Command to prepend string to each line?

Question

Looking for something like this? Any ideas?

cmd | prepend "[ERRORS] "

[ERROR] line1 text
[ERROR] line2 text
[ERROR] line3 text
... etc

Answer 1

cmd | while read line; do echo "[ERROR] $line"; done

has the advantage of only using bash builtins so fewer processes will be created/destroyed so it should be a touch faster than awk or sed.

Answer 2

Try this:

cmd | awk '{print "[ERROR] " $0}'

Cheers

Answer 3

cmd | sed 's/.*/[ERROR] &/'

Answer 4

I wanted a solution that handled stdout and stderr, so I wrote prepend.sh and put it in my path:

#!/bin/bash

prepend_lines(){
  local prepended=$1
  while read line; do
    echo "$prepended" "$line"
  done
}

tag=$1

shift

"$@" > >(prepend_lines "$tag") 2> >(prepend_lines "$tag" 1>&2)

Now I can just run prepend.sh "[ERROR]" cmd ..., to prepend "[ERROR]" to the output of cmd, and still have stderr and stdout separate.

Answer 5

With all due credit to @grawity, I'm submitting his comment as an answer, as it seems the best answer here to me.

sed 's/^/[ERROR] /' cmd