Say you have a simple loop
while read line
do
printf "${line#*//}\n"
done < text.txt
Is there an elegant way of printing the current iteration with the output? Something like
0 The 1 quick 2 brown 3 fox
I am hoping to avoid setting a variable and incrementing it on each loop.
To do this, you would need to increment a counter on each iteration (like you are trying to avoid).
count=0
while read -r line; do
printf '%d %s\n' "$count" "${line*//}"
(( count++ ))
done < test.txt
EDIT: After some more thought, you can do it without a counter if you have bash version 4 or higher:
mapfile -t arr < test.txt
for i in "${!arr[@]}"; do
printf '%d %s' "$i" "${arr[i]}"
done
The mapfile builtin reads the entire contents of the file into the array. You can then iterate over the indices of the array, which will be the line numbers and access that element.
mapfile can be awfully memory hungry if you're dealing with large files. Without knowing the OP's file size, I'd recommend going with something that runs through a pipe, and so is filesize agnostic.You don't often see it, but you can have multiple commands in the condition clause of a while loop. The following still requires an explicit counter variable, but the arrangement may be more suitable or appealing for some uses.
while ((i++)); read -r line
do
echo "$i $line"
done < inputfile
The while condition is satisfied by whatever the last command returns (read in this case).
Some people prefer to include the do on the same line. This is what that would look like:
while ((i++)); read -r line; do
echo "$i $line"
done < inputfile
for does something similar: for ((i=0, j=14; i<=20; i++, j+=11)). Also, you can read two (or more) lines from a file at a time: while read -r oddline; read -r evenline; do echo "odd: [$oddline] even: [$evenline]"; done < filenameYou can use a range to go through, it can be an array, a string, a input line or a list.
In this example, i use a list of numbers [0..10] is used with an increment of 2, as well.
#!/bin/bash
for i in {0..10..2}; do
echo " $i times"
done
The output is:
0 times
2 times
4 times
6 times
8 times
10 times
To print the index regardless of the loop range, you have to use a variable "COUNTER=0" and increase it in each iteration "COUNTER+1".
my solution prints each iteration, the FOR traverses an inputline and increments by one each iteration, also shows each of words in the inputline:
#!/bin/bash
COUNTER=0
line="this is a sample input line"
for word in $line; do
echo "This i a word number $COUNTER: $word"
COUNTER=$((COUNTER+1))
done
The output is:
This i a word number 0: this
This i a word number 1: is
This i a word number 2: a
This i a word number 3: sample
This i a word number 4: input
This i a word number 5: line
to see more about loops: enter link description here
to test your scripts: enter link description here
n=0
cat test.txt | while read line; do
printf "%7s %s\n" "$n" "${line#*//}"
n=$((n+1))
done
This will work in Bourne shell as well, of course.
If you really want to avoid incrementing a variable, you can pipe the output through grep or awk:
cat test.txt | while read line; do
printf " %s\n" "${line#*//}"
done | grep -n .
or
awk '{sub(/.*\/\//, ""); print NR,$0}' test.txt
sub(/.*\/\//, ""); is not the same as ${line#*//}
cat is useless and read without -r will mangle backslashes in the input. This is a really inefficient way to reimplement nl.Update: Other answers posted here are better, especially those of @Graham and @DennisWilliamson.
Something very like this should suit:
tr -s ' ' '\n' <test.txt | nl -ba
You can add a -v0 flag to the nl command if you want indexing from 0.
n1 command in FreeBSD, OSX or Illumos. Where is it from?//, as ${line#*//} does. But you could certainly pipe the while loop through nl, just as I did with grep -n . in my answer.