Question

我在4张桌子，1张主桌子和3张小桌子上工作。主表称为“文档”，它是来自其他3个小表的引用和ID，它们是：重要性，自然，access_level。

e.g.

(Table1) document

document_id | title | file_name | importance_id | nature_id | access_level_id
-----------------------------------------------------------------------------
      1     | Food  | food.docx |       1       |     1     |       1        


(Table2) importance

importance_id | name
--------------------
       1      | High


(Table3) nature

nature_id | name
----------------
     1    | General


(Table4) nature

access_level_id | name
----------------------
        1       | Public

我想要做的是显示所有文档（使用SELECT），并将其ID替换为相应的名称..像这样：

document_id | title | file_name | importance |  nature  | access_level
-----------------------------------------------------------------------------
      1     | Food  | food.docx |    High    |  General |    Public

我尝试使用此查询，但效果不佳：

SELECT `document_id`, `title`, `file_name`, `date_of_archiving`, `duration_of_life`, importance.name as importance, nature.name as nature, access_level.name as access_level
FROM document, importance, nature, access_level
WHERE importance.importance_id = document.document_id
AND nature.nature_id = document.nature_id
AND access_level.access_level_id = document.access_level_id;

那我怎么能做到这一点？

提前致谢：）

Answer 1

使用JOINS而不是您正在使用的过时语法。调试更清晰，更容易。并保存键入使用表别名。如下：

SELECT 
  document_id, 
  title, 
  file_name, 
  date_of_archiving, 
  duration_of_life, 
  i.NAME AS [importance],
  n.name as [nature], 
  al.name as [access_level]
FROM 
  DOCUMENT D
  JOIN importance i ON i.Importance_id = D.importance_id
  JOIN nature n ON n.nature_id = D.nature_id
  JOIN access_level al on al.access_level_id = D.access_level_id

Answer 2

如果希望行不返回空值

，请使用内部联接

SELECT 
  document_id, 
  title, 
  file_name, 
  date_of_archiving, 
  duration_of_life, 
  i.NAME AS [importance],
  n.name as [nature], 
  al.name as [access_level]
FROM 
  DOCUMENT D
  INNER JOIN importance i ON i.Importance_id = D.importance_id
  INNER JOIN nature n ON n.nature_id = D.nature_id
  INNER JOIN access_level al on al.access_level_id = D.access_level_id

Answer 3

使用此代码：

SELECT
  Document_id,
  Title,
  File_name,
  i.name AS Importance,
  n.name AS Nature,
  s.name AS Access_level
FROM Document d
INNER JOIN importance i ON i.importance_id = d.importance_id
INNER JOIN nature n     ON n.nature_id = d.nature_id
INNER JOIN nature s     ON s.access_level_id = d.access_level_id

如何用SQL中的相应值替换ID？

3 个答案: