MySQL Inner Join表基于列值

Question

假设我有一个表'stats'具有以下结构:
tableName | id | pageViews
tableName列对应于数据库中的单独表.

针对"统计信息"运行查询时,内部联接对tableName列结果的最佳方法是获取每个表的数据？
我想在foreach中运行动态选择然后合并结果.例如:

foreach($tableNames as $tableName) {
    $sql = "SELECT      *
            FROM        stats s
            INNER JOIN  $tableName tbl ON s.id = tbl.id
            WHERE       tableName = '$tableName'";
}

Answer 1

要拥有所有表的统计信息,可以使用UNION,具有2个或更多选择,每个表一个:

( SELECT s.*
       , table1.title AS name      --or whatever field you want to show
  FROM stats s
    JOIN $tableName1 table1
      ON s.id = table1.id
  WHERE tableName = '$tableName1'
)
UNION ALL
( SELECT s.*
       , table2.name AS name      --or whatever field you want to show
  FROM stats s
    JOIN $tableName2 table2
      ON s.id = table2.id
  WHERE tableName = '$tableName2'
)
UNION ALL
( SELECT s.*
       , table3.lastname AS name      --or whatever field you want to show
  FROM stats s
    JOIN $tableName3 table3
      ON s.id = table3.id
  WHERE tableName = '$tableName3'
)
;

使用Winfred的想法与LEFT JOINs.它会产生不同的结果,例如,其他表中的每个字段都在其自己的列中输出(并且会出现许多NULL).

SELECT s.*
     , table1.title      --or whatever fields you want to show
     , table2.name
     , table3.lastname   --etc
FROM stats s
  LEFT JOIN $tableName1 table1
    ON s.id = table1.id
      AND s.tableName = '$tableName1'
  LEFT JOIN $tableName2 table2
    ON s.id = table2.id
      AND s.tableName = '$tableName2'
  LEFT JOIN $tableName3 table3
    ON s.id = table3.id
      AND s.tableName = '$tableName3'
--this is to ensure that omited tables statistics don't appear
WHERE s.tablename IN
   ( '$tableName1'
   , '$tableName2'
   , '$tableName3'
   )
;