条件处理的智能方式

Question

if (lineStyle == 5 || lineStyle == 21 || lineStyle == 82 || lineStyle == 83 || lineStyle == 3) {
    lineStyleString = "DOUBLE";
} else if (lineStyle == 6 || lineStyle == 35 || lineStyle == 39 || lineStyle == 30) {
    lineStyleString = "DOTTED" ;
} else if (lineStyle == 26 || lineStyle == 27  || lineStyle == 28  || lineStyle == 29 || lineStyle == 1) {
    lineStyleString = "SOLID";
} else if(lineStyle == -1) {
    lineStyleString = "NONE";
}

我们如何在Java中巧妙地处理这些代码呢？切换大小写、枚举或密钥对值模式？

Answer 1

你的情况看起来更随机。

Switch在这里看起来不错

switch(lineStyle) {
    case 5:
    case 21:
    case 82:
    case 83:
    case 3: 
     lineStyleString = "DOUBLE";   
     break;
    .. // add more cases
}

或者我更喜欢创建实用方法

public static boolean contains(int expecxted, int... vals) {
        for (int i = 0; i < vals.length; i++) {
            if (expecxted == vals[i]) {
                return true;
            }
        }
        return false;
    }

你可以像这样使用它

if (contains(lineStyle, 5,21,82,83,3)) {
    lineStyleString = "DOUBLE";
} else if(contains(lineStyle,6,35,39,30)){
   lineStyleString = "DOTTED";
}

Answer 2

一个缩进良好的交换机案例需要30行行(Netbeans建议自己进行转换，以便我可以计数)

所以我认为这种方式更好(9行)：

if (Arrays.asList(5, 21, 82, 83, 3).contains(lineStyle)) {
     lineStyleString = "DOUBLE";
} else if (Arrays.asList(6, 35, 39, 30).contains(lineStyle)) {
     lineStyleString = "DOTTED";
} else if (Arrays.asList(26, 27, 28, 29, 1).contains(lineStyle)) {
     lineStyleString = "SOLID";
}else if (lineStyle == -1) {
     lineStyleString = "NONE";
}

Answer 3

我更喜欢这样的枚举：

enum LineStyle {

  DOUBLE(3, 5, 21, 82, 83),
  DOTTED(6, 30, 35, 39),
  SOLID(1, 26, 27, 28, 29),
  NONE(-1);

  private final Set<Integer> types;

  private LineStyle(Integer... types) {
    this.types = Stream.of(types).collect(Collectors.toSet());
  }

  public static LineStyle of(int lineStyle) {
    return Stream.of(LineStyle.values())
      .filter(ls -> ls.types.contains(lineStyle))
      .findFirst().orElse(null);
  }
}

然后您可以简单地调用：LineStyle ls = LineStyle.of(lineStyle);