Question

有3个不同的用户输入字段，如文本框和＆amp;在UI上下载。用户可以使用三个输入字段中的任何一个来搜索表，或者通过提供输入字段的不同组合来缩小结果集。我使用if else语句来使用这三个字段来实现我的逻辑，但问题是代码看起来非常如果在UI中添加了另外一个字段，那么很多if-else语句会更难看，那么我必须检查if else语句的更多组合。如果有人能够了解如何在这些场景中遵循最佳实践或更好的方法，我真的很感激。

我的代码：

      public boolean filterResults(UserInputs userInputs){

            String firstName = userInputs.getFirstName();
            String lastName = userInputs.getLastName();
            String phoneNumber = userInputs.getNumber();

                  //**search only by First Name**
                 if((firstName!=null && firstName!="") 
                    && (lastName==null || lastName=="" )
                   && phoneNumber==null || phoneNumber ==""){

                    //logic to narrow down the result set based on condition met input values
                     list =DAO.getService(firstName);
                      return true;
                       }else if((lastName!=null && lastName!="")// **search only by Last Name**
                                  && (firstName== null || firstName == "")
                                   && (phoneNumber==null || phoneNumber == "")){

                             //logic to narrow down the result set based on condition met input values
 list =DAO.getService(lastName);                          return true;
                       }else if((phoneNumber!=null && phoneNumber!="")// **Search Only by PhoneNumber**
                                               && (lastName==null || lastName=="" )
                                                && (firstName== null || firstName == "")){
        //logic to narrow down the result set based on condition met input values
                    list =DAO.getService(phoneNumber);                                                return true;
                               }else if((firstName!=null && firstName!="")// **Search by all params**.
                                      &&(lastName!=null && lastName!="")
                                       && (phoneNumber!=null && phoneNumber!="")){
        //logic to narrow down the result set based on condition met input values
                      list =DAO.getService(firstName, lastName, phoneNumber);
                                                       return true;
                            }else if((firstName!=null && firstName!="")//**Search by first&last name.**
                                && (lastName!=null || lastName!="" )
                                && (phoneNumber==null || phoneNumber =="")){
                           //logic to narrow down the result set based on condition met input values
                                 list =DAO.getService(firstName, lastName);

                                 return true;
                                 }else if (.......){// **Search by last name & phone number.**
                                       return true;
                                   }else if (.........){// **Search by phone number & first name.**
                                    return true;}



                     return false;

            }

Answer 1

现在，您的逻辑等同于这一行：

return (firstName != null && !firstName.isEmpty()) ||
    (lastName != null && !lastName.isEmpty()) ||
    (phoneNumber != null && !phoneNumber.isEmpty());

Answer 2

您似乎正在为搜索条件的每种可能组合实施搜索方法。你不应该这样做。相反，过滤元素的方法应该测试每个标准：

public List<Element> search(List<Element> allElements, UserInputs criteria) {
    List<Element> acceptedElements = new ArrayList<>();
    for (Element e : allElements) {
        if (isAccepted(e, criteria)) {
            acceptedElements.add(e);
        }
    }
    return acceptedElements;
}

private boolean isAccepted(Element e, UserInputs criteria) {
    String firstName = criteria.getFirstName();
    if (isCriterionFilled(firstName) && !hasFirstName(e, firstName)) {
        return false;
    }

    String lastName = criteria.getLastName();
    if (isCriterionFilled(lastName) && !hasLastName(e, lastName)) {
        return false;
    }

    String phone = criteria.getPhone();
    if (isCriterionFilled(phone) && !hasPhone(e, phone)) {
        return false;
    }

    return true;
}

替代if else语句或建议最佳方法

2 个答案: