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How do I exit BASH while loop using modulus operator?

Running bash loopWhile loop in ShellCan you help me to understand this explanation of shell quoting?For loop syntax bash scriptExit terminal after running a bash scriptRedirection operator priority in BashDisplay images in a loop using bashMeaning of exit 0, exit 1 and exit 2 in a bash scriptBash - add zero to single digit in while loopRunning a Bash while loop over all similar files

1

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

share|improve this question

edited 1 hour ago

Roosevelt Mendieta

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the assignment specifies bash, then you might consider using its built-in arithmetic expansion syntax e.g. (( INPUT % 5 == 0 ))
  
  – steeldriver
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the loop does not end when entering 25 @steeldriver
  
  – Roosevelt Mendieta
  1 hour ago
  

  
  
  
  

add a comment | 

1

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

share|improve this question

edited 1 hour ago

Roosevelt Mendieta

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If the assignment specifies bash, then you might consider using its built-in arithmetic expansion syntax e.g. (( INPUT % 5 == 0 ))
  
  – steeldriver
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the loop does not end when entering 25 @steeldriver
  
  – Roosevelt Mendieta
  1 hour 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

share|improve this question

edited 1 hour ago

Roosevelt Mendieta

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

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

share|improve this question

edited 1 hour ago

Roosevelt Mendieta

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

share|improve this question

edited 1 hour ago

Roosevelt Mendieta

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

asked 2 hours ago

Roosevelt MendietaRoosevelt Mendieta

3415

3 Answers
3

active

oldest

votes

2

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

share|improve this answer

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't work, when I enter 40 the code exits
  
  – Roosevelt Mendieta
  1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
  
  – Kulfy
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i version controlled the code back to it's original state @Kulfy
  
  – Roosevelt Mendieta
  1 hour ago
  

  
  
  
  

add a comment | 

3

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.)

share|improve this answer

edited 1 hour ago

answered 1 hour ago

sudodussudodus

25.6k33078

add a comment | 

2

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"
$

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325

add a comment | 

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2

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

share|improve this answer

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't work, when I enter 40 the code exits
  
  – Roosevelt Mendieta
  1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
  
  – Kulfy
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i version controlled the code back to it's original state @Kulfy
  
  – Roosevelt Mendieta
  1 hour ago
  

  
  
  
  

add a comment | 

2

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

share|improve this answer

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't work, when I enter 40 the code exits
  
  – Roosevelt Mendieta
  1 hour 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RooseveltMendieta It seems that you changed the original code in your question. So, PerlDuck might need to modify explanation of the answer.
  
  – Kulfy
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  i version controlled the code back to it's original state @Kulfy
  
  – Roosevelt Mendieta
  1 hour 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

share|improve this answer

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

#!/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

share|improve this answer

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

answered 2 hours ago

PerlDuckPerlDuck

7,89611636

3

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.)

share|improve this answer

edited 1 hour ago

answered 1 hour ago

sudodussudodus

25.6k33078

add a comment | 

3

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.)

share|improve this answer

edited 1 hour ago

answered 1 hour 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.)

share|improve this answer

edited 1 hour ago

answered 1 hour ago

sudodussudodus

25.6k33078

#!/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

share|improve this answer

edited 1 hour ago

answered 1 hour ago

sudodussudodus

25.6k33078

answered 1 hour ago

sudodussudodus

25.6k33078

answered 1 hour ago

sudodussudodus

25.6k33078

2

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"
$

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325

add a comment | 

2

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"
$

share|improve this answer

edited 1 hour ago

answered 1 hour 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"
$

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325

$ 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

$ [ $((25%5)) -eq 0 ] && echo "Zero"
Zero
$ [ $((26%5)) -eq 0 ] && echo "Zero"
$

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325

answered 1 hour ago

Sergiy KolodyazhnyySergiy Kolodyazhnyy

74.8k9155325