} {print $0"\t""\t"$5}' file
-F'-[[:space:]][[:space:]]+' says that your fields are separated by a - followed by 2 or more spaces, which they aren't.
NR==0{foo} says "do foo for line number 0" but there is no line number 0 in any input.
awk 2
awk -F'[-\\]' 'NR==0{print; next} {print $0"\t""\t"$5}' file
-F'[-\\]' appears to be trying to set FS to a minus sign or a backslash, but you already told us your fields are tab-separated, not backslash-separated.
When setting FS this way it goes through a few different phases of interpretation, converting a shell string to an awk string, converting an awk string to a regexp, and using the regexp as a field separator, so you need several layers of escaping, not just 1, to make a backslash literal. If unsure, keep adding backslashes until the warnings and errors go away.

Answer 2

You may use this awk:
awk -F'\t' '{n=split($1, a, /-/); print $0 FS FS a[(n > 4 ? 5 : n)]}' file
00-0000-L-F-Male \\path\to xxx xxx Male
00-0001-L-F-Female \\path\to xxx xxx Female
a[(n > 4 ? 5 : n)] expression gets 5th element from array if there are 5 or more elements in array otherwise it gets last element.

Answer 3

Presuming your file is '\t' separated with one-tab per field and you want an empty field before the Male/Female output, you can use:
awk -F"\t" '{ split($1,arr,"-"); print $0 "\t\t" arr[5] }' filetabs.txt
Example Use/Output
Where filetabs.txt contains your sample data with tab field-separators you would get:
$ awk -F"\t" '{ split($1,arr,"-"); print $0 "\t\t" arr[5] }' filetabs.txt
00-0000-L-F-Male \\path\to xxx xxx Male
00-0001-L-F-Female \\path\to xxx xxx Female

Answer 4

With perl one liner which supports lazy match we can try following code. Written and tested in shown samples only.
perl -pe 's/^((?:.*?-)+)([^[:space:]]+)([[:space:]]+.*)$/\1\2\3\t\t\2/' Input_file
OR above could be written as following also:
perl -pe 's/^((?:.*?-)+)(\S+)(\s+.*)$/\1\2\3\t\t\2/' Input_file
Explanation: Adding detailed explanation for used regex above. Here is the Online Demo for used regex in code.
^( ##From starting of the value creating one capturing group here.
(?: ##Opening non-capturing group here.
.*?- ##Using lazy match till - here.
)+ ##Closing non-capturing group here with matching 1 OR more occurrences of this.
) ##Closing 1st capturing group here.
([^[:space:]]+) ##Creating 2nd capturing group and matching all non-spaces in it.
([[:space:]]+.*)$ ##Creating 3rd capturing group which matches spaces till end of the value.