If function returning integers?

[jordanb93]

This code snippet is being used to calculate a time into a normal time in the H.M format. The numbers are something like 1.93 which means 1 hour and 93 minutes. I'm trying to convert this into 2.33 so it is more easily read/understood.

| eval x1=substr(Total_Time_Refresh,1,1) 
| eval x2=substr(Total_Time_Refresh,3) 
| eval x4=(x2/100)
| eval x3 = if(x4>=(0.60),((x4/.6)+x1),(x4+x1))

Unfortunately anytime the substring x4 is greater than .60 it returns a whole number 2 or 3 only.

Anyone have any ideas as too why it is doing this?

| eval x1=substr(Total_Time_Refresh,1,1) 
| eval x2=substr(Total_Time_Refresh,3) 
| eval x3=(x2/100)
| eval x5=((x3/0.6)+x1)
| eval x6=(x3+x1)
| eval x4=if(x3>=.6,x5,x6)

Edit: I spread out the code to troubleshoot it and the error lies in '| eval x5=((x3/0.6)+x1)'
It seems to be rounding the number to a whole number if it is over '2'. I don't know the reason why though.

[jordanb93 · Accepted solution]

Solution:

| eval x5=exact((x3/0.6)+x1)

Adding the exact function to the equation forces it to not round the answer.

Edit:

| eval x1=substr(Total_Time_Refresh,1,1) 
| eval x2=substr(Total_Time_Refresh,3) 
| eval x3=(x2/100)
| eval x5=if(x3>.60,((x3-.60)+2),x3+1)
| eval Fixed_Refresh_Time = substr(x5,1,4)

This solution seems to be working but only will work on refreshes that are shorting than 3hrs which is over 99.9% of refreshes in my case and have alerts to go off if they are over 3hr.

[jordanb93]

Although this solves the integer issue the time is now off. Still searching for solution that returns the correct time.