Solved

c# determine if a 50 character string is a number

Posted on 2014-03-14
10
268 Views
Last Modified: 2014-03-18
I am creating an excel document and need a value in a cell to be stored as a string.  If it is a number I need to put an apostrophe in front of it.  I was using decimal.tryparse but that won't work with a 50 digit number.  How can I determine if a string is a number no matter how many characters or digits are there?  I also need it to work with a negative sign, commas and a decimal point.

(Please do not ask me to change the way I am creating the excel, that is not important and is not the question.)

Thanks!

decimal out_dec = 0;
//this will always evaluate to false
var isNumber decimal.TryParse("12345678901234567890123456789012345678901234567890", out out_dec)

Open in new window

0
Comment
Question by:jackjohnson44
  • 5
  • 4
10 Comments
 
LVL 74

Expert Comment

by:käµfm³d 👽
Comment Utility
You could check each digit:

public bool IsNumeric(string value)
{
    for (int i = 0; i < value.Length; i++)
    {
        if (i == 0 && value[i] == '-') continue;

        if (!char.IsDigit(value[i]))
        {
            return false;
        }
    }

    return true;
}

Open in new window

0
 

Author Comment

by:jackjohnson44
Comment Utility
That won't work with decimal points, and commas.  I need a way to determine if it is a real number, not just 0-9, comma, decimal point, and negative sign

for example
100,000 is a number
10,00 is not a number
1.2.3 is not a number
etc...
0
 
LVL 74

Expert Comment

by:käµfm³d 👽
Comment Utility
OK, I overlooked that bit. Modified:

public bool IsNumeric(string value)
{
    bool decimalFound = false;

    for (int i = 0; i < value.Length; i++)
    {
        if (i == 0 && value[i] == '-') continue;

        if (value[i] == '.')
        {
            if (decimalFound)
            {
                return false;
            }

            decimalFound = true;
        }
        else if (!char.IsDigit(value[i]))
        {
            return false;
        }
    }

    return true;
}

Open in new window

0
 
LVL 34

Accepted Solution

by:
Dan Craciun earned 250 total points
Comment Utility
If you can use regex, you can use this:
@"^-?[0-9]*(,[0-9]{3})*\.?[0-9]*$"

Open in new window

Works correctly for the following cases:
100,000
10,00
1.2.3
50
100000.58796
-50
-23412.6535
-5.3.2

Open in new window


Borrowing the above function name, you could use something like this:
public bool IsNumeric(string value)
{
    string pattern = @"^-?[0-9]*(,[0-9]{3})*\.?[0-9]*$";

    Regex r = new Regex(pattern);
	
	Match m = r.Match(value)
	
	if (m.Success) {
	
		return true;
		
	}

    return false;
}

Open in new window

HTH,
Dan
0
 
LVL 74

Expert Comment

by:käµfm³d 👽
Comment Utility
Dan Craciun's pattern also accepts (in case it's relevant):

[nothing]
1. [no fractional part]
100000000,000.00 [partially formatted]
,000.00 [partially formatted]
0
How your wiki can always stay up-to-date

Quip doubles as a “living” wiki and a project management tool that evolves with your organization. As you finish projects in Quip, the work remains, easily accessible to all team members, new and old.
- Increase transparency
- Onboard new hires faster
- Access from mobile/offline

 
LVL 34

Expert Comment

by:Dan Craciun
Comment Utility
OK, thanks for that.

1. and ,000.00 are easy to solve. Still needs work on the 10000,000.00 part :)

@"^-?[0-9]+(,[0-9]{3})*(\.[0-9]+)?$"

Open in new window

0
 
LVL 74

Expert Comment

by:käµfm³d 👽
Comment Utility
Try {1,3} rather than +   ; )
0
 
LVL 34

Expert Comment

by:Dan Craciun
Comment Utility
Yeah, but that's only if the numbers were always grouped by ",". Since that was not the case, I need to use an if (if there is at least an , try to match groups of 3, if not, search for digits only):
@"^-?(?(?=.*,)([0-9]{1,3}(,[0-9]{3})*)|([0-9]*))(\.[0-9]+)?$"

Open in new window

works correctly for:
100,000
10,00
1.2.3
50
100000.58796
-50
-23412.6535
-5.3.2
1.
10000000,000.00
,000.00

Open in new window

0
 
LVL 74

Assisted Solution

by:käµfm³d 👽
käµfm³d   👽 earned 250 total points
Comment Utility
You don't need the if...a simple alternation will suffice.

e.g.

^-?(?:\d{1,3}(?:,\d{3})+|\d+)(?:\.\d+)?$

Open in new window

0
 
LVL 34

Expert Comment

by:Dan Craciun
Comment Utility
Yup, thank you for that. I'm still new at regular expressions and when I don't understand something I fall back to the programmer's way of thinking :)

BTW, I recently found out that  \d is looking for a lot more characters than [0-9], so in the case of long strings it's faster to use [0-9], although I agree it's easier to read \d.

Here are all the characters \d is looking for: http://www.fileformat.info/info/unicode/category/Nd/list.htm
0

Featured Post

Highfive Gives IT Their Time Back

Highfive is so simple that setting up every meeting room takes just minutes and every employee will be able to start or join a call from any room with ease. Never be called into a meeting just to get it started again. This is how video conferencing should work!

Join & Write a Comment

This document covers how to connect to SQL Server and browse its contents.  It is meant for those new to Visual Studio and/or working with Microsoft SQL Server.  It is not a guide to building SQL Server database connections in your code.  This is mo…
Exception Handling is in the core of any application that is able to dignify its name. In this article, I'll guide you through the process of writing a DRY (Don't Repeat Yourself) Exception Handling mechanism, using Aspect Oriented Programming.
Illustrator's Shape Builder tool will let you combine shapes visually and interactively. This video shows the Mac version, but the tool works the same way in Windows. To follow along with this video, you can draw your own shapes or download the file…
This video gives you a great overview about bandwidth monitoring with SNMP and WMI with our network monitoring solution PRTG Network Monitor (https://www.paessler.com/prtg). If you're looking for how to monitor bandwidth using netflow or packet s…

744 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

10 Experts available now in Live!

Get 1:1 Help Now