Splitting file based on first column's first character and length

Question

I want to split a .txt into two, with one file having all lines where the first column's first character is "A" and the total of characters in the first column is 6, while the other file has all the rest. Searching led me to find the awk command and ways to separate files based on the first character, but I couldn't find any way to separate it based on column length.

I'm not familiar with awk, so what I tried (to no avail) was awk -F '|' '$1 == "A*****" {print > ("BeginsWithA.txt"); next} {print > ("Rest.txt")}' FileToSplit.txt.

Any help or pointers to the right direction would be very appreciated.

EDIT: As RavinderSingh13 reminded, it would be best for me to put some samples/examples of input and expected output.

So, here's an input example:

#FileToSplit.txt#
2134|Line 1|Stuff 1
31516784|Line 2|Stuff 2
A35646|Line 3|Stuff 3
641|Line 4|Stuff 4
A48029|Line 5|Stuff 5
A32100|Line 6|Stuff 6
413|Line 7|Stuff 7

What the expected output is:

#BeginsWith6.txt#
A35646|Line 3|Stuff 3
A48029|Line 5|Stuff 5
A32100|Line 6|Stuff 6

#Rest.txt#
2134|Line 1|Stuff 1
31516784|Line 2|Stuff 2
641|Line 4|Stuff 4
413|Line 7|Stuff 7

Answer 1

What you want to do is use a regex and length function. You don't show your input, so I will leave it to you to set the field separator. Given your description, you could do:

awk '/^A/ && length($1) == 6 { print > "file_a.txt"; next } { print > "file_b.txt" }' file

Which would take the information in file and if the first field begins with "A" and is 6 characters in length, the record is written to file_a.txt, otherwise the record is written to file_b.txt (adjust names as needed)

Answer 2

A non-regex awk solution:

awk -F'|' '{print $0>(index($1,"A")==1 && length($1)==6 ? "file_a.txt" : "file_b.txt")}' file