java - 降低圈复杂度，多个if语句java

Question

我还有很多没有包含的 else-if 条件。我如何重构它以降低圈复杂度？

if (ONE.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setOne(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}
else if (TWO.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setTwo(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}
else if (THREE.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setThree(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}
else if (FOUR.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setFour(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}
else if (FIVE.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setFive(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}
else if (SIX.equalsIgnoreCase(eachTag.getNodeName()))
{
    myclassDto.setSix(formatter
            .getElementValueAfterNullCheckWithTrim((Element) eachTag));
}

如何在 java 中降低此函数的圈复杂度？

Answer 1

你的代码会更容易阅读，同时它也同样“复杂”(虽然实际上并没有那么复杂)，如果你:

• 使用 switch 语句
• 预先提取 getNodeName() 的值

• 预先提取getElementValueAfterNullCheckWithTrim()返回的值

  String value = formatter.getElementValueAfterNullCheckWithTrim((Element) eachTag);
  String nodeName = eachTag.getNodeName();
  
  switch (nodeName) {
      case ONE:
          myclassDto.setOne(value);
          break;
      case TWO:
          myclassDto.setTwo(value);
          break;
      ...
  }
  

编辑: 您可能想要重构 DTO 以使其更易于使用，例如

  myclassDto.setValue(nodeName, value)