LEFT JOIN用法有什么问题？

Question

我对使用LEFT JOIN有一点疑问。

我有这个查询：

SELECT
    LD.id                                   AS disease_id,
    LD.disease_name_en                      AS disease_name_en,
    LDN.disease_name                        AS disease_name,
    LDN.description                         AS description,
    LD.image_link                           AS image_link
FROM LsDiseases                             AS LD
LEFT JOIN LsDiseaseName                    AS LDN
      ON LDN.disease_id = LD.id

WHERE 
     LD.id = 1 
AND
     LDN.language_id = 1

返回这样的输出：

disease_id           disease_name_en          disease_name             description      image_link                                                                                                                                                                                                                                                     
----------------------------------------------------------------------------------------------------------------------------------------
1                    Newcastle disease        Newcastle disease        TEST             https://firebasestorage.googleapis.com/XXX

如您所见，这些字段是：

• disease_name_en
• disease_name
• 描述

来自LsDiseaseName ， LsDiseaseName是一个包含将记录转换为LsDisease表的表。

好吧，行得通。 我用一个LEFT JOIN，因为我的想法是：“如果存在翻译成LsDiseaseName表（用于LsDisease表的记录），这些领域将进行增值，如果翻译不存在，只返回信息到LsDisease表和LsDiseaseName的前一个字段将使用空值进行赋值。

问题在于changint这个WHERE子句：

LDN.language_id = 3

使用我没有翻译语言的ID

我得到一个空的结果集。 那么，我的推理出了什么问题？ 我想念什么？

Answer 1

在WHERE子句中放置LDN.Language_id = 3会将左联接的含义更改为内部联接（因为它不允许使用空值）

将该子句移至join语句

LEFT JOIN LsDiseaseName                    AS LDN
      ON LDN.disease_id = LD.id
      AND LDN.Language_id = 3

Answer 2

将此行移到左连接中

LEFT JOIN LsDiseaseName                    AS LDN
      ON LDN.disease_id = LD.id
      AND LDN.language_id = 3

Answer 3

实际上，首先要应用联接，然后过滤id = 3的最终结果集，这就是获取空结果集的原因。

您应该在加入自身时进行语言过滤。

SELECT
    LD.id                                   AS disease_id,
    LD.disease_name_en                      AS disease_name_en,
    LDN.disease_name                        AS disease_name,
    LDN.description                         AS description,
    LD.image_link                           AS image_link
FROM LsDiseases                             AS LD
LEFT JOIN LsDiseaseName                    AS LDN
      ON LDN.disease_id = LD.id AND LDN.Language_id = 3

WHERE 
     LD.id = 1