Error in while loop

 
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# 1  
Old 09-01-2011
Error in while loop

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1. The problem statement, all variables and given/known data:

I am getting the following error while executing for while loo:

count: command not found
[: -le: unary operator expected

2. Relevant commands, code, scripts, algorithms:

I have created the following code lines:

#!/bin/sh
count =1
while [ $count -le 10 ]
do
echo $count
count = `expr $count + 1`
done


3. The attempts at a solution (include all code and scripts):

Wrote the above code.

4. Complete Name of School (University), City (State), Country, Name of Professor, and Course Number (Link to Course):

Amity, Noida, India, Prof. Sanjeev Thakur, MCA

Note: Without school/professor/course information, you will be banned if you post here! You must complete the entire template (not just parts of it).

Last edited by manishdivs; 09-01-2011 at 02:15 AM.. Reason: to complete the info
# 2  
Old 09-01-2011
Error due to additional spaces b/w count variables .. Corrected one below
Code:
#!/bin/sh
count=1
while [ $count -le 10 ]
do
echo $count
count=`expr $count + 1`
done

# 3  
Old 09-01-2011
Thanks for the reply

Still getting the following error msg:

./while_loop.ksh_new: line 3: count: command not found
./while_loop.ksh_new: line 5: [: -le: unary operator expected

cat ./while_loop.ksh_new
#!/bin/sh
count =1
while [ $count -le 10 ]
do
echo $count
count=`expr $count + 1`
done
bash-3.2$
# 4  
Old 09-01-2011
again u have given additional space while assigning value to a varaible count =1 .. Try to use the code which i have corrected ..
# 5  
Old 09-01-2011
Worked now..Thanks a lot for the quick help.. Smilie
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