AWK拆分为多个分隔符行

Question

I'm trying to split a file using AWK one-line but the code below that I came with is not working properly.

awk '
BEGIN { idx=0; file="original_file.split." }
/^REC_DELIMITER.(HIGH|TOP)$/ { idx++ }
/^REC_DELIMITER.TOP$/,/^REC_DELIMITER.(HIGH|TOP)$/ { print > file sprintf("%03d", idx) }
' original_file

Test file is "original_file":

REC_DELIMITER.TOP
lineA1
lineA2
lineA3
REC_DELIMITER.HIGH
lineB1
lineB2
lineB3
REC_DELIMITER.TOP
lineC1
lineC2
lineC3
REC_DELIMITER.HIGH
lineD1
lineD2
lineD3

AWK code above is for REC_DELIMITER.TOP and it is giving me these files:

original_file.split.001:
REC_DELIMITER.TOP

original_file.split.003:
REC_DELIMITER.TOP

however, I'm trying to get this:

original_file.split.001:
REC_DELIMITER.TOP
lineA1
lineA2
lineA3

original_file.split.003:
REC_DELIMITER.TOP
lineC1
lineC2
lineC3

There will be other record delimiters, and when needed, we can run for them like REC_DELIMITER.HIGH, this way getting files like below:

original_file.split.002:
REC_DELIMITER.HIGH
lineB1
lineB2
lineB3

original_file.split.004:
REC_DELIMITER.HIGH
lineD1
lineD2
lineD3

Any help guys is very appreciate, I have been trying to get this working past few days and AWK code above is the best I was able to get. I need now help from AWK masters. :)

Thank you!

Answer 1

You can try something like this:

awk '
/REC_DELIMITER\.TOP/ {
    a=1
    b=0
    file = sprintf (FILENAME".split.%03d",++n)
}    
/REC_DELIMITER\.HIGH/ {
    b=1
    a=0
    file = sprintf (FILENAME".split.%03d",++n)
}  
a {
    print $0 > file
}    
b {
    print $0 > file
}' file

Answer 2

You need something like this (untested):

awk -v dtype="TOP" '
BEGIN { dbase = "^REC_DELIMITER\\."; delim = dbase dtype "$" }
$0 ~ dbase { inBlock=0 }
$0 ~ delim { inBlock=1; idx++ }
inBlock { print > sprintf("original_file.split.%03d", idx) }
' original_file

Answer 3

awk -vRS=REC_DELIMITER '/^.TOP\n/{print RS $0 > sprintf("original_file.split.%03d",n)};!++n' original_file

(Give or take an extra newline at the end.)

Generally, when input is supposed to be treated as a series of multi-line records with a special line as delimiter, the most direct approach is to set RS (and often ORS) to that delimiter.

Normally you'd want to add newlines to its beginning and/or end, but this case is a little special so it's easier without them.

Edited to add: You need GNU Awk for this. Standard Awk considers only the first character of RS.

Answer 4

I made some changes so the different delimiters go to the their own file, even when they occur later in the file. make a file like splitter.awk with the contents below, the chmod +x it and run it with ./splitter.awk original_file

#!/usr/bin/awk -f
BEGIN {
  idx=0;
  file="original_file.split.";
  out=""
}
{
  if($0 ~ /^REC_DELIMITER.(TOP|HIGH)/){
    if (!cnt[$0]) {
      cnt[$0] = ++idx;
    }
    out=cnt[$0];
  }
  print >  file sprintf("%03d", out)
}

Answer 5

I'm not very used to AWK, however, plasticide's answer put me towards right direction and I finally got AWK script working as requirements.

In below code, first IF turn echo to 0 if a demilier is found. Second IF turn echo to 1 if the wanted delimiter is found, then the want ones are are split from file.

I know regex could be something like /^(REC_(DELIMITER\\.(TOP|HIGH|LOW)|NO_CATEGORY)$/ but since regex is created dynamically via shellscript that reads from an specific file a list of delimiters, it will look more like in AWK below.

awk 'BEGIN {
  idx=0; echo=1; file="original_file.split."
}
{
  #All the delimiters to consider in given file
  if($0 ~ /^(REC_DELIMITER.TOP|REC_DELIMITER.HIGH|REC_DELIMITER.LOW|REC_NO_CATEGORY)$/) {
    echo=0
  }
  #Delimiters that should actually be pulled
  if($0 ~ /^(REC_DELIMITER.HIGH|REC_DELIMITER.LOW)$/ {
    idx++; echo=1
  }
  #Print to a file is match wanted delimmiter
  if(echo) {
    print > file idx
  }
}' original_file

Thank you all. I really appreciate it very much.