How do I exit BASH while loop using modulus operator?
So practically for my assignment I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0, ex: (25 % 5 = 0 break loop) Where in my attempt below have I gone wrong?
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
else
echo "you entered right"
break
fi
done
command-line bash scripts
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
add a comment |
So practically for my assignment I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0, ex: (25 % 5 = 0 break loop) Where in my attempt below have I gone wrong?
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
else
echo "you entered right"
break
fi
done
command-line bash scripts
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
2
If the assignment specifiesbash, then you might consider using its built-in arithmetic expansion syntax e.g.(( INPUT % 5 == 0 ))
– steeldriver
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago
add a comment |
So practically for my assignment I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0, ex: (25 % 5 = 0 break loop) Where in my attempt below have I gone wrong?
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
else
echo "you entered right"
break
fi
done
command-line bash scripts
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
So practically for my assignment I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0, ex: (25 % 5 = 0 break loop) Where in my attempt below have I gone wrong?
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
else
echo "you entered right"
break
fi
done
command-line bash scripts
command-line bash scripts
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
edited 7 hours ago
Roosevelt Mendieta
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
asked 8 hours ago
Roosevelt MendietaRoosevelt Mendieta
3915
3915
2
If the assignment specifiesbash, then you might consider using its built-in arithmetic expansion syntax e.g.(( INPUT % 5 == 0 ))
– steeldriver
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago
add a comment |
2
If the assignment specifiesbash, then you might consider using its built-in arithmetic expansion syntax e.g.(( INPUT % 5 == 0 ))
– steeldriver
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago
2
2
If the assignment specifies
bash, then you might consider using its built-in arithmetic expansion syntax e.g. (( INPUT % 5 == 0 ))– steeldriver
7 hours ago
If the assignment specifies
bash, then you might consider using its built-in arithmetic expansion syntax e.g. (( INPUT % 5 == 0 ))– steeldriver
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago
add a comment |
3 Answers
3
active
oldest
votes
Move the break from the else part to the if part:
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
break
else
echo "you entered right"
fi
done
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
add a comment |
It works for me according to @steeldriver's tips,
make sure you use
bash
#!/bin/bash
use the bash syntax for arithmetic evaluation
((...))
Otherwise the shellscript can remain the same,
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if (( INPUT % 5 == 0 )) ; then
echo "you entered right"
break
else
echo "you entered wrong"
fi
done
Edit: You have modified the question. This answer corresponds to a previous version of the question. (It is not clear to me, if you want to break the loop, when there is no remainder or when there is a remainder.)
answered 7 hours ago
sudodussudodus
25.6k33078
add a comment |
Since this is bash script we're talking about, you may want to use read -p and arithmetic evaluation ((...))
$ while read -p "Enter number:" input ; do (( input%5 == 0 )) && { echo "Wrong"; break;} || echo "alright"; done
Enter number:11
alright
Enter number:7
alright
Enter number:10
Wrong
Portably, you might want to use [ aka test
$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Move the break from the else part to the if part:
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
break
else
echo "you entered right"
fi
done
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
add a comment |
Move the break from the else part to the if part:
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
break
else
echo "you entered right"
fi
done
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
add a comment |
Move the break from the else part to the if part:
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
break
else
echo "you entered right"
fi
done
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
Move the break from the else part to the if part:
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if [ `expr $INPUT % 5` -eq 0 ]; then
echo "you entered wrong"
break
else
echo "you entered right"
fi
done
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
answered 8 hours ago
PerlDuckPerlDuck
7,90611636
7,90611636
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
add a comment |
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
this doesn't work, when I enter 40 the code exits
– Roosevelt Mendieta
7 hours ago
2
2
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@RooseveltMendieta Isn't it what you want? I need to break out of a true while loop when the user inputs a number that gives a modulus remainder of 0. 40%5 is also 0.
– Kulfy
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@Kulfy i was thinking of division in my head instead of modulus, how embarrassing lol yes this solution in fact does work and is exactly what I needed. I need to go to sleep i've been up to late working on this assignment.
– Roosevelt Mendieta
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
@RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
– Kulfy
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
i version controlled the code back to it's original state @Kulfy
– Roosevelt Mendieta
7 hours ago
add a comment |
It works for me according to @steeldriver's tips,
make sure you use
bash
#!/bin/bash
use the bash syntax for arithmetic evaluation
((...))
