Shell Programming and Scripting

When I did,
echo $SHELL in cmd prompt of putty,
its displaying /bin/sh

And in my shell script.,

I hav started with.,
#!/bin/sh

and

i=1;

while [ $i -lt $VAR ] ;
do
.
.
.

i=$[$i + 1];
#i=`expr $i +1`;
#i=`bc $i + 1`
#i=`($i+1 | bc)`


done

I am trying to increment the var, its running to infinite loop with o/p:


inside loop
./run1.sh: +: not found
inside loop
./run1.sh: +: not found
inside loop
./run1.sh: +: not found
inside loop
./run1.sh: +: not found
inside loop
./run1.sh: +: not found
inside loop
./run1.sh: +: not found
.
.
.
.



Also I tried with other ways of incrementing the var, as indicated with bold text.

I request you to guide to debug this error.

Thanks.,

Try

Code:

i=$(($i+1))

I tried with the above syntax,

still it is throwing the error:

./run1.sh: syntax error at line 44: `i=$' unexpected


Regards.,

Can you please try this code alone and let me know what is its ouptut:

Code:

#!/bin/sh

i=0
VAR=10

while [ $i -lt $VAR ]
do
i=$(($i+1))
echo $i
done

If it works then there's probably another error in you script (please post your complete script in this case), else you work with a non-posix sh or something.

I tried with the sample script provided. The same error I am getting as an output. ..

Code:

$ ./rr.sh
./rr.sh: syntax error at line 8: `i=$' unexpected

And also in your reply, you said about " non-posix sh " , whether it means coding inside bash script?
Actually I tried with writing the things starting with #!/bin/bash also, but din't succeeded doing that either.

Is there any other way of incrementing the var in sh script?.


Thanks.,

change (ksh/bash specific)

Code:

i=$(($i+1))

to (Bourne)

Code:

i=`expr $i + 1`

av_vinay,
please try my sample code, but this time change the i=$(($i+1)) line to i="$(($i+1))"
If it won't work, change the shebang line from #!/bin/sh to #!/bin/bash

Let me know how it goes.