MYSQL-选择并将多个表连接到一个JSON

Question

您能告诉我如何解决此查询吗？我有3个数据库表。

1. 网络。
2. 产品。
3. 评论。

在添加“ comments”表（按预期方式工作之前，查询看起来像这样：]

SELECT networks.*, products.product, products.type 
FROM products 
JOIN networks ON products.id=networks.product_Id

现在我尝试使用附加表对其进行修改，但是它不起作用：

SELECT 
    networks.*, 
    products.product, 
    products.type, 
    comments.id AS comment_Id, 
    comments.newLine, 
    comments.lineComment AS comment, 
    comments.topComment, 
    comments.bottomComment 
JOIN networks ON products.id=networks.product_Id, 
ON comments.id=networks.comment_Id

如何解决此查询？

感谢帮手：）

Answer 1

我认为这是您要使用的语法：

SELECT 
    n.*, 
    p.product, 
    p.type,
    c.id AS comment_id, 
    c.newLine, 
    c.lineComment AS comment, 
    c.topComment, 
    c.bottomComment 
FROM products p
INNER JOIN networks n ON p.id = n.product_id
INNER JOIN comments c ON c.id = n.comment_id

即：多个联接的语法为FROM ... JOIN ... ON ... JOIN ... ON ...。

请注意，使用表别名会缩短查询的时间，并使读和写更加容易。

Answer 2

您加入了条件为[[products.id = networks.product_Id的networks表但是，对于注释表，您指定了条件[[comments.id = networks.comment_Id，但您忘记了加入comments表。

尝试

 SELECT networks.*, 
    products.product, 
    products.type, 
    comments.id AS comment_Id, 
    comments.newLine, 
    comments.lineComment AS comment, 
    comments.topComment, 
    comments.bottomComment 
    JOIN networks ON products.id=networks.product_Id
    JOIN comments ON comments.id=networks.comment_Id