Otherwise the shellscript can remain the same,
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if (( INPUT % 5 == 0 )) ; then
echo "you entered right"
break
else
echo "you entered wrong"
fi
done
Edit: You have modified the question. This answer corresponds to a previous version of the question. (It is not clear to me, if you want to break the loop, when there is no remainder or when there is a remainder.)
answered 7 hours ago
sudodussudodus
25.6k33078
add a comment |
It works for me according to @steeldriver's tips,
make sure you use
bash
#!/bin/bash
use the bash syntax for arithmetic evaluation
((...))
Otherwise the shellscript can remain the same,
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if (( INPUT % 5 == 0 )) ; then
echo "you entered right"
break
else
echo "you entered wrong"
fi
done
Edit: You have modified the question. This answer corresponds to a previous version of the question. (It is not clear to me, if you want to break the loop, when there is no remainder or when there is a remainder.)
answered 7 hours ago
sudodussudodus
25.6k33078
add a comment |
It works for me according to @steeldriver's tips,
make sure you use
bash
#!/bin/bash
use the bash syntax for arithmetic evaluation
((...))
Otherwise the shellscript can remain the same,
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if (( INPUT % 5 == 0 )) ; then
echo "you entered right"
break
else
echo "you entered wrong"
fi
done
Edit: You have modified the question. This answer corresponds to a previous version of the question. (It is not clear to me, if you want to break the loop, when there is no remainder or when there is a remainder.)
answered 7 hours ago
sudodussudodus
25.6k33078
It works for me according to @steeldriver's tips,
make sure you use
bash
#!/bin/bash
use the bash syntax for arithmetic evaluation
((...))
Otherwise the shellscript can remain the same,
#!/bin/bash
while true
do
echo "Please input anything here: "
read INPUT
if (( INPUT % 5 == 0 )) ; then
echo "you entered right"
break
else
echo "you entered wrong"
fi
done
Edit: You have modified the question. This answer corresponds to a previous version of the question. (It is not clear to me, if you want to break the loop, when there is no remainder or when there is a remainder.)
answered 7 hours ago
sudodussudodus
25.6k33078
edited 7 hours ago
answered 7 hours ago
sudodussudodus
25.6k33078
answered 7 hours ago
sudodussudodus
25.6k33078
answered 7 hours ago
sudodussudodus
25.6k33078
25.6k33078
add a comment |
add a comment |
Since this is bash script we're talking about, you may want to use read -p and arithmetic evaluation ((...))
$ while read -p "Enter number:" input ; do (( input%5 == 0 )) && { echo "Wrong"; break;} || echo "alright"; done
Enter number:11
alright
Enter number:7
alright
Enter number:10
Wrong
Portably, you might want to use [ aka test
$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
add a comment |
Since this is bash script we're talking about, you may want to use read -p and arithmetic evaluation ((...))
$ while read -p "Enter number:" input ; do (( input%5 == 0 )) && { echo "Wrong"; break;} || echo "alright"; done
Enter number:11
alright
Enter number:7
alright
Enter number:10
Wrong
Portably, you might want to use [ aka test
$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
add a comment |
Since this is bash script we're talking about, you may want to use read -p and arithmetic evaluation ((...))
$ while read -p "Enter number:" input ; do (( input%5 == 0 )) && { echo "Wrong"; break;} || echo "alright"; done
Enter number:11
alright
Enter number:7
alright
Enter number:10
Wrong
Portably, you might want to use [ aka test
$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
Since this is bash script we're talking about, you may want to use read -p and arithmetic evaluation ((...))
$ while read -p "Enter number:" input ; do (( input%5 == 0 )) && { echo "Wrong"; break;} || echo "alright"; done
Enter number:11
alright
Enter number:7
alright
Enter number:10
Wrong
Portably, you might want to use [ aka test
$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
edited 7 hours ago
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
answered 7 hours ago
Sergiy KolodyazhnyySergiy Kolodyazhnyy
74.8k9155325
74.8k9155325
add a comment |
add a comment |
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2
If the assignment specifies
bash, then you might consider using its built-in arithmetic expansion syntax e.g.(( INPUT % 5 == 0 ))– steeldriver
7 hours ago
the loop does not end when entering 25 @steeldriver
– Roosevelt Mendieta
7 hours ago