Lum – How to sed output be formatted like printf’s formatted printing

columnssedtext processing

Can sed replace text with a string formatted like printf's formatted printing?

The following sed command replaces a line starting with the current value of "$domain" with several values specified in variables.

/bin/sed  "s/\(^${domain} *${limittype} * ${limititem}.*\)/$EXPL#\1\n${domain} ${limittype} ${limititem} ${value}/" /etc/security/limits.conf

However the output will not be properly aligned because the length of the values of domain etc. are not the same.

So the output would be something like

#oracle   hard   nproc    131072
oracle hard nproc 666

While valid, it is difficult to read. I would prefer to get something like

#oracle   hard   nproc    131072
oracle   hard   nproc    666

The best I can come with to get the desired output is:

/bin/sed  "s/\(^${domain}\)\( *\)\(${limittype}\)\( *\)\(${limititem}\)\( *\)\(.*\)/$EXPL#\1\2\3\4\5\6\7\n${domain}\2${limittype}\4${limititem}\6${value}/" /etc/security/limits.conf

But I believe there must a more elegant way to do this.

The sed one liners document contains some examples that use a specified number of characters, e.g.

sed -e :a -e 's/^.\{1,78\}$/ &/;ta'  # set at 78 plus 1 space

But this is in the regexp section not in the replacement section.

Best Answer

This uses the extended regex syntax -r, which clears up a lot of the clutter. Also, because you already know some of the field values, you don't actually need to back-reference them, again reducing clutter (and overhead).

& is a special replacement value: it hold the entire matched pattern. Using the &, again reduces clutter. As it is not a back-reference, it has significantly less overhead.

I've used ( +) vs. ( *). The + assumes that there is at least one space between input fields. Just change it to the * it that is not the case.

EXPL=
dom=oracle
typ=hard
itm=nproc
val=666

echo "oracle   hard   nproc    131072" |
  sed -r "s/^$dom( +)$typ( +)$itm( +).*/$EXPL#&\n$dom\1$typ\2$itm\3$val/"

output

#oracle   hard   nproc    131072
oracle   hard   nproc    666

Related Solutions

Lum – Easily get a particular column from output without sed or awk

I'm not sure why

ls -hal / | awk '{print $5, $9}'

seems to you to be much more disruptive to your thought processes than

ls -hal / | cut -d'\s' -f5,9

would have been, had it worked. Would you really have to write that down? It only takes a few awk lines before adding the {} becomes automatic. (For me the hardest issue is remembering which field number corresponds to which piece of data, but perhaps you don't have that problem.)

You don't have to use all of awk's features; for simply outputing specific columns, you need to know very little awk.

The irritating issue would have been if you'd wanted to output the symlink as well as the filename, or if your filenames might have spaces in them. (Or, worse, newlines). With the hypothetical regex-aware cut, this is not a problem (except for the newlines); you would just replace -f5,9 with -f5,9-. However, there is no awk syntax for "fields 9 through to the end", and you're left with having to remember how to write a for loop.

Here's a little shell script which turns cut-style -f options into an awk program, and then runs the awk program. It needs much better error-checking, but it seems to work. (Added bonus: handles the -d option by passing it to the awk program.)

#!/bin/bash
prog=\{
while getopts f:d: opt; do
  case $opt in
    f) IFS=, read -ra fields <<<"$OPTARG"
       for field in "${fields[@]}"; do
         case $field in
           *-*) low=${field%-*}; high=${field#*-}
                if [[ -z $low  ]]; then low=1; fi
                if [[ -z $high ]]; then high=NF; fi
                ;;
            "") ;;
             *) low=$field; high=$field ;;
         esac
         if [[ $low == $high ]]; then
           prog+='printf "%s ", $'$low';'
         else
           prog+='for (i='$low';i<='$high';++i) printf "%s ", $i;'
         fi
       done
       prog+='printf "\n"}'
       ;;
    d) sep="-F$OPTARG";;
    *) exit 1;;
  esac
done
if [[ -n $sep ]]; then
  awk "$sep" "$prog"
else
  awk "$prog"
fi

Quick test:

$ ls -hal / | ./cut.sh -f5,9-
7.0K bin 
5.0K boot 
4.2K dev 
9.0K etc 
1.0K home 
8.0K host 
33 initrd.img -> /boot/initrd.img-3.2.0-51-generic 
33 initrd.img.old -> /boot/initrd.img-3.2.0-49-generic 
...

Best Answer

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Lum – Easily get a particular column from output without sed or awk

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