Extracting file name with spaces using awk

Question

I have this line of output:

M file.txt

Getting fed into this awk program:

awk '{printf("\t%s: %s %s\n", FNR, $1, $2)}'

which returns

1: M file.txt

This works and is the usual case. However, if the file has spaces and the line of output looks something like this:

M "tmux split-window subcommand.md"

awk gives give this:

1: M "tmux

But the desired output is this:

1: M "tmux split-window subcommand.md"

How can I fix this?

Answer 1

To prepend the line number to the current line using awk:

awk '{ printf("%d: %s\n", FNR, $0) }'

The zeroth field, $0, is the complete line.

Answer 2

If you use a delimiter, it will not consider the spaces in between your quotations as separate words.

So we can use any delimiter that you feel will not show up in your input string.

echo 'M "tmux split-window subcommand.md"' | awk -F: '{printf("\t%s: %s %s\n", FNR, $1, $2)}' 1: M "tmux split-window subcommand.md"

Answer 3

try this:

  #!/usr/bin/awk -f
  BEGIN {
        #FPAT = "([^,]+)|(\"[^\"]+\")" # seperator is ,
        #FPAT = "([^,]*)|(\"[^\"]*\")" # seperator is , and allow empty field -> notice the "+" changed to "*"
        FPAT = "([^ ]+)|(\"[^\"]*\")" # seperator is <space> -> notice the "," changed to " "
  }
  {
    printf("%d: ", FNR);
        for (i = 1; i <= NF; i++) {
            printf("%s ", $i)
        }
    printf("\n");
  }

I got it from GNU awk's online manual. Here is it. https://www.gnu.org/software/gawk/manual/html_node/Splitting-By-Content.html