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.
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
You 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
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.
Related
I need an output of current count vs total count in single line. I would like to know if this could be done Via Bash using 'for' 'while' loop.
Expecting an output that should only update the count and should not display multiple lines
File Content
$ cat ~/test.rtf
hostname1
hostname2
hostname3
hostname4
#!/bin/sh
j=1
k=$(cat ~/test.rtf | wc -l)
for i in $(cat ~/test.rtf);
do
echo "Working on line ($j/$k)"
echo "$i"
#or any other command for i
j=$((j+1))
done
EX:
Working on line (2/4)
Not like,
Working on line (2/4)
Working on line (3/4)
Assumptions:
OP wants to generate n lines of output that overwrite each other on successive passes through the loop
in OP's sample code there are two echo calls so in this case n=2
General approaches:
issue a clear at the beginning of each pass through the loop so as to clear the current output and reposition the cursor at the 'top' of the console/window
use tput to manage movement of the cursor (and clearing/overwriting of previous output)
Sample input:
$ cat test.rtf
this is line1
then line2
and line3
and last but not least line4
real last line5
clear approach:
j=1
k=$(wc -l < test.rtf)
while read -r line
do
clear
echo "Working on line ($j/$k)"
echo "${line}"
((j++))
done < test.rtf
tput approach:
j=1
k=$(wc -l < test.rtf)
EraseToEOL=$(tput el) # grab terminal specific code for clearing from cursor to EOL
clear # optional: start with a new screen otherwise use current position in console/window for next command ...
tput sc # grab current cursor position
while read -r line
do
tput rc # go (back) to cursor position stored via 'tput sc'
echo "Working on line ($j/$k)"
echo "${line}${EraseToEOL}" # ${EraseToEOL} forces clearing rest of line, effectively erasing a previous line that may have been longer then the current line of output
((j++))
done < test.rtf
Both of these generate the same result:
Something along these lines:
file=~/test.rtf
nl=$(wc -l "$file")
nl=${nl%%[[:blank:]]*}
i=0
while IFS= read -r line; do
i=$((i+1))
echo "Working on line ($i/$nl)"
done < "$file"
Your main question is how to avoid each the counter to be written to new lines. The newlines are \n characters, which is appended by echo. You want \r, like
for ((i=0; i<10; i++)); do
printf "Counter $i\r"
sleep 1
done
echo
When you echo something from the line you are working on, you will use \n again. I will use cut as an example of processing the inputline. Use the output string in the same printf command like
j=1
k=$(cat ~/test.rtf | wc -l)
while IFS= read -r line; do
printf "Working on line (%s): %s\r" "$j/$k" $(cut -c1-10 <<< "${line}")
sleep 1
((j++))
done < ~/test.rft
The problem with the above solution is that you will see output from previous lines when your last output is shorter than the previous one. When you know the maximum length that your processing of the line will show, you can force additional spaces:
j=1
k=$(cat ~/test.rtf | wc -l)
while IFS= read -r line; do
printf "Working on line (%5.5s): %-20s\r" "$j/$k" "$(cut -c1-20 <<< "${line}")";
sleep 1
((j++))
done < ~/test.rft
I'm a bit confused by the done < $1 notation.
I'm trying to write a program "sumnums" that reads in a file called "nums" that has a couple rows of numbers. Then it should print out the rows of the numbers followed by a sum of all the numbers.
Currently I have:
#!/bin/bash
sum=0;
while read myline
do
echo "Before for; Current line: \"$myline\""
done
for i in $myline; do
sum=$(expr $sum + $i)
done < $1
echo "Total sum is: $sum"
and it outputs the list of the numbers from nums correctly then says
./sumnums: line 10: $1: ambiguous redirect, then outputs Total sum is: 0.
So somehow it isn't adding. How do I rearrange these lines to fix the program and get rid of the "ambiguous redirect"?
