// only int
public static boolean isNumber(int num) 
{
    return (num >= 48 && c <= 57); // 0 - 9
}

// is type of number including . - e E 
public static boolean isNumber(String s) 
{
    boolean isNumber = true;
    for(int i = 0; i < s.length() && isNumber; i++) 
    {
        char c = s.charAt(i);
        isNumber = isNumber & (
            (c >= '0' && c <= '9') || (c == '.') || (c == 'e') || (c == 'E') || (c == '')
        );
    }
    return isInteger;
}

// is type of number 
public static boolean isInteger(String s) 
{
    boolean isInteger = true;
    for(int i = 0; i < s.length() && isInteger; i++) 
    {
        char c = s.charAt(i);
        isInteger = isInteger & ((c >= '0' && c <= '9'));
    }
    return isInteger;
}

public static boolean isNumeric(String s) 
{
    try
    {
        Double.parseDouble(s);
        return true;
    }
    catch (Exception e) 
    {
        return false;
    }
}

你可以使用NumberFormat#parse:

try
{
     NumberFormat.getInstance().parse(value);
}
catch(ParseException e)
{
    // Not a number.
}

这里有两种可能有效的方法。(不使用异常)。 注意:Java默认是值传递，String的值是String对象数据的地址。 所以，当你在做

stringNumber = stringNumber.replaceAll(" ", "");

您已将输入值更改为没有空格。 如果你愿意，可以去掉这条线。

private boolean isValidStringNumber(String stringNumber)
{
    if(stringNumber.isEmpty())
    {
        return false;
    }

    stringNumber = stringNumber.replaceAll(" ", "");

    char [] charNumber = stringNumber.toCharArray();
    for(int i =0 ; i<charNumber.length ;i++)
    {
        if(!Character.isDigit(charNumber[i]))
        {
            return false;
        }
    }
    return true;
}

这里是另一个方法，以防你想允许浮动 据称，这种方法允许表单中的数字通过 1123123123123123年.123 我刚做好，我想还需要进一步测试以确保它能正常工作。

private boolean isValidStringTrueNumber(String stringNumber)
{
    if(stringNumber.isEmpty())
    {
        return false;
    }

    stringNumber = stringNumber.replaceAll(" ", "");
    int countOfDecimalPoint = 0;
    boolean decimalPointPassed = false;
    boolean commaFound = false;
    int countOfDigitsBeforeDecimalPoint = 0;
    int countOfDigitsAfterDecimalPoint =0 ;
    int commaCounter=0;
    int countOfDigitsBeforeFirstComma = 0;

    char [] charNumber = stringNumber.toCharArray();
    for(int i =0 ; i<charNumber.length ;i++)
    {
        if((commaCounter>3)||(commaCounter<0))
        {
            return false;
        }
        if(!Character.isDigit(charNumber[i]))//Char is not a digit.
        {
            if(charNumber[i]==',')
            {
                if(decimalPointPassed)
                {
                    return false;
                }
                commaFound = true;
                //check that next three chars are only digits.
                commaCounter +=3;
            }
            else if(charNumber[i]=='.')
            {
                decimalPointPassed = true;
                countOfDecimalPoint++;
            }
            else
            {
                return false;
            }
        }
        else //Char is a digit.
        {
            if ((commaCounter>=0)&&(commaFound))
            {
                if(!decimalPointPassed)
                {
                    commaCounter--;
                }
            }

            if(!commaFound)
            {
                countOfDigitsBeforeFirstComma++;
            }

            if(!decimalPointPassed)
            {
                countOfDigitsBeforeDecimalPoint++;
            }
            else
            {
                countOfDigitsAfterDecimalPoint++;
            }
        }
    }
    if((commaFound)&&(countOfDigitsBeforeFirstComma>3))
    {
        return false;
    }
    if(countOfDecimalPoint>1)
    {
        return false;
    }

    if((decimalPointPassed)&&((countOfDigitsBeforeDecimalPoint==0)||(countOfDigitsAfterDecimalPoint==0)))
    {
        return false;
    }
    return true;
}

一个有效的方法避免尝试捕获和处理负数和科学符号。

Pattern PATTERN = Pattern.compile( "^(-?0|-?[1-9]\\d*)(\\.\\d+)?(E\\d+)?$" );

public static boolean isNumeric( String value ) 
{
    return value != null && PATTERN.matcher( value ).matches();
}

如果你们用下面的方法来检查:

public static boolean isNumeric(String str) {
    NumberFormat formatter = NumberFormat.getInstance();
    ParsePosition pos = new ParsePosition(0);
    formatter.parse(str, pos);
    return str.length() == pos.getIndex();
}

然后输入非常长的字符串会发生什么，比如我调用这个方法:

System.out.println(isNumeric("94328948243242352525243242524243425452342343948923"));

结果是“真”，也是一个太大的数字! 如果你使用regex来检查，同样的事情也会发生! 所以我宁愿使用“解析”方法来检查，就像这样:

public static boolean isNumeric(String str) {
    try {
        int number = Integer.parseInt(str);
        return true;
    } catch (Exception e) {
        return false;
    }
}

结果就如我所料!

我们可以尝试用("")即空格替换给定字符串中的所有数字，如果在此之后字符串的长度为零，那么我们可以说给定字符串只包含数字。 例子:

boolean isNumber(String str){
        if(str.length() == 0)
            return false; //To check if string is empty
        
        if(str.charAt(0) == '-')
            str = str.replaceFirst("-","");// for handling -ve numbers
    
        System.out.println(str);
        
        str = str.replaceFirst("\\.",""); //to check if it contains more than one decimal points
        
        if(str.length() == 0)
            return false; // to check if it is empty string after removing -ve sign and decimal point
        System.out.println(str);
        
        return str.replaceAll("[0-9]","").length() == 0;
    }