方案一：Try...Catch(执行效率不高)
复制代码 代码如下:
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)
复制代码 代码如下:
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*$");
}
b)
复制代码 代码如下:
using System;
using System.Text.RegularExpressions;
public bool IsNumber(String strNumber)
{
Regex objNotNumberPattern=new Regex("[^0-9.-]");
Regex objTwoDotPattern=new Regex("[0-9]*[.][0-9]*[.][0-9]*");
Regex objTwoMinusPattern=new Regex("[0-9]*[-][0-9]*[-][0-9]*");
String strValidRealPattern="^([-]|[.]|[-.]|[0-9])[0-9]*[.]*[0-9]+$";
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);
}
方案三：遍历
a)
复制代码 代码如下:
public bool isnumeric(string str)
{
char[] ch=new char[str.Length];
ch=str.ToCharArray();
for(int i=0;i {
if(ch[i]<48 || ch[i]>57)
return false;
}
return true;
}
b)
复制代码 代码如下:
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)
复制代码 代码如下:
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)
复制代码 代码如下:
//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[])
{
Expression = new string((char[]) Expression);
}
else
{
return false;
}
}
TypeCode code1 = convertible1.GetTypeCode();
if ((code1 != TypeCode.String) && (code1 != TypeCode.Char))
{
return Utils.IsNumericTypeCode(code1);
}
string text1 = convertible1.ToString(null);
try
{
long num2;
if (!StringType.IsHexOrOctValue(text1, ref num2))
{
double num1;
return DoubleType.TryParse(text1, ref num1);
}
flag1 = true;
}
catch (Exception)
{
flag1 = false;
}
return flag1;
}//子函数
// return Utils.IsNumericTypeCode(code1);
internal static bool IsNumericTypeCode(TypeCode TypCode)
{
switch (TypCode)
{
case TypeCode.Boolean:
case TypeCode.Byte:
case TypeCode.Int16:
case TypeCode.Int32:
case TypeCode.Int64:
case TypeCode.Single:
case TypeCode.Double:
case TypeCode.Decimal:
{
return true;
}
case TypeCode.Char:
case TypeCode.SByte:
case TypeCode.UInt16:
case TypeCode.UInt32:
case TypeCode.UInt64:
{
break;
}
}
return false;
}
//-----------------
//StringType.IsHexOrOctValue(text1, ref num2))
internal static bool IsHexOrOctValue(string Value, ref long i64Value)
{
int num1;
int num2 = Value.Length;
while (num1 < num2)
{
char ch1 = Value[num1];
if (ch1 == '&')
{
ch1 = char.ToLower(Value[num1 + 1], CultureInfo.InvariantCulture);
string text1 = StringType.ToHalfwidthNumbers(Value.Substring(num1 + 2));
if (ch1 == 'h')
{
i64Value = Convert.ToInt64(text1, 0x10);
}
else if (ch1 == 'o')
{
i64Value = Convert.ToInt64(text1, 8);
}
else
{
throw new FormatException();
}
return true;
}
if ((ch1 != ' ') && (ch1 != '/u3000'))
{
return false;
}
num1++;
}
return false;
}
//----------------------------------------------------
// DoubleType.TryParse(text1, ref num1);
internal static bool TryParse(string Value, ref double Result)
{
bool flag1;
CultureInfo info1 = Utils.GetCultureInfo();
NumberFormatInfo info3 = info1.NumberFormat;
NumberFormatInfo info2 = DecimalType.GetNormalizedNumberFormat(info3);
Value = StringType.ToHalfwidthNumbers(Value, info1);
if (info3 == info2)
{
return double.TryParse(Value, NumberStyles.Any, info2, out Result);
}
try
{
Result = double.Parse(Value, NumberStyles.Any, info2);
flag1 = true;
}
catch (FormatException)
{
flag1 = double.TryParse(Value, NumberStyles.Any, info3, out Result);
}
catch (Exception)
{
flag1 = false;
}
return flag1;
}
方案五：直接引用vb运行库(执行效率不高)
方法：首先需要添加Visualbasic.runtime的引用
代码中Using Microsoft.visualbasic;
程序中用Information.isnumeric("ddddd");
以上就是C#判断一个String是否为数字类型的全部内容，推荐大家使用正则表达式的方法，比较简单且效率高。