Assuming your filename is in $1 (that is, that your script was called with ./yourscript nums):
#!/bin/bash
[[ $1 ]] || set -- nums ## use $1 if already set; otherwise, override with "nums"
sum=0
while read -r i; do ## read from stdin (which is redirected by < for this loop)
sum=$(( sum + i )) ## ...treat what we read as a number, and add it to our sum
done <"$1" ## with stdin reading from $1 for this loop
echo "Total sum is: $sum"
If $1 doesn't contain your filename, then use something that does contain your filename in its place, or just hardcode the actual filename itself.
Notes:
<"$1" is applied to a while read loop. This is essential, because read is (in this context) the command that actually consumes content from the file. It can make sense to redirect stdin to a for loop, but only if something inside that loop is reading from stdin.
$(( )) is modern POSIX sh arithmetic syntax. expr is legacy syntax; don't use it.
awk to the rescue!
awk '{for(i=1;i<=NF;i++) sum+=$i} END{print "Total sum is: " sum}' file
bash is not the right tool for this task.
I am trying to do some math on 2nd column of a txt file , but some lines are not numbers , i only want to operate on the lines which have numbers .and keep other line unchanged
txt file like below
aaaaa
1 2
3 4
How can I do this?
Doubling the second column in any line that doesn't contain any alphabetic content might look a bit like the following in native bash:
#!/bin/bash
# iterate over lines in input file
while IFS= read -r line; do
if [[ $line = *[[:alpha:]]* ]]; then
# line contains letters; emit unmodified
printf '%s\n' "$line"
else
# break into a variable for the first word, one for the second, one for the rest
read -r first second rest <<<"$line"
if [[ $second ]]; then
# we extracted a second word: emit it, doubled, between the first word and the rest
printf '%s\n' "$first $(( second * 2 )) $rest"
else
# no second word: just emit the whole line unmodified
printf '%s\n' "$line"
fi
fi
done
This reads from stdin and writes to stdout, so usage is something like:
./yourscript <infile >outfile
thanks all ,this is my second time to use this website ,i find it is so helpful that it can get the answer very quickly
I also find a answer below
#!/bin/bash
FILE=$1
while read f1 f2 ;do
if[[$f1 != *[!0-9]*]];then
f2=`echo "$f2 -1"|bc` ;
echo "$f1 $f2"
else
echo "$f1 $f2"
fi
done< %FILE
Hi I am trying to print/echo line numbers that are multiple of 5. I am doing this in shell script. I am getting errors and unable to proceed. below is the script
#!/bin/bash
x=0
y=$wc -l $1
while [ $x -le $y ]
do
sed -n `$x`p $1
x=$(( $x + 5 ))
done
When executing above script i get below errors
#./echo5.sh sample.h
./echo5.sh: line 3: -l: command not found
./echo5.sh: line 4: [: 0: unary operator expected
Please help me with this issue.
For efficiency, you don't want to be invoking sed multiple times on your file just to select a particular line. You want to read through the file once, filtering out the lines you don't want.
#!/bin/bash
i=0
while IFS= read -r line; do
(( ++i % 5 == 0 )) && echo "$line"
done < "$1"
Demo:
$ i=0; while read line; do (( ++i % 5 == 0 )) && echo "$line"; done < <(seq 42)
5
10
15
20
25
30
35
40
A funny pure Bash possibility:
#!/bin/bash
mapfile ary < "$1"
printf "%.0s%.0s%.0s%.0s%s" "${ary[#]}"
This slurps the file into an array ary, which each line of the file in a field of the array. Then printf takes care of printing one every 5 lines: %.0s takes a field, but does nothing, and %s prints the field. Since mapfile is used without the -t option, the newlines are included in the array. Of course this really slurps the file into memory, so it might not be good for huge files. For large files you can use a callback with mapfile:
#!/bin/bash
callback() {
printf '%s' "$2"
ary=()
}
mapfile -c 5 -C callback ary < "$1"
We're removing all the elements of the array during the callback, so that the array doesn't grow too large, and the printing is done on the fly, as the file is read.
