Mysql的选择查询数据接合

Question

这是我的第一查询

SELECT 
    tax_productid, 
    CONCAT(GROUP_CONCAT(tax_code SEPARATOR " "), " ", sum(tax_iva), "%") AS tax 
FROM tb_tax t group by tax_productid order by tax_productid, prd_type desc; 

返回施加在产品的适当税（多税类型可以应用在单一的产品）

输出数据等。

tax_productid  tax 
========================= 
1     tax1 tax2 10% 
2     tax1 tax2 20% 
3     tax1 tax2 30% 

现在有一个另一个表product和查询就像

SELECT * FROM product ORDER BY productName; 

回报10产品记录。

我想要得到的产品明智的纳税排行榜，但如果没有税收秀空

SELECT 
    t1.*, 
    CONCAT(GROUP_CONCAT(tax_code SEPARATOR " "), " ", sum(tax_iva), "%") AS tax 
FROM 
    product t1, tb_tax t2 
WHERE 
    t2.tax_product = t1.id; 

此查询只返回3行。我想列出所有产品，如果表tb_tax没有产品条目，则返回null。所以我该如何实现这一点。

Answer 1

您的查询使用INNER JOIN，你应该使用LEFT JOIN这种情况：

SELECT 
    t1.*, 
    t2.tax  
FROM 
    product t1 
LEFT JOIN (
    SELECT 
     tax_productid, 
     CONCAT(GROUP_CONCAT(tax_code SEPARATOR " "), " ", sum(tax_iva), "%") AS tax 
    FROM tb_tax 
    GROUP BY tax_productid 
) t2 ON t2.tax_productid = t1.id 

Answer 2

使用LEFT JOIN。

SELECT 
    t1.*, 
    CONCAT(GROUP_CONCAT(tax_code SEPARATOR " "), " ", sum(tax_iva), "%") AS tax 
FROM 
    product t1 
LEFT JOIN 
    tb_tax t2 
ON 
    t2.tax_product = t1.id;