我通常使用以下习语来检查String是否可以转换为整数。

public boolean isInteger( String input ) {
    try {
        Integer.parseInt( input );
        return true;
    }
    catch( Exception e ) {
        return false;
    }
}

是我的错觉,还是你觉得这有点粗俗?还有什么更好的办法吗?


看看我的回答(带有基准,基于CodingWithSpike之前的回答),看看为什么我改变了立场,接受了Jonas Klemming对这个问题的回答。我认为大多数人都会使用这个原始代码,因为它实现起来更快,也更易于维护,但是当提供非整数数据时,它就慢了几个数量级。


当前回答

另一个选择:

private boolean isNumber(String s) {
    boolean isNumber = true;
    for (char c : s.toCharArray()) {
        isNumber = isNumber && Character.isDigit(c);
    }
    return isNumber;
}

其他回答

这是对Jonas代码的修改,用于检查字符串是否在转换为整数的范围内。

public static boolean isInteger(String str) {
    if (str == null) {
        return false;
    }
    int length = str.length();
    int i = 0;

    // set the length and value for highest positive int or lowest negative int
    int maxlength = 10;
    String maxnum = String.valueOf(Integer.MAX_VALUE);
    if (str.charAt(0) == '-') { 
        maxlength = 11;
        i = 1;
        maxnum = String.valueOf(Integer.MIN_VALUE);
    }  

    // verify digit length does not exceed int range
    if (length > maxlength) { 
        return false; 
    }

    // verify that all characters are numbers
    if (maxlength == 11 && length == 1) {
        return false;
    }
    for (int num = i; num < length; num++) {
        char c = str.charAt(num);
        if (c < '0' || c > '9') {
            return false;
        }
    }

    // verify that number value is within int range
    if (length == maxlength) {
        for (; i < length; i++) {
            if (str.charAt(i) < maxnum.charAt(i)) {
                return true;
            }
            else if (str.charAt(i) > maxnum.charAt(i)) {
                return false;
            }
        }
    }
    return true;
}

如果你不关心潜在的溢出问题,这个函数将比使用Integer.parseInt()快20-30倍。

public static boolean isInteger(String str) {
    if (str == null) {
        return false;
    }
    int length = str.length();
    if (length == 0) {
        return false;
    }
    int i = 0;
    if (str.charAt(0) == '-') {
        if (length == 1) {
            return false;
        }
        i = 1;
    }
    for (; i < length; i++) {
        char c = str.charAt(i);
        if (c < '0' || c > '9') {
            return false;
        }
    }
    return true;
}

当解释比表现更重要时

我注意到许多讨论都围绕着某些解决方案的效率,但没有一个讨论为什么字符串不是整数。而且,每个人似乎都认为数字“2.00”不等于“2”。从数学和人类的角度来说,它们是平等的(尽管计算机科学说它们不是,而且有充分的理由)。这就是为什么“Integer.”上面的parseInt”解决方案是弱的(取决于您的需求)。

无论如何,为了使软件更智能、更人性化,我们需要创造出能够像我们一样思考并能解释失败原因的软件。在这种情况下:

public static boolean isIntegerFromDecimalString(String possibleInteger) {
possibleInteger = possibleInteger.trim();
try {
    // Integer parsing works great for "regular" integers like 42 or 13.
    int num = Integer.parseInt(possibleInteger);
    System.out.println("The possibleInteger="+possibleInteger+" is a pure integer.");
    return true;
} catch (NumberFormatException e) {
    if (possibleInteger.equals(".")) {
        System.out.println("The possibleInteger=" + possibleInteger + " is NOT an integer because it is only a decimal point.");
        return false;
    } else if (possibleInteger.startsWith(".") && possibleInteger.matches("\\.[0-9]*")) {
        if (possibleInteger.matches("\\.[0]*")) {
            System.out.println("The possibleInteger=" + possibleInteger + " is an integer because it starts with a decimal point and afterwards is all zeros.");
            return true;
        } else {
            System.out.println("The possibleInteger=" + possibleInteger + " is NOT an integer because it starts with a decimal point and afterwards is not all zeros.");
            return false;
        }
    } else if (possibleInteger.endsWith(".")  && possibleInteger.matches("[0-9]*\\.")) {
        System.out.println("The possibleInteger="+possibleInteger+" is an impure integer (ends with decimal point).");
        return true;
    } else if (possibleInteger.contains(".")) {
        String[] partsOfPossibleInteger = possibleInteger.split("\\.");
        if (partsOfPossibleInteger.length == 2) {
            //System.out.println("The possibleInteger=" + possibleInteger + " is split into '" + partsOfPossibleInteger[0] + "' and '" + partsOfPossibleInteger[1] + "'.");
            if (partsOfPossibleInteger[0].matches("[0-9]*")) {
                if (partsOfPossibleInteger[1].matches("[0]*")) {
                    System.out.println("The possibleInteger="+possibleInteger+" is an impure integer (ends with all zeros after the decimal point).");
                    return true;
                } else if (partsOfPossibleInteger[1].matches("[0-9]*")) {
                    System.out.println("The possibleInteger=" + possibleInteger + " is NOT an integer because it the numbers after the decimal point (" + 
                                partsOfPossibleInteger[1] + ") are not all zeros.");
                    return false;
                } else {
                    System.out.println("The possibleInteger=" + possibleInteger + " is NOT an integer because it the 'numbers' after the decimal point (" + 
                            partsOfPossibleInteger[1] + ") are not all numeric digits.");
                    return false;
                }
            } else {
                System.out.println("The possibleInteger=" + possibleInteger + " is NOT an integer because it the 'number' before the decimal point (" + 
                        partsOfPossibleInteger[0] + ") is not a number.");
                return false;
            }
        } else {
            System.out.println("The possibleInteger="+possibleInteger+" is NOT an integer because it has a strange number of decimal-period separated parts (" +
                    partsOfPossibleInteger.length + ").");
            return false;
        }
    } // else
    System.out.println("The possibleInteger='"+possibleInteger+"' is NOT an integer, even though it has no decimal point.");
    return false;
}
}

测试代码:

String[] testData = {"0", "0.", "0.0", ".000", "2", "2.", "2.0", "2.0000", "3.14159", ".0001", ".", "$4.0", "3E24", "6.0221409e+23"};
int i = 0;
for (String possibleInteger : testData ) {
    System.out.println("");
    System.out.println(i + ". possibleInteger='" + possibleInteger +"' isIntegerFromDecimalString=" + isIntegerFromDecimalString(possibleInteger));
    i++;
}

这更短,但更短并不一定更好(它不会捕获超出范围的整数值,正如danatel的评论所指出的那样):

input.matches("^-?\\d+$");

就我个人而言,由于实现被保存在一个helper方法中,并且正确性胜过长度,所以我将使用类似于您所拥有的东西(减去捕获基Exception类而不是NumberFormatException)。

另一个选择:

private boolean isNumber(String s) {
    boolean isNumber = true;
    for (char c : s.toCharArray()) {
        isNumber = isNumber && Character.isDigit(c);
    }
    return isNumber;
}