How can I read line by line using for loop?
what I've tried:
hh=$(echo -e "\n1\n2\n3\n4\n")
IFS=$'\n';
for r in "$hh"; do echo $r; done
1 2 3 4
echo -e "$hh"
1
2
3
4
Use a while loop:
$ while read -r r; do echo $r; done <<< "$hh"
1
2
3
4
The correct answer is, you don't read line-by-line with a for loop. Use a while loop instead with the read builtin:
while IFS= read -r line; do
echo "$line"
done <<< "$hh"
Though using while read is definitely can solve this but if you really want to use for loop then you need to use IFS=$'\n' to read your input string in the bash's for loop:
hh=$(echo -e "\nname n1\nval n2\n3\nfoo n4\n")
IFS=$'\n' && for r in $hh; do echo "r='$r'"; done
r='name n1'
r='val n2'
r='3'
r='foo n4'
Remove the quotes from around $hh and the original code works fine:
hh=$(echo -e "\n1\n2\n3\n4\n")
IFS=$'\n'
for r in $hh; do echo "Value: $r"; done
# output:
Value: 1
Value: 2
Value: 3
Value: 4
It's only the expansion of the $hh variable in the for loop that DOES NOT need quoting.
This code works.
hh=$(echo -e "\n1\n2\n3\n4\n")
echo "Starting string"
echo -e "$hh"
IFS=$'\n';
echo "Original Code"
for r in "$hh"; do echo r is $r; done
echo "Fixed Code 1"
IFS=$'\n';
for r in "$hh"; do echo r is "$r"; done
echo "Fixed Code 2"
IFS=$'\n';
for r in $hh ; do echo r is $r ; done
And produces,
Starting string
1
2
3
4
Original Code
r is 1 2 3 4
Fixed Code 1
r is
1
2
3
4
Fixed Code 2
r is 1
r is 2
r is 3
r is 4
Related
I need a bash script to find the sum of the absolute value of integers separated by spaces. For instance, if the input is:
1 2 -3
the script should print 6 to standard output
I have:
while read x ; do echo $(( ${x// /+} )) ; done
which gives me
0
Without over complicated things, how would I include an absolute value of each x in that statement so the output would be:
6
With Barmar's idea:
echo "1 2 -3" | tr -d - | tr ' ' '+' | bc -l
Output:
6
You have almost done it, but the -s must have been removed from the line read:
while read x; do x=${x//-}; echo $(( ${x// /+} )); done
POSIX friendly implementation without running a loop and without spawning a sub-shell:
#!/usr/bin/env sh
abssum() {
IFS='-'
set -- $*
IFS=' '
set -- $*
IFS=+
printf %d\\n $(($*))
}
abssum 1 2 -3
Result:
6
Please tell why printing odd numbers in bash script with the following code gives the error:
line 3: {1 % 2 : syntax error: operand expected (error token is "{1 % 2 ")
for i in {1 to 99}
do
rem=$(( $i % 2 ))
if [$rem -neq 0];
then
echo $i
fi
done
This is working example:
for i in {1..99}
do
rem=$(($i % 2))
if [ "$rem" -ne "0" ]; then
echo $i
fi
done
used for loop have a typo in minimum and maximum number, should be {1..99} instead of {1 to 99}
brackets of the if statement needs to be separated with whitespace character on the left and on the right side
Comparision is done with ne instead of neq, see this reference.
As already pointed out, you can use this shell checker if you need some clarification of the error you get.
Not really sure why nobody included it, but this works for me and is simpler than the other 'for' solutions:
for (( i = 1; i < 100; i=i+2 )); do echo $i ; done
To print odd numbers between 1 to 99
seq 1 99 | sed -n 'p;n'
With GNU seq, credit to gniourf-gniourf
seq 1 2 99
Example
$ seq 1 10 | sed -n 'p;n'
1
3
5
7
9
if you reverse it will print even
$ seq 1 10 | sed -n 'n;p'
2
4
6
8
10
One liner:
for odd in {1..99..2}; do echo "${odd}"; done
Or print in a cluster.
for odd in {1..99..2}; do echo -n " ${odd} "; done
Likewise, to print even numbers only:
for even in {2..100..2}; do echo "${even}"; done
OR
for even in {2..100..2}; do echo -n " ${even} "; done
Replace {1 to 99} by {1..99}.
for (( i=1; i<=100; i++ ))
do
((b = $i % 2))
if [ $b -ne 0 ]
then
echo $i
fi
done
for i in {1..99}
do
rem=`expr $i % 2`
if [ $rem == 1 ]
then
echo "$i"
fi
done
for i in {0..49}
do
echo $(($i*2+1))
done
I'm working on a Bash script which that takes in two integers and outputs all the numbers in between the two. It would look something like this:
Input:
bash testScript 3 10
3
4
5
6
7
8
9
10
This is some code that I wrote that I thought would work but I haven't had much luck getting it to work yet.
read myvar
read myvar2
while [ $myvar -le myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
//timer in-between numbers
sleep .5
done
Bash supports c style for loops using a double parenthesis construct:
$ for ((x=3; x<=10; x++)); { echo $x; }
3
4
5
6
7
8
9
10
Or, brace expansion:
$ for i in {3..6}; do echo $i; done
3
4
5
6
Problem with brace expansion is you need to use eval to use variables...
