error in shell script code..ksh

[amonamon]

I am using ksh..


I calculated variable with bc and if I do echo $g I get: 167.500

I would like to do check for variable $g

in case it has value: 167.500 assign 168 to variable $h..or in case it has value 165.000 assign 165 to variable $h..here is what I did so far:

in=$1
a=`echo "$in/5"|bc`
b=`expr $a + 1`

g=`echo "(($b * 123) / 100)+1" | bc`

if [ $g==*000 ]
then
h=`echo "(($b * 123) / 100)" | bc`
else
h=`echo "(($b * 123) / 100)+1" | bc`
fi

value $h should be one that I look for..

any help would be helpfull..

[Peter Nikitka]

Hi,

the syntax you are using is not correct.
>>
if [ $g==*000 ]
then
<<

You could use a 'case', but if you want a simple rounding to integer of
$i*1.23/5
my suggestion is:

i=$1
h=$(print $i | awk '{printf("%.0f\n",$1*1.23/5)}')
print $h

mfG Peter

[SANTOSH S. MHASKAR]

Hi,

-------
in case it has value: 167.500 assign 168 to variable $h..or in case it has value 165.000 assign 165 to variable $h
-----------

how( and what do u mean by above) u r going to achive above by using

-------------------
if [ $g==*000 ]
then
h=`echo "(($b * 123) / 100)" | bc`
else
h=`echo "(($b * 123) / 100)+1" | bc`
fi
------------------------

I think u have to round the figure to next integer (eg. 16.500 to 17 and 16.49 to 16)
is it this u want?

[amonamon]

ok here is again what I do have:

# in this U enter $1

a=`echo "$1/5"|bc`
b=`expr $a + 1`
c=`echo "$b*5"|bc`


g=`echo "(($b * 123) / 100)+1" | bc`


# so by calculating g goes if statement...

if [ $g==*000 ]
then
h=`echo "(($b * 123) / 100)" | bc`
else
h=`echo "(($b * 123) / 100)+1" | bc`
fi


# this means if for example $g is 167.500 value of h will be:
h=`echo "(($b * 123) / 100)+1" | bc`

and if value od $g is 156.000 vale of h will be:
h=`echo "(($b * 123) / 100)" | bc`

or more examples:

is g is 14.145 h will be:
h=`echo "(($b * 123) / 100)+1" | bc`

or if g is 45.000 h will be:
h=`echo "(($b * 123) / 100)" | bc`

my problem is in syntax of if statement in this:

if [ $g==*000 ]

[amonamon]

problem with rounding I want to solve with this is starement

so as it says rounding will be:

(eg. 16.500 to 17 and 16.49 to 17 again)
16.000 will be rounded to 16!..

I know specific rounding..
It is just problem with if stat.to recognise last 3 digits..
//if last 3 digits are 0 if is true:
if [ $g==*000 ] #but this does not work

[Peter Nikitka]

Hi,

if you insist on such a construct:
...
case $g in
*000) # if part
;;
## possibly wanted:
#*[024]00) # elif round to floor
#;;
#*[68]00) # elif round to ceiling
#;;
*) # else part
;;
esac

mfG Peter

[amonamon]

well it is not that I insist..that was my solution..and it is alllwwayss very nice to hear better solution..

thank U veryyy much

[Peter Nikitka]

Hi,

to come back to my first solution:
Really easy for checks (modify the formula as required, if I guessed wrong):
awk '{printf("%.0f\n",$1*1.23/5)}'

Now enter you input values (stdin) and the calculated values you see on stdout.

mfG Peter

[amonamon]

by the way this case is not working is always reads *) statement

problem is he does never go in *000) check..even is $g is 123.000

[amonamon]

Thanks Peter your solution for rounding is OK..but my requirement for rounding is specific..see above post

12.000 rounded: 12
12.043 rounded: 13
23.999 rounded: 24
etc.

[Peter Nikitka]

Hi,

then I suggest do look at the value from an integer point of view:

awk '{printf("%d\n",$1*1.23/5)}'

mfG Peter

[amonamon]

whatever..problem is not now with rounding..it is about if or case statment...
after that we can deal with rounding..

[Peter Nikitka · Accepted solution]

Sorry,

I did not notice, that you want to round up - my last solution just drops the fractional part.
So better add .5 and use usual rounding:
awk '{printf("%.0f\n",$1*1.23/5+0.5)}'


BTW: the case solution does not work due to the format your '$g' has: I don't see any fractional output in
print $g

mfG Peter

[amonamon]

thanks for help guys..

[amonamon]

thanks for case statement..:)