基于列值的MySQL内部联接表 [英] MySQL Inner Join table based on column value

问题描述

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

Suppose I have a table 'stats' with the following structure:
tableName | id | pageViews
The tableName column corresponds to separate tables in the database.

对"stats"运行查询时，对tableName列结果进行内部联接以获取每个表的数据的最佳方法是什么?
我当时正在考虑在foreach中运行动态选择，然后合并结果.例如:

When running a query against "stats", what would be the best way to inner join against the tableName column result to get each table's data?
I was thinking of running dynamic selects in a foreach and then merging the results. E.g.:

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

推荐答案

要获取所有表的统计信息，可以使用具有2个或更多选择的UNION，每个表一个:

To have all tables' statistics, you can use a UNION, with 2 or more selects, one for each table:

( 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'
)
;

---

在LEFT JOIN中使用Winfred的想法.它产生不同的结果，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).

---

Using Winfred's idea with LEFT JOINs. It produces different results, e.g. every field from the other tables is output in it's own column (and many NULLs occur).

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'
   )
;

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