RangeValidator Formula Correct to get 50% higher

Question

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I have a formula calculation that should find the 50% increase on a value. If a value is 100 and the 50% increase will be 151. I am trying to see if my code of getting this is correct. I tested this and the RangeValidator fires even when the number is the same or less than 50%. Please let me know if the code is correct or not.

C#

int itxtVal13 = Convert.ToInt32(TextBox1.Text);
        int iminVal13 = (itxtVal13 * 0);
        int imaxVal13 = (itxtVal13 + itxtVal13 * 150 / 100);
        RangeValidatorLYFTE4050.MinimumValue = iminVal13.ToString();
        RangeValidatorLYFTE4050.MaximumValue = imaxVal13.ToString();

Posted 22-Oct-14 3:50am

Computer Wiz99

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Sergey Alexandrovich Kryukov 22-Oct-14 10:17am

What, now arithmetic? :-)
—SA

1 solution

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CHill60 · Accepted Answer · 2014-10-22T03:59:00

Solution 1

If you have itxtVal13 = 100 and put a breakpoint on the line

C#

RangeValidatorLYFTE4050.MinimumValue = iminVal13.ToString();

then examine the value of imaxVal13 what do you see?

You have calculated 250% of itxtVal13 not 150%

Either change the calculation to

int imaxVal13 = itxtVal13 + (itxtVal13 * 50 / 100);

OR to

int imaxVal13 =(itxtVal13 * 150 / 100);

See the difference?

[Edit]
The breakpoint on the line mentioned was just so that imaxVal13 had already been calculated. Your corrected code (this issue only) will look like:

C#

int itxtVal13 = Convert.ToInt32(TextBox1.Text);
int iminVal13 = (itxtVal13 * 0);  // Which = 0 by the way!!
int imaxVal13 = itxtVal13 * (150 / 100);
RangeValidatorLYFTE4050.MinimumValue = iminVal13.ToString();
RangeValidatorLYFTE4050.MaximumValue = imaxVal13.ToString();

Posted 22-Oct-14 3:59am

CHill60

Updated 22-Oct-14 4:43am

v2

Comments

Computer Wiz99 22-Oct-14 10:38am

CHill60, So in my code I am deleting this line, int iminVal13 = (itxtVal13 * 0);, and replacing this line, RangeValidatorLYFTE4050.MinimumValue = iminVal13.ToString();, with the line you gave me? RangeValidatorLYFTE4050.MinimumValue = iminVal13.ToString();

CHill60 22-Oct-14 10:43am

That's not quite what I said - I've updated my solution

Computer Wiz99 22-Oct-14 11:35am

That works. Thanks. Had to change some other ones when testing.

Computer Wiz99 22-Oct-14 11:40am

Is there a way to have a validator to fire and give a value like 1 to be saved in the database? So that next time the user logs in and the form is populated the error will show and that user can print a report with that error on it?

CHill60 22-Oct-14 12:00pm

Well that's not really the role of the validator - but there is nothing to stop you doing something like that in the code behind - don't get confused between a validation control acting on the client side (browser) and code being executed on the server

Computer Wiz99 22-Oct-14 12:04pm

How do you mean?

CHill60 22-Oct-14 12:09pm

It's not my strongest subject to explain, so I'll direct you to these articles http://aspnet.4guysfromrolla.com/articles/112305-1.aspx[^]