这是一个Java 8版本的Jonas Klemming回答:

public static boolean isInteger(String str) {
    return str != null && str.length() > 0 &&
         IntStream.range(0, str.length()).allMatch(i -> i == 0 && (str.charAt(i) == '-' || str.charAt(i) == '+')
                  || Character.isDigit(str.charAt(i)));
}

测试代码:

public static void main(String[] args) throws NoSuchAlgorithmException, UnsupportedEncodingException {
    Arrays.asList("1231231", "-1232312312", "+12313123131", "qwqe123123211", "2", "0000000001111", "", "123-", "++123",
            "123-23", null, "+-123").forEach(s -> {
        System.out.printf("%15s %s%n", s, isInteger(s));
    });
}

测试代码的结果:

        1231231 true
    -1232312312 true
   +12313123131 true
  qwqe123123211 false
              2 true
  0000000001111 true
                false
           123- false
          ++123 false
         123-23 false
           null false
          +-123 false

你可以:

检查字符串是否为数字 检查它是否太长而不能被解析为long 检查结果的长值是否小到可以用整型表示

(假设您出于某种原因必须自己实现此功能:您可能应该首先查看com.google.common.primitives.Ints.tryParse(String)，并查看它是否足够适合您的目的(正如另一个答案所建议的那样)。)

// Credit to Rastislav Komara’s answer: https://stackoverflow.com/a/237895/1725151
private static final Pattern nonZero = Pattern.compile("^-?[1-9]\\d*$");

// See if `str` can be parsed as an `int` (does not trim)
// Strings like `0023` are rejected (leading zeros).
public static boolean parsableAsInt(@Nonnull String str) {
    if (str.isEmpty()) {
        return false;
    }
    if (str.equals("0")) {
        return true;
    }
    if (canParseAsLong(str)) {
        long value = Long.valueOf(str);
        if (value >= Integer.MIN_VALUE && value <= Integer.MAX_VALUE) {
            return true;
        }
    }
    return false;
}

private static boolean canParseAsLong(String str) {
    final int intMaxLength = 11;
    return str.length() <= intMaxLength && nonZero.matcher(str).matches();
}

这个方法也可以转换为返回Optional<Integer>，这样你就不必在客户端代码中解析两次字符串了(一次是为了检查是否可行，第二次是为了“真正地”执行它)。例如:

if (canParseAsLong(str)) {
    long value = Long.valueOf(str);
    if (value >= Integer.MIN_VALUE && value <= Integer.MAX_VALUE) {
        return Optional.of((int) value);
    }
}

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;
}

如何:

return Pattern.matches("-?\\d+", input);

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;
}

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++;
}