Another funny possibility, in the spirit of glenn jackmann's solution, yet without a counter (and still pure Bash):
#!/bin/bash
while read && read && read && read && IFS= read -r line; do
printf '%s\n' "$line"
done < "$1"
Use sed.
sed -n '0~5p' $1
This prints every fifth line in the file starting from 0
Also
y=$wc -l $1
wont work
y=$(wc -l < $1)
You need to create a subshell as bash will see the spaces as the end of the assignment, also if you just want the number its best to redirect the file into wc.
Dont know what you were trying to do with this ?
x=$(( $x + 5 ))
Guessing you were trying to use let, so id suggest looking up the syntax for that command. It would look more like
(( x = x + 5 ))
Hope this helps
There are cleaner ways to do it, but what you're looking for is this.
#!/bin/bash
x=5
y=`wc -l $1`
y=`echo $y | cut -f1 -d\ `
while [ "$y" -gt "$x" ]
do
sed -n "${x}p" "$1"
x=$(( $x + 5 ))
done
Initialize x to 5, since there is no "line zero" in your file $1.
Also, wc -l $1 will display the number of line counts, followed by the name of the file. Use cut to strip the file name out and keep just the first word.
In conditionals, a value of zero can be interpreted as "true" in Bash.
You should not have space between your $x and your p in your sed command. You can put them right next to each other using curly braces.
You can do this quite succinctly using awk:
awk 'NR % 5 == 0' "$1"
NR is the record number (line number in this case). Whenever it is a multiple of 5, the expression is true, so the line is printed.
You might also like the even shorter but slightly less readable:
awk '!(NR%5)' "$1"
which does the same thing.
I'm trying to parse a csv file I made with Google Spreadsheet. It's very simple for testing purposes, and is basically:
1,2
3,4
5,6
The problem is that the csv doesn't end in a newline character so when I cat the file in BASH, I get
MacBook-Pro:Desktop kkSlider$ cat test.csv
1,2
3,4
5,6MacBook-Pro:Desktop kkSlider$
I just want to read line by line in a BASH script using a while loop that every guide suggests, and my script looks like this:
while IFS=',' read -r last first
do
echo "$last $first"
done < test.csv
The output is:
MacBook-Pro:Desktop kkSlider$ ./test.sh
1 2
3 4
Any ideas on how I could have it read that last line and echo it?
Thanks in advance.
You can force the input to your loop to end with a newline thus:
#!/bin/bash
(cat test.csv ; echo) | while IFS=',' read -r last first
do
echo "$last $first"
done
Unfortunately, this may result in an empty line at the end of your output if the input already has a newline at the end. You can fix that with a little addition:
!/bin/bash
(cat test.csv ; echo) | while IFS=',' read -r last first
do
if [[ $last != "" ]] ; then
echo "$last $first"
fi
done
Another method relies on the fact that the values are being placed into the variables by the read but they're just not being output because of the while statement:
#!/bin/bash
while IFS=',' read -r last first
do
echo "$last $first"
done <test.csv
if [[ $last != "" ]] ; then
echo "$last $first"
fi
That one works without creating another subshell to modify the input to the while statement.
Of course, I'm assuming here that you want to do more inside the loop that just output the values with a space rather than a comma. If that's all you wanted to do, there are other tools better suited than a bash read loop, such as:
tr "," " " <test.csv
cat file |sed -e '${/^$/!s/$/\n/;}'| while IFS=',' read -r last first; do echo "$last $first"; done
If the last (unterminated) line needs to be processed differently from the rest, #paxdiablo's version with the extra if statement is the way to go; but if it's going to be handled like all the others, it's cleaner to process it in the main loop.
You can roll the "if there was an unterminated last line" into the main loop condition like this:
while IFS=',' read -r last first || [ -n "$last" ]
do
echo "$last $first"
done < test.csv