在查找特定事例时，如何简化过多 IF 语句的使用-解网

问：

我有代码，用户将从 5 个下拉菜单中选择不同的温度计，这些可以是：“无”、“名称”、“经理”、“访问级别”、“居民”和“位置”，这些将与搜索以获取输入值的 jtextfields 匹配。但是，为了获得这些值，我基本上做了很多if语句来确定值的组合，例如Name，Manager和Location一起将给出SQL语句：query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND propertyManager = ? AND propertyLocation = ?";

这导致了 if 语句的混乱，有什么方法可以简化它吗？

它有效，但是要实际创建它，我必须为不同的组合做数百行，并且以后很难扩展。但是，这是我当前的一些代码：

`if (searchingForName == true) {
          if (searchingForManager == true) {
              if (searchingForAccessLevel == true) { // final lvl 1
                   query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND propertyManager = ? AND accessLevelRequired = ?"; 
                   query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND propertyManager = ? AND accessLevelRequired = ?"; 
              }else if ((searchType1.equals("Inhabitant") || searchType2.equals("Inhabitant") || searchType3.equals("Inhabitant"))) { // final lvl 2
                query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND propertyManager = ? AND propertyInhabitant = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND propertyManager = ? AND propertyInhabitant = ?"; 
            }else if ((searchType1.equals("Location") || searchType2.equals("Location") || searchType3.equals("Location"))) { // finaly lvl 3
                query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND propertyManager = ? AND propertyLocation = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND propertyManager = ? AND propertyLocation = ?"; 
            }
          }  else if ((searchType1.equals("AccessLevel") || searchType2.equals("AccessLevel") || searchType3.equals("AccessLevel"))) { // sub lvl 2
               if ((searchType1.equals("Inhabitant") || searchType2.equals("Inhabitant") || searchType3.equals("Inhabitant"))) { //final lvl 1
                query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND accessLevelRequired = ? AND propertyInhabitant = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND accessLevelRequired = ? AND propertyInhabitant = ?"; 
            }else if ((searchType1.equals("Location") || searchType2.equals("Location") || searchType3.equals("Location"))) { // final lvl 2
                query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND accessLevelRequired = ? AND propertyLocation = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND accessLevelRequired = ? AND propertyLocation = ?"; 
            }
          } else if ((searchType1.equals("Inhabitant") || searchType2.equals("Inhabitant") || searchType3.equals("Inhabitant"))) { // sub lvl 3
               if ((searchType1.equals("Location") || searchType2.equals("Location") || searchType3.equals("Location"))) { // final lvl 1
                query = "SELECT * FROM propertyInfo WHERE propertyName = ? AND propertyInhabitant = ? AND propertyLocation = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyName = ? AND propertyInhabitant = ? AND propertyLocation = ?"; 
          }
               
      //}
      }else if ((searchType1.equals("Manager") || searchType2.equals("Manager") || searchType3.equals("Manager"))) { // main lvl 2
          
          
          if ((searchType1.equals("AccessLevel") || searchType2.equals("AccessLevel") || searchType3.equals("AccessLevel"))) { // sub lvl 1
              
              
            if ((searchType1.equals("Inhabitant") || searchType2.equals("Inhabitant") || searchType3.equals("Inhabitant"))) {// final lvl 1
                   query = "SELECT * FROM propertyInfo WHERE propertyManager = ? AND accessLevelRequired = ? AND propertyInhabitant = ?"; 
                   query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE   propertyManager = ? AND accessLevelRequired = ? AND propertyInhabitant = ?"; 
                   
            }else if ((searchType1.equals("Location") || searchType2.equals("Location") || searchType3.equals("Location"))) { // final lvl 2
                query = "SELECT * FROM propertyInfo WHERE propertyManager = ? AND accessLevelRequired = ? AND propertyLocation = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyManager = ? AND accessLevelRequired = ? AND propertyLocation = ?"; 
            }
          }  else if ((searchType1.equals("Inhabitant") || searchType2.equals("Inhabitant") || searchType3.equals("Inhabitant"))) { // sub lvl 2
                if ((searchType1.equals("Location") || searchType2.equals("Location") || searchType3.equals("Location"))) { // final lvl 1
                query = "SELECT * FROM propertyInfo WHERE propertyManager = ? AND propertyInhabitant = ? AND propertyLocation = ?"; 
                query2 = "SELECT COUNT(*) AS recordCount FROM propertyInfo WHERE  propertyManager = ? AND propertyInhabitant = ? AND propertyLocation = ?"; 
            }
          }
        `  } 
      }

Java 简化

// Define the base query
String query = "SELECT * FROM propertyInfo WHERE 1 = 1";

// Initialize a list to store query parameters
List<Object> parameters = new ArrayList<>();

// Check if Name is selected
if (searchingForName) {
    query += " AND propertyName = ?";
    parameters.add(nameValue);
}

// Check if Manager is selected
if (searchingForManager) {
    query += " AND propertyManager = ?";
    parameters.add(managerValue);
}

// Check if AccessLevel is selected
if (searchingForAccessLevel) {
    query += " AND accessLevelRequired = ?";
    parameters.add(accessLevelValue);
}

// Check if Inhabitant is selected
if (searchingForInhabitant) {
    query += " AND propertyInhabitant = ?";
    parameters.add(inhabitantValue);
}

// Check if Location is selected
if (searchingForLocation) {
    query += " AND propertyLocation = ?";
    parameters.add(locationValue);
}

对于每个选定的参数（例如，Name、Manager 等），您可以动态地将相应的条件添加到查询字符串中，并将参数值添加到参数列表中。最后，您可以使用带有参数的预准备语句执行查询。

祝你有一个好的！

在查找特定事例时，如何简化过多 IF 语句的使用

How to simplify the use of too many IF statements when finding specific cases

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