AWK - error in script with count variable OR error in AWK Implementation / version?

Question

• Awk version: GNU Awk 3.1.1
• Platform: OpenVMS 8.4-L1
• Awk knowledge: limited

Script (Awk one liner). NOTE: OpenVMS parameter parsing requires quotes

gawk "-F" ";" "BEGIN {x = 0} ; x++ {print $0,$1"";""x}" in.file > out.file

**in.file**
file.log;2000
file.log;1999
file.log;1998
file.log;1997

**out.file**
file.log;1999 file.log;2
file.log;1998 file.log;3
file.log;1997 file.log;4

I wanted the following instead

file.log;2000 file.log;1
file.log;1999 file.log;2
file.log;1998 file.log;3
file.log;1997 file.log;4

If I change the awk command above to: ...{x = -1}... I get

**out.file**
file.log;2000 file.log;0
file.log;1998 file.log;2
file.log;1997 file.log;3

What I am wondering is

1. Can someone else using a different platform test this to see if it produces the same output?
2. Could it be the awk version I am running that is at fault OR is there something wrong with my awk script that I am not understanding

It really seems to me that it should produce a file.log;1999 file.log;1 line but it doesn't. I'm at a loss and need some pointers / awk education

Thanks in advance :-)

Answer 1

You need to use ++x in place of x++. The x++ initially evaluates to 0 (but increments x to 1), and 0 is false, so the action is not taken; nothing is printed for the first line.

When you initialize x to -1, the first line is printed because x++ evaluates to -1 which is true; the second is skipped because x is 0, etc.

You don't need the BEGIN block. Variables are auto-vivified with zero (or the empty string) as the value. And in fact you don't need to trigger the printing based on whether x is zero or not; you simply want to always print an incremented value of x.

So, on a Mac (Unix-like) system, I can run:

$ gawk "-F" ";" '{print $0,$1";"++x}' file.in
file.log;2000 file.log;1
file.log;1999 file.log;2
file.log;1998 file.log;3
file.log;1997 file.log;4
$

Translated to VMS conventions, that would be:

gawk "-F" ";" "{print $0,$1"";""++x}" file.in

Answer 2

For understanding what happened, you can put additional output.

$ gawk -F\; 'BEGIN {x = 0} {print "step 1: "x} x++ {print "step 2: "x; print $0,$1";"x} {print "step 3: "x"\n"}' in.file
step 1: 0
step 3: 1

step 1: 1
step 2: 2
file.log;1999 file.log;2
step 3: 2

step 1: 2
step 2: 3
file.log;1998 file.log;3
step 3: 3

step 1: 3
step 2: 4
file.log;1997 file.log;4
step 3: 4

So there is the effect of the post-increment.

x {something} is same as {if(x){something}}. So value of x will be use as condition for calling block {something}.

On first line, on step 1, x++ will return 0 and increment 'x', which will skip step 2. But on step 3 'x' will be 1.

On second line, on step 1, x++ will return 1 and increment 'x'. On step 2, x will be already '2'. That's what you get.

You can fix it like this:

gawk -F\; 'BEGIN {x = 0} {print $0,$1";"++x}' in.file

Answer 3

"";"" is the problem. Try. 'BEGIN ... $1";"x ... instead. What you posted amounts to. print $0,$1; x} I.e. throw x awayt. Hth

Answer 4

Since you don't know AWK too well, may I suggest to skip it altogether and switch to PERL?

But first things first. What problem are you really trying to solve? Seems to me you obtained a list of file names (DIR/COL=2/OUT=x.x) You want to use that list to generate a rename where the file with the highest version number becomes number 1, the next down 2 and so on. Is that correct?

I hope there no need to worry about overlap perhaps thanks to a version limit in place.

DCL defaults can do that just fine. - Using PERL here just as a handy way to create a bunch of files - Using FILE_ID to show which file is which Here it goes.

$perl -e "open X,qq(>file.log;$_) for 1997..2000"
$dir/file
FILE.LOG;2000        (59705,105,0)
FILE.LOG;1999        (46771,399,0)
FILE.LOG;1998        (42897,980,0)
FILE.LOG;1997        (24538,519,0)
$rena/log file.log.* tmp.log;
%RENAME-I-RENAMED, FILE.LOG;2000 renamed to TMP.LOG;1
%RENAME-I-RENAMED, FILE.LOG;1999 renamed to TMP.LOG;2
%RENAME-I-RENAMED, FILE.LOG;1998 renamed to TMP.LOG;3
%RENAME-I-RENAMED, FILE.LOG;1997 renamed to TMP.LOG;4
$rena tmp.log;* file.log/log
%RENAME-I-RENAMED, TMP.LOG;4 renamed to FILE.LOG;4
%RENAME-I-RENAMED, TMP.LOG;3 renamed to FILE.LOG;3
%RENAME-I-RENAMED, TMP.LOG;2 renamed to FILE.LOG;2
%RENAME-I-RENAMED, TMP.LOG;1 renamed to FILE.LOG;1
$dir/file file.log;*
FILE.LOG;4           (24538,519,0)
FILE.LOG;3           (42897,980,0)
FILE.LOG;2           (46771,399,0)
FILE.LOG;1           (59705,105,0)

ok? No helpers needed. Just two commands relying on the ";" magic to stop inheritance.

Now let's see how to do this in Perl directly.

$ dir/file
FILE.LOG;2000        (59705,105,0)
FILE.LOG;1999        (46771,399,0)
FILE.LOG;1998        (42897,980,0)
FILE.LOG;1997        (24538,519,0)
$ perl -e "for (<file.log;*>) {$i++; $old = $_; s/;\d+/;$i/; rename $old, $_}"
$ dir/file
FILE.LOG;4           (24538,522,0)
FILE.LOG;3           (42897,983,0)
FILE.LOG;2           (46771,402,0)
FILE.LOG;1           (59705,108,0)

Broken down in steps:

<xxx> = glob("xxx") = glob(qq(xxx) = wildcard lookup with interpolated string.
for (<file.log;*>)        # Loop over all files matching pattern, output into $_
{                         # code block
$i++;                     # increment for each iteration. start at 1
$old = $_;                # save the fetched file name
s/;\d+/;$i/;              # substitute a semicolon followed by numbers 
rename $old, $_           # Actual rename. Try with PRINT first.
}                         # end of code block

ok?