php - 搜索引擎的 SQL 查询 (PHP/MySQL)

Question

我正在为我的网络项目 (PHP/MySQL) 开发一个搜索引擎。 用户应该能够通过特定条件找到酒店

1. 地点
2. 设施
3. 语言

表格如下

1. tblHotels *hotel_id* PK hotel_name
2. tblLocations *location_id* PK location_name
3. tblFacilities *facility_id* PK facility_name
4. tblLanguages *language_id* PK language_name
5. tblHotelLocations *location_id* FK *hotel_id* FK
6. tblHotelFacilities *facility_id* FK *hotel_id* FK
7. tblHotelLanguages *language_id* FK *hotel_id* FK

例如，用户想要搜索位于旧金山 (*location_id*) 的酒店 免费 Wi-Fi (*facility_id*) 和泳池 (*facility_id*) 工作人员会说英语 (*language_id*) 和西类牙语 (*language_id*)

显然，用户可以更改搜索条件，例如将早餐添加到设施中或将德语添加到语言中。 在这种情况下，搜索结果应反射(reflect)所有条件均成立的酒店。

或者他可能不选择任何设施(语言等)，在这种情况下，只应返回根本没有设施的酒店。

是否可以通过一个查询来完成？我很确定它与 INNER JOIN 有某种联系。但我只是卡住了 :( 有什么线索吗？

提前致谢!

更新:附加数据库架构

/*
-- Query: desc tblHotels
-- Date: 2011-11-02 09:37
*/

INSERT INTO `tblhotels` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('hotel_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tblhotels` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('hotel_name','varchar(45)','YES','UNI',NULL,'');


/*
-- Query: desc tblFacilities
-- Date: 2011-11-02 09:37
*/

INSERT INTO `tblfacilities` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('facility_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tblfacilities` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('facility_name','varchar(45)','YES','UNI',NULL,'');

/*
-- Query: desc tblLocations
-- Date: 2011-11-02 09:37
*/

INSERT INTO `tbllocations` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('location_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tbllocations` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('location_name','varchar(45)','YES','UNI',NULL,'');


/*
-- Query: desc tblLanguages
-- Date: 2011-11-02 09:37
*/

INSERT INTO `tbllanguages` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('language_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tbllanguages` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('language_name','varchar(45)','YES','UNI',NULL,'');


/*
-- Query: desc tblHotelFacilities
-- Date: 2011-11-02 09:39
*/

INSERT INTO `tblhotelfacilities` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('tblHotelFacilities_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tblhotelfacilities` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('facility_id','int(11)','YES','',NULL,'');

INSERT INTO `tblhotelfacilities` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('hotel_id','int(11)','YES','',NULL,'');


/*
-- Query: desc tblHotelLocations
-- Date: 2011-11-02 09:39
*/

INSERT INTO `tblhotellocations` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('tblHotelLocations_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tblhotellocations` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('location_id','int(11)','YES','',NULL,'');

INSERT INTO `tblhotellocations` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('hotel_id','int(11)','YES','',NULL,'');


/*
-- Query: desc tblHotelLanguages
-- Date: 2011-11-02 09:39
*/

INSERT INTO `tblhotelanguages` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('tblHotelLanguages_id','int(11)','NO','PRI',NULL,'auto_increment');

INSERT INTO `tblhotelanguages` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('language_id','int(11)','YES','',NULL,'');

INSERT INTO `tblhotelanguages` (`Field`,`Type`,`Null`,`Key`,`Default`,`Extra`) VALUES ('hotel_id','int(11)','YES','',NULL,'');

Answer 1

Select *
from tbHotels h, tblFacilities f, tblHotelFacilities hf
where f.facilitiesName IN (?,?,?) AND hf.facilityId = f.id 
AND hf.hotelId = h.id

这是一个仅针对设施的查询，您可以添加位置和语言....

?的可以这样加，

假设您有一个包含用户输入的设施名称的 ArrayList。

String query = " Select * " +
"from tbHotels h, tblFacilities f, tblHotelFacilities hf"+
"where f.facilitiesName IN <facilities> AND hf.facilityId = f.id "+
"AND f.hotelId = hf.hotelId";

String x = "";
int length = ArrayListOfFacilityNames.length();
while(length>0) {
  x=x+"?, ";
  length = length-1;
}
//remove last "," from string x

x=x.substring(0, x.length -1);

    //paste this string of facilities in main query 

   query = query.replace("<facilities>",x);

   // now ur query contains number of question marks equivalent to number of facilities      entered by user

   //make prepares statement with String query

   Iterator fnames = ArrayListOfFacilitiesNames.Iterator();
   int i=0;
   while(fnames.hasNext())
   {
     String name = fname.next();
     preparesStatement.setString(i,name)
     i++;
   }