Using an if block in awk

user_3058494 0
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I'm processing a file in awk.

I want to pass along the rows in the file that have blanks in column positions 25 through 34 and I want to do work on the rows that have blanks in column positions 10 through 19. Specifically I want to replace the blanks in columns positions 10 through 19 with 0s. That way the output file will have the original rows with blanks in 25-34 untouched. and the rows with blanks in 10-19 with have been replaced with '0's. So the output file will be the same as the input file only with zeros in the relevant rows in positions 10-19. The file looks like this:

###########################################
#########          ########################
###########################################
###########################################
###########################################
###########################################
###########################################
###########################################
#########          #####          #########
###########################################
###########################################
###########################################

I know I have to use an if block but I've never used one before in awk. The syntax below is what I think I need but please help me with the details. Specifically what I'm using to specify 'blanks' in the if statements.

I apologize ahead of time for the bad syntax. This is my first time using an If block in awk. I know the syntax doesn't work, which is one of the reasons I'm posting this.

cat scr2 | awk 'BEGIN {
    pos1=substr($0,25,10); 
    pos2=substr($0,10,10);

      if (pos1 = ^[[:blank:]]$) 
         printf $0 
      else if (pos2 == ^[[:blank:]]$)
         {val=substr($0,25,10)} 
         gsub(/ /,0,val){$0=substr($0,1,24) val substr($0,35)} 1}'`

The sample output would be :

###########################################
#########0000000000########################
###########################################
###########################################
###########################################
###########################################
###########################################
###########################################
#########          #####          #########
###########################################
###########################################
###########################################

So the row with blanks only at positions 10-19 gets changed and the row with blanks at both 10-19 and 25-34 get left alone.

awk linux
4 Answers

With your shown samples, please try following awk code, written and tested in GNU awk, should work in any awk.

awk '
substr($0,10,10) ~ /^ +$/ && substr($0,20) !~ / / {
  $0=substr($0,1,9) "0000000000" substr($0,20)
}
1
' Input_file

Explanation: Simple explanation would be, checking 2 conditions in main program of awk. 1st to make sure position 10th to 20th contains only space AND 2nd rest of the line's values are NOT having spaces in it, if this is the case then enter zeroes in place of spaces and print edited/non-edited lines.

user_58665800

Another option is using match() to fill the RSTART built-in variable specifying the start of the block of spaces. You can then use substr() in a regex comparison to verify the remainder of the line is comprised only of '#' characters. For example:

awk '{if (match($0,/[ ]{10}/) && RSTART == 10 && substr($0,20) ~ /^#*$/) sub(/[ ]{10}/,"0000000000")}1' file

The above will match() each line with 10-spaces beginning at column 10 and replace them with 10 '0's.

Example Use/Output

With your input in the file named file, you would have:

$ awk '{if (match($0,/[ ]{10}/) && RSTART == 10 && substr($0,20) ~ /^#*$/) sub(/[ ]{10}/,"0000000000")}1' lines
###########################################
#########0000000000########################
###########################################
###########################################
###########################################
###########################################
###########################################
###########################################
#########          #####          #########
###########################################
###########################################
###########################################
user_34221020

I'd use sed here:

sed -E 's/^(.{9}) {10}(.{5}[^ ]{10})/\10000000000\2/' file
user_75520

You can do it in awk without any if statement:

awk '{print gensub(/^(.{9}) {10}([^ ]{24})/, "\\10000000000\\2", "g")}' file

This will replace 10 blanks by 10 0 in positions 10 to 19 only on the lines where there are no blanks in positions 20 to 43, which is what you want, I guess.

user_17827820
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