A common GNU utility for this is seq but it is not POSIX, so may not be on every *nix. If you want to write a similar function in Bash, it would look like this:
my_seq ()
# function similar to seq but with integers
# my_seq [first [incr]] last
{
incr=1
start=1
if [[ $# -gt 2 ]]; then
start=$1
incr=$2
end=$3
elif [[ $# -gt 1 ]]; then
start=$1
end=$2
else
end=$1
fi
for ((x=start; x<=end; x+=incr)); { echo $x; }
}
Then call that with 1, 2 or 3 arguments:
$ my_seq 30 10 60
30
40
50
60
with brace expansion
$ echo {3..10} | tr ' ' '\n'
or for variables with eval
$ a=3; b=10; eval echo {$a..$b} | ...
or, if you have awk
$ awk -v s=3 -v e=10 'BEGIN{while(s<=e) print s++}'
Use positional parameters:
myvar=$1
myvar2=$2
while [ $myvar -le $myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
#timer in-between numbers
sleep .5
done
Output:
scottsmudger#ns207588:~ $ bash test 1 5
1
2
3
4
5
See http://www.tldp.org/LDP/abs/html/othertypesv.html
Your posted codes are not aligned properly, not sure if it's your actual codes. But only problem other than the alignment is you missed a $ for myvar2 in the while statment.
read myvar
read myvar2
while [ $myvar -le $myvar2 ]
do
echo $myvar
myvar=$(($myvar+1))
#//timer in-between numbers
sleep .5
done
For example,
I have output every second like
user#pc:
A
B
C
D
E
I want the output refresh from A to E in the same line(no new lines being created)
Thanks!
You can use the carriage return character "\r". Check the difference between:
for x in $(seq 10); do printf "$x"; sleep 1; done
and
for x in $(seq 10); do printf "$x\r"; sleep 1; done
Depending upon where your input is coming from.
I would employ the below techniques.
For files:
#!/bin/bash
while read line
do
echo -e "\e[1A" # moving the cursor back to the previously printed line
sleep 1s;
echo -ne "$line\e[K" # \e[K cleans the residues of the previous output.
done <file_name
echo
For arrays:
#!/bin/bash
arr=(12 11 10 9 8 7 6 5 4 3 2 1 0)
for i in ${arr[#]}
do
echo -e "\e[1A"
sleep 1s;
echo -ne "Waiting time : "$i" Seconds\e[K"
done
echo
so I'm dealing with column averaging problems
I have several columns of number,
1 2 3 4 5
2 3 4 5 6
8 4 5 6 7
9 8 7 6 5
1 9 9 9 2
What I want to do is to take an average of each column so that they will yield
4 5 5 6 5
The summing part is not a problem, I'm able to get the sum of each column. However, somehow I'm having trouble in counting the number of columns (variable $x)
Here is my code
while read line; do
x = 0
read -a array <<< "$line"
for i in "${!array[#]}"
do
column[${i}]=$((${column[${i}]}+${array[$i]}))
((x++))
done
done < $TMP
for sum in ${column[#]}
do
average=`expr $sum / $x`
remainder=`expr $sum % $x`
mult=`expr $remainder \* 10`
fracvalue=`expr $mult / $x`
echo $x
echo $average
echo $sum
echo $remainder
echo $mult
echo $fracvalue
done
The last several lines are for my testing purposes, here the $X keeps showing 1 instead of 5. That's why it screws up all the other variable. Does anybody know where's the flaw in this code? Really appreciate your help. Thanks
Another approach:
#!/bin/bash
declare -ia sum # declare array of integers
declare -i lines=0 # declare integer to count lines
# read file to array and sum values
while read -ra array; do
for ((i=0; i<${#array[#]};i++)); do
sum[$i]+=${array[$i]}
done
lines+=1
done < file
# print averages
for ((j=0; j<$i;j++)); do
echo -n "$((${sum[$j]}/$lines)) "
done
Output:
4 5 5 6 5
The problem is in the line
x = 0
should be
x=0 and make ((x+1)) as let "x+=1" it should work.
Maybe your file doesn't have the same number of lines in each column.
while read line; do
read -a array <<< "$line"
for i in "${!array[#]}"
do
column[${i}]=$((${column[${i}]}+${array[$i]}))
((x[${i}]++))
done
done < $TMP
for i in ${!column[#]}
do
sum=${column[$i]}
x=${x[${i}]}
#your calcul
done
The initial issue, aside from a few syntax peculiarities, was the resetting of x on each read which caused you to lose your line count. Additionally, your array is an indexed array, not an associative array.
Making the small adjustments and declareing the variables at the beginning (to hint to bash they are integers or arrays), it works as planned. note: there is no need for a here-string to assign the array, just use regular array syntax. Also note, for indexed arrays, there is no need to dereference the variable within [ ], simply ${array[i]} is fine.
While the use of expr is fine (old, slow, but portable and fine), you could use bash arithmetic syntax for your parameter calculations at the end. (you already have arrays). Finally, you are better served initializing each column[$i]=0 on the first read loop. You can do that with a simple flag:
#!/bin/bash
declare -i ncols=0
declare -i x=0
declare -a column
while read -r line; do
array=( $line )
[ $ncols -eq 0 ] && ncols=${#array[#]}
for ((i = 0; i < ncols; i++))
do
[ $x -eq 0 ] && column[$i]=0
column[$i]=$(( ${column[i]} + ${array[i]}))
done
((x++))
done < "$1"
for sum in ${column[#]}
do
average=`expr $sum / $x`
remainder=`expr $sum % $x`
mult=`expr $remainder \* 10`
fracvalue=`expr $mult / $x`
echo $x
echo $average
echo $sum
echo $remainder
echo $mult
echo $fracvalue
done
exit 0
Output
$ bash colavg.sh dat/colavg.txt
5
4
21
1
10
2
5
5
26
1
10
2
5
5
28
3
30
6
5
6
30
0
0
0
5
5
25
0
0
0