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我有以下代码部分：

第一种方法：

public ArrayList<String> getTestFilesForOrderNumber(Integer testStepOrderNumber, Integer testOrderNumber,
      String version)
  {
    NodeList list = documentTest.getElementsByTagName(XmlElements.TESTSTEPS);
    ArrayList<String> files = new ArrayList<>();
    for (int i = 0; i < list.getLength(); i++)
    {
      Element e = (Element) list.item(i);
      if (e.getAttribute(XmlAttributes.VERSION).equalsIgnoreCase(version))
      {
        list = e.getElementsByTagName(XmlElements.TESTSTEP);
        for (int l = 0; l < list.getLength(); l++)
        {
          e = (Element) list.item(l);
          if (e.getAttribute(XmlAttributes.ORDER_NUMBER).equals(testStepOrderNumber.toString()))
          {
            NodeList testsList = e.getElementsByTagName(XmlElements.TEST);
            for (int j = 0; j < testsList.getLength(); j++)
            {
              Element e2 = (Element) testsList.item(j);
              if (e2.getAttribute(XmlAttributes.ORDER_NUMBER).equals(testOrderNumber.toString()))
              {
                NodeList fileList = e2.getElementsByTagName(XmlElements.FILE);
                for (int k = 0; k < fileList.getLength(); k++)
                {
                  Element e3 = (Element) fileList.item(k);
                  if (e3.getParentNode().getNodeName().equals(XmlElements.FILES))
                  {
                    files.add(e3.getFirstChild().getNodeValue());
                  }
                }
              }
            }
          }
        }
      }
    }

    return files;
  }

第二种方法：

public String getTestDescription(String langCode, Integer testStepOrderNumber, Integer testOrderNumber, String version)
  {
    NodeList list = documentTest.getElementsByTagName(XmlElements.TESTSTEPS);
    for (int i = 0; i < list.getLength(); i++)
    {
      Element e = (Element) list.item(i);
      if (e.getAttribute(XmlAttributes.VERSION).equalsIgnoreCase(version))
      {
        list = e.getElementsByTagName(XmlElements.TESTSTEP);
        for (int l = 0; l < list.getLength(); l++)
        {
          e = (Element) list.item(l);
          if (e.getAttribute(XmlAttributes.ORDER_NUMBER).equals(testStepOrderNumber.toString()))
          {
            NodeList testsList = e.getElementsByTagName(XmlElements.TEST);
            for (int j = 0; j < testsList.getLength(); j++)
            {
              Element e2 = (Element) testsList.item(j);
              if (e2.getAttribute(XmlAttributes.ORDER_NUMBER).equals(testOrderNumber.toString()))
              {
                NodeList testList = e2.getElementsByTagName(langCode.toUpperCase());
                for (int k = 0; k < testList.getLength(); k++)
                {
                  Element e3 = (Element) testList.item(k);
                  if (e3.getParentNode().getNodeName().equals(XmlElements.DESCRIPTION))
                  {
                    return testList.item(k).getFirstChild().getNodeValue();
                  }
                }
              }
            }
          }
        }
      }
    }

    return "ERROR_NO_TEST_DESC";
  }

现在，如您所见，第一个 for 循环完全相同，但方法返回值和方法参数却不同。现在我不知道如何减少这个重复的代码。

先感谢您。

4

1 回答 1

2

似乎使用 XPath 可能会有所帮助。请查看w3schools 的教程和java 中的一些使用示例

于 2013-10-04T08:03:50.407 回答