0 
public static enum pie {
    APPLE_PIE1(1, 250),
    PUMPKIN_PIE1(2, 300),
    OTHER_PIE1(3, 350),
    APPLE_PIE2(4, 400),
    OTHER_PIE2(5, 450),
    PUMPKIN_PIE2(6, 500),
    APPLE_PIE3(7, 550),
    ;
    private static Map<Integer, pie> pie = new HashMap<Integer, pie>();
    static {
        for(pie pie : pie.values()) {
            pie.put(pie.getId(), pie);
        }
    }

    public static pie forId(int id) {
        return pie.get(id);
    }

    private pie(int id, double exp) {
        this.id = id;
        this.exp = exp;
    }

    public int id;
    public double exp;


    System.out.println("My favorite type of pie is " + pie.toString().toLowerCase().replaceAll("_", " ").replaceAll("2", "").replaceAll("3", "").replaceAll("4", "") + ".");

示例代码被撕掉了，但是我如何让 '.replaceAll(" number ", "")' 用更简单的代码替换所有小于或等于 4 的数字？创建一个新的 .replaceAll 3 次是多余的。如果我做...：

    double number[] = {1, 2, 3, 4};

    pie.toString().toLowerCase().replaceAll("_", " ").replaceAll(number.toString(), "") + ".");

...我会得到一个令人讨厌的错误。

这是错误...

SEVERE: An error occurred in an executor service! The server will be halted immediately.
java.util.regex.PatternSyntaxException: Unclosed character class near index 9
[D@12452e8
        ^
at java.util.regex.Pattern.error(Unknown Source)
at java.util.regex.Pattern.clazz(Unknown Source)
at java.util.regex.Pattern.sequence(Unknown Source)
at java.util.regex.Pattern.expr(Unknown Source)
at java.util.regex.Pattern.compile(Unknown Source)
at java.util.regex.Pattern.<init>(Unknown Source)
at java.util.regex.Pattern.compile(Unknown Source)
at java.lang.String.replaceAll(Unknown Source)
4

1 回答 1

0

要修复您的错误，您应该替换

double number[] = {1, 2, 3, 4};

经过

String number = "[1, 2, 3, 4]";

但我不会使用 replaceAll 来做到这一点。相反，我会label以与其他字段类似的方式向您的枚举添加一个字段，并调用getLabel()而不是toString()，或覆盖toString()以返回标签：

public static enum pie {
    APPLE_PIE1(1, 250, "apple pie"),
    PUMPKIN_PIE1(2, 300, "pumpkin pie"),
    ...

    private int id;
    private double exp;
    private String label;

    private pie(int id, double exp, String label) {
        this.id = id;
        this.exp = exp;
        this.label = label;
    }

    public String getLabel() {
        return label;
    }

    // optional: I wouldn't do it as it would make debugging harder
    @Override
    public String toString() {
        return label;
    }
}
于 2013-01-03T23:09:17.953 回答