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Problem with dividing

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# 1  
Old 11-30-2009
Problem with dividing
Hi All,

I have two variables like below..

Code:
ETIMEHR =03            #END TIME HOUR
ETIMEMIN=02            #END TIME MIN

Now I'm converting the min to hour by deviding it by 60
Code:
ETIMEMINCONV=`echo "scale=2; $ETIMEMIN/60" | bc`

but when i print ETIMEMINCONV instead of getting converted value i'm getting an error like "syntax error on line 1"

I'm using this inside for loop.. so for some values it is working and for others it is giving the above said error...

is there any other alternative for this?

thanks&regards,
Deepak.
# 2  
Old 11-30-2009
Can you show more (like "the loop")? For which values is it not working? Can you run the script with -x (set -x) to start debugging the problem?

(I trust that the space before the = in "ETIMEHR =03" is not really there in your actual script)
# 3  
Old 11-30-2009
But no need of ETIMEHR here know.

i didnt get the debug part which you told me above since i'm not a unix programmer.

and yeah in my script set -x is there but how i can find out or debug where i'm getting the error?
# 4  
Old 11-30-2009
I did a test...

Code:
$ cat Test
set -x

X=1

for I in 1 2 3 4 5 6 7 8; do
  X=$(echo "scale=2; $X/$I" | bc)
  echo $X $I
done

$ ./Test
+ X=1
+ + bc
+ echo scale=2; 1/1
X=1.00
+ echo 1.00 1
1.00 1
+ + echo scale=2; 1.00/2
+ bc
X=.50
+ echo .50 2
.50 2
...

If bc gives an error you should see it among this output. Just before that you should see the value of ETIMEMIN (i.e. "echo scale=2 .....") that was used in the "echo ... | bc" command.

If you are able to show the code for the loop (and anything else needed within it) then it would made it easier.
# 5  
Old 11-30-2009
Smilie i dont have the exact code with me now.. It is there in my office system.

i will do as per your suggestion tomorrow and let you know.. Thanks a lot for your advice.

---------- Post updated at 05:36 PM ---------- Previous update was at 05:29 AM ----------

hi scottn

Code:
ENDDAY=11/25/09
+ + echo 11/25/09
ENDMM=11/25/09
+ + awk -F/ {print $2}
+ echo 11/25/09
ENDDD=25
+ + awk -F/ {print $3}
+ cut -c1-2
+ echo 11/25/09
ENDYY=09
+ + grep -i $PATH
+ awk -F; {print $21}
+ awk -F  {print $2}
+ awk -F: {print $1}
+ sort -u
ETIMEHR=06
+ + awk -F; {print $21}
+ awk -F: {print $2}
+ grep -i $PATH
+ sort -u
+ awk -F  {print $2}
ETIMEMIN=02
+ + bc
/60cho scale=2;02
ETIMEMINCONV=syntax error on line 1, 
+ print  syntax error on line 1,  
 syntax error on line 1,

this is what the error i'm getting.. see there it is taking /60 at the beginning instead of 02/60... pls help me..

---------- Post updated at 07:20 PM ---------- Previous update was at 05:36 PM ----------

For some values i'm getting the proper ans as shown below..

Code:
 + grep -i $PATH
+ awk -F; {print $21}
+ cut -d  -f1
+ sed -e s/\$/ /g
ENDDAY=11/30/09
+ + echo 11/30/09
ENDMM=11/30/09
+ + awk -F/ {print $2}
+ echo 11/30/09
ENDDD=30
+ + awk -F/ {print $3}
+ echo 11/30/09
+ cut -c1-2
ENDYY=09
+ + awk -F; {print $21}
+ awk -F: {print $1}
+ grep -i $PATH
+ sort -u
+ awk -F  {print $2}
ETIMEHR=09
+ + grep $PATH
+ awk -F; {print $21}
+ awk -F  {print $2}
+ awk -F: {print $2}
+ sort -u
ETIMEMIN=51
+ X=51
+ + bc
+ echo scale=2; (51/60)
ETIMEMINCONV=.85
+ print  .85
 .85

so please tell me what is wrong with this...
Last edited by smarty86; 11-30-2009 at 11:17 PM..
# 6  
Old 12-01-2009
Perhaps your version of bc has problem with the leading zero. What happens if you use:
Code:
ETIMEMINCONV=`echo "scale=2; ${ETIMEMIN##0}/60" | bc`

# 7  
Old 12-01-2009
hey i dont know what was wrong.. i just multiplied first by 1000 to the value of ETIMEMIN then instead of deviding by 60 i used 60000..so it is working fine now Smilie

Thanks for all your reply...
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