How to print lines where it matchs specific number in bash?

Question

I have a text file called file.txt like below,

01_ABC_0000  AA
02_CDE_0000  BB
03_EFG_0000  CC
04_ABC_0001  DD
05_CDE_0001  EE
06_EFG_0001  FF

where it should separated into two different files, like file0.txt

01_ABC_0000  AA
02_CDE_0000  BB
03_EFG_0000  CC

and file1.txt

04_ABC_0001  DD
05_CDE_0001  EE
06_EFG_0001  FF

what i have been trying, cat file00.txt | awk '{print $1}' | sed 's/.*\(....\)/\1/') to get the only numbers from first word but I am not able use this to go forward separating it into the two files.

Any help is much appreciated.

EDIT: Sorry, I have edited the question where the first field is 01_ABC_0000 something like this, If I use field separator as underscore it doesn't work as expected.

Answer 1

1st solution: Considering your entries are sorted with values of 2nd column(0000, 00001 and so on). With your shown samples, please try following awk program.

awk -v count="1" -F'_| +' '
prev!=$2{
  count++
  close(outputFile)
  outputFile=("file"count".txt")
  prev=$2
}
{
  print > (outputFile)
}
'  Input_file

---

---

2nd solution: Using sort + awk combination solution in case entries are not sorted.

awk -F'_| +' '{print $2,$0}' Input_file | 
sort -nk1                               | 
awk -v count="1" -F'_| +' '
{
  sub(/^[^[:space:]]+[[:space:]]+/,"")
}
prev!=$2{
  count++
  close(outputFile)
  outputFile=("file"count".txt")
  prev=$2
}
{
  print > (outputFile)
}
'