Hi All,
I have problem in the following shell script (problem in 2and3 line i guess)
#!/bin/sh
set value1 = 90;
set value2 = 70;
if [ $value1 > $value2 ]; then
echo "$value1 is normal"
else
echo "$value2 is abnormal"
fi

when executed
output:
$ value_test.sh (Enter)
is abnormal
Neither it's printing the $value1,nor it's displaying correct output.but no errors.
It is not atall executing the loop properly and jumps to else directly.

I guess the problem is with Set value command..
i guess it's wrong.... pls let me know the correct command.

Thanks

Btw: There is no need to have semicolons at the end of variable declaration in bash and there is a [ C O D E ] tag to put code into a box so you don't have to mark it red

Just as namishtiwari wrote, or define the values as integer and you can got with arithmetic signs (else it will think you might want to redirect an output):

Hi zaxxon and namishtiwari thanks for your quick turn over,

For namishtiwari
If I edit code as per u say… im getting the following output:
$ value_test.sh
Is normal.

Here it’s executing the loop successfully nut it’s not displaying $value1 value
The out put should be like this right…
$ value_test.sh
90 Is normal.

Please see $value is missing.

For zaxxon,
If u edit code as per u say…
Im getting the following error…
$ value_test.sh
value_test.sh: typeset: not found
value_test.sh: typeset: not found
is normal

But here also it’s executing the loop.

Please help me.

Thanks

Strange you dont have typeset... nvm, try this:

Hi Zaxxon,
Still im not getting the desired output..
here is the code
#!/bin/sh
value1=90
value2=70
if (( ${value1} > ${value2})); then
echo "${value1} is normal"
else
echo "${value2} is abnormal"
fi

and output:
$ ./value_test.sh
./value_test.sh: 90: not found
70 is abnormal

I DONT KNOW WATS WRONG...
PLSSSSSSS HELP.

THANKS

I am getting the right output as you are saying see this--

namish@france => ./sample1
90 is normal

No need to edit anything just copy the code.

Thanks
namish