方案一：Try...Catch(执行效率不高)
/// <summary>
/// 名称：IsNumberic
/// 功能：判断输入的是否是数字
/// 参数：string oText：源文本
/// 返回值：　bool true:是　false:否
/// </summary>
/// <param name="oText"></param>
/// <returns></returns>
private bool IsNumberic(string oText)
{
try
         {
int var1=Convert.ToInt32 (oText);
return true;
         }
catch
{
return false;
}
}

方案二：正则表达式(推荐)
a)
using System;
using System.Text.RegularExpressions;

public bool IsNumber(String strNumber)
{
Regex objNotNumberPattern=new Regex("[^0-9.-]");
Regex objTwoDotPattern=new Regex("[0-9]*[.][0-9]*[.][0-9]*");
Regex objTwoMinusPattern=new Regex("[0-9]*[-][0-9]*[-][0-9]*");
String strValidRealPattern="^([-]|[.]|[-.]|[0-9])[0-9]*[.]*[0-9]+$";
String strValidIntegerPattern="^([-]|[0-9])[0-9]*$";
Regex objNumberPattern =new Regex("(" + strValidRealPattern +")|(" + strValidIntegerPattern + ")");

return !objNotNumberPattern.IsMatch(strNumber) &&
!objTwoDotPattern.IsMatch(strNumber) &&
!objTwoMinusPattern.IsMatch(strNumber) &&
objNumberPattern.IsMatch(strNumber);
}

b)
public static bool IsNumeric(string value)
{
return Regex.IsMatch(value, @"^[+-]?\d*[.]?\d*$");
}
public static bool IsInt(string value)
{
return Regex.IsMatch(value, @"^[+-]?\d*$");
}
public static bool IsUnsign(string value)
{
return Regex.IsMatch(value, @"^\d*[.]?\d*$");
}
方案三：遍历
a)
public bool isnumeric(string str)
{
    char[] ch=new char[str.Length];
    ch=str.ToCharArray();
    for(int i=0;i<ch.Length;i++)
    {
        if(ch[i]<48 || ch[i]>57)
            return false;
    }
    return true;
}

b)
public bool IsInteger(string strIn) {
bool bolResult=true;
if(strIn=="") {
bolResult=false;
}
else {
foreach(char Char in strIn) {
if(char.IsNumber(Char))
continue;
else {
bolResult=false;
break;
}
}
}
return bolResult;
}

c)
public static bool isNumeric(string inString)
{
inString=inString.Trim();
bool haveNumber=false;
bool haveDot=false;
for(int i=0;i<inString.Length;i++)
{
if (Char.IsNumber(inString[i]))
{
haveNumber=true;
}
else if(inString[i]=='.')
{
if (haveDot)
{
return false;
}
else
{
haveDot=true;
}
}
else if(i==0)
{
if(inString[i]!='+'&&inString[i]!='-')
{
return false;
}
}
else
{
return false;
}
if(i>20)
{
return false;
}
}
return haveNumber;
}
}

方案四：改写vb的IsNumeric源代码(执行效率不高)

//主调函数
public static bool IsNumeric(object Expression)
{
      bool flag1;
      IConvertible convertible1 = null;
      if (Expression is IConvertible)
      {
            convertible1 = (IConvertible) Expression;
      }
      if (convertible1 == null)
      {
            if (Expression is char[])