Question

我正在刷新一些MySQL技能（希望如此）。

我有这个问题：

SELECT stops.name, 
    stops.stop_id,  
    stop_times.trip_id,
    stop_times.departure_time,
    trips.route_id
FROM stops, routes, stop_times, trips
WHERE stops.stop_id = stop_times.stop_id AND trips.route_id = 2 AND stop_times.trip_id = 'SCHED255'
ORDER BY stops.stop_id ASC

这会返回正确的信息，但我会多次出现每条记录。我缺少什么想法？ TIA

Answer 1

FROM - 条款中有4个表格。因此，您至少需要3 JOINS才能避免多次出现数据。

基本规则：n tables =（n-1）JOINS。

+++++++++++ EDIT +++++++++++

在你的情况下它应该是这样的：

/*WITH EXCPLICIT JOINS*/
SELECT stops.name, stops.stop_id, stop_times.trip_id, stop_times.departure_time, trips.route_id
FROM stops s JOIN routes r ON s.col_name = r.col_name /*col_name is the name of the column that is defined as Contraint between these two*/
             JOIN stop_times st ON st.stop_id = s.stop_id
             JOIN trips t ON t.route_id = r.route_id
    AND t.route_id = 2 
    AND st.trip_id = 'SCHED255'
ORDER BY s.stop_id ASC

/*THE SAME THING WITH IMPLICIT JOINS*/
SELECT stops.name, stops.stop_id, stop_times.trip_id, stop_times.departure_time, trips.route_id
FROM stops AS s, routes AS r, stop_times AS st, trips AS t
WHERE s.col_name = r.col_name /*JOIN of stops and trips*/
    AND st.stop_id = s.stop_id /*JOIN of stop_times and stops*/
    AND t.route_id = r.route_id /*JOIN of trips and rutes*/
    AND t.route_id = 2 /*Other condition*/
    AND st.trip_id = 'SCHED255' /*Other condition*/
ORDER BY s.stop_id ASC

Answer 2

带有连接的查询可能看起来像这样...

SELECT s.name
     , s.stop_id
     , st.trip_id
     , st.departure_time
     , t.route_id
  FROM stops s
  JOIN routes r
    ON r.some_column = something
  JOIN stop_times st
    ON st.stop_id = s.stop_id 
  JOIN trips t
    ON t.some_column = some_other_table.some_column
 WHERE t.route_id = 2 
   AND st.trip_id = 'SCHED255'
 ORDER 
    BY s.stop_id ASC

MySQL - 为什么这个SELECT查询会提供多条记录

2 个答案: