C#判断一个String是否为数字类型

方案一：Try...Catch(执行效率不高)

{
    try
    {
        int var1=Convert.ToInt32 (oText);
        return true;
    }
    catch
    {
        return false;
    }
}</div>

方案二：正则表达式(推荐)

a)

{
    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*$");
}</div>

b)

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);
}</div>

方案三：遍历

a)

{
    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;
}</div>

b)

    bool bolResult=true;
    if(strIn=="") {
        bolResult=false;
    }
    else {
        foreach(char Char in strIn) {
            if(char.IsNumber(Char))
                continue;
            else {
                bolResult=false;
                break;
            }
        }
    }
    return bolResult;
}</div>

c)

{
    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;
}</div>

方案四：改写vb的IsNumeric源代码(执行效率不高)