I need to write a shell script that will output prime numbers within a range..

what I tried is to check whether a number is prime or not

set -vx
echo "Enter the number"
read N

I=$N
J=2

until [ $J == $N ]
do
if [ `expr $I % $J` == 0]
then
echo "The number is not prime"
exit
else
J=`expr $J + 1`
fi

done
echo "The number is not prime"


I wrote the program that will output prime numbers within a range in C

the program in C is


int n,m,i,j,p
printf ("Enter the first number")
scanf ("%d",&n )
printf ("Enter the second number")
scanf ("%d",&m )

for ( i=m;i<=n;i++)
{
p=0
for(j=2;j<=i/2;j++)
{
if(i%j==0)
{
p=1
break;
}
printf ("%d", i );
else
{
printf ("%d, is not prime" );
}
}
}



I dont know how to write these conditions in scripting using for loop
Can you help me to solve this.

Look at "help for". The form like "for (( num=1; num<$max; num+=2));do " might be useful for you. You don't need to check even numbers of course, and the max number you need to check is the square root of the highest number you want to check.

can you tell me what is this " help for " Is this a website

it means type help for at the command prompt

genix@xion:~$ help for
for: for NAME [in WORDS ... ;] do COMMANDS; done
The `for' loop executes a sequence of commands for each member in a
list of items. If `in WORDS ...;' is not present, then `in "$@"' is
assumed. For each element in WORDS, NAME is set to that element, and
the COMMANDS are executed.
for ((: for (( exp1; exp2; exp3 )); do COMMANDS; done
Equivalent to
(( EXP1 ))
while (( EXP2 )); do
COMMANDS
(( EXP3 ))
done
EXP1, EXP2, and EXP3 are arithmetic expressions. If any expression is
omitted, it behaves as if it evaluates to 1.

1 members found this post helpful.

Thank you friend...

I tried with for loop as you said

But my script has certain problems
Can you guys help with this

problems
1)
the number 2 is not counting as a prime number
2)
the numbers like 9,15...etc are counting as prime numbers

echo "Enter the number"
read N
echo "Enter the second number"
read M


for (( i=$N+1; i <= $M-1; i++ ))
do
for (( j=2; j <= $i-1; j++ ))
do
if [ `expr $i % $j` = 0 ]
then
p=1
break
else
p=0
echo $i

fi
done
done

Try like this:

echo "Enter the number"
read N
echo "Enter the second number"
read M


for (( i=$N+1; i <= $M-1; i++ ))
do
p=0
for (( j=2; j <= $i-1; j++ ))
do
if [ `expr $i % $j` = 0 ]
then
p=1
break
fi
done
if [ `expr $p` = 0 ]
then
echo $i
fi
done

if [ `expr $p` = 0 ]
should be
if [[ `expr $p` -eq 0 ]]

use symbolic comparators ( = < > ) for strings, strings (-eq -lt -le ..) for numbers (yeah, i know i looks weird, but them's the rules)

Well when you compare 2 strings, what is actually compared (from my knowledge, although I am not very experienced with bash scripting, tell me if I'm wrong) is the ascii value of the letters, one by one.... so theoretically it's still a number comparison, and when you compare 2 numbers with "=", it should provide the same result... Anyway, it seems to work, I haven't tested it throughoutly...

Read and enjoy: http://www.tldp.org/LDP/abs/html/comparison-ops.html

Ok thanks for the clarification.

Conclusion: although both methods work (in this case), the -eq method is more correct because it compares the actual numbers, not their ascii code.

Thank you,dear "Valkyrie_of_valhalla"....I got the answer.You rock!!!!!!!!!

Check bellow link for a shell script which finds prime numbers within
a range and between the range

http://bashscript.blogspot.com/2009/...e-numbers.html

using bash for prime numbers is like
trying to break a sledgehammer with a nut!

hard work.