【转】C#判断一个字符串是否是数字

//判断一个string是否可以为数字

//方案一：Try...Catch(执行效率不高)
///
/// 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
///
///
///
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;
}