MySQL phpMyAdmin 从外键构建 JOIN

Question

我正在使用 LAMP 并通过 phpMyAdmin 维护 MySQl。每次我想在 InnoDB 表上构建多表查询时，我都厌倦了输入 JOIN。

有谁知道从外键生成 JOIN 语句的某些功能、插件、方法或资源？

Answer 1

我最终编写了一个查询，从所有外键构建 JOIN 语句：

CREATE VIEW Joins_All AS 
SELECT 
    concat(" JOIN ",TABLE_NAME," ON ",
    TABLE_NAME,".",COLUMN_NAME," = ",
    REFERENCED_TABLE_NAME,".", REFERENCED_COLUMN_NAME) 
    AS myJoin 
  FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
  WHERE REFERENCED_TABLE_NAME IS NOT NULL 

UNION

SELECT  
    concat(" JOIN ",REFERENCED_TABLE_NAME," ON ",
    TABLE_NAME,".",COLUMN_NAME," = ",
    REFERENCED_TABLE_NAME,".", REFERENCED_COLUMN_NAME) 
    AS myJoin 
  FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
  WHERE REFERENCED_TABLE_NAME IS NOT NULL 

ORDER BY myJoin  ASC

它会生成这样的东西：

在 Area_Tare.Id_Area = Area.Id_Area 上加入区域

加入 Area_Tare ON Area_Tare.Id_Area = Area.Id_Area

在 Insect.Id_Insect_Class = Insect_Class.Id_Insect_Class 上加入昆虫

在 Insect_Insumo_Formula_Ingredient.Id_Insect = Insect.Id_Insect 上加入昆虫

我保存了这个文件并复制/粘贴适当的行，如果我需要它们，或者我用所需表的名称过滤视图

这可以很容易地在 php 中进行调整（我目前不会这样做）以创建基于外键的 GUI 查询构建器。任何决定这样做的人，请随时使用/修改此代码。