当没有找到记录时，sql显示0

Question

我正在尝试在找不到该特定ID的记录时显示零。

图像是我想要实现的目标。现在我只得到他找到至少一条记录的行（所以3,4和5）。

有人可以告诉我我做错了什么。这是我的尝试（许多之一）：

编辑：

SELECT  statusses.statusses_namenl,
        IFNULL(COUNT(projectrecipes.fk_projectrecipes_projectid),0) AS CntRows
FROM recipes
RIGHT JOIN projectrecipes
ON projectrecipes.fk_projectrecipes_recipeid = recipes.id
LEFT JOIN statusses
ON recipes.fk_recipe_status = statusses.id
WHERE projectrecipes.fk_projectrecipes_projectid = 213
GROUP BY recipes.fk_recipe_status

编辑：

这是我得到的结果：

Answer 1

我认为你不需要使用CASE，也可以使用RIGHT join，所以如果Join中没有匹配，那么结果也会出现。尝试下面它应该工作。

SELECT  recipes.fk_recipe_status, COUNT(projectrecipes.fk_projectrecipes_projectid)
FROM projectrecipes
RIGHT JOIN recipes
ON projectrecipes.fk_projectrecipes_recipeid = recipes.id
WHERE projectrecipes.fk_projectrecipes_projectid = 213
GROUP BY recipes.fk_recipe_status

Answer 2

试试这个：

SELECT  recipes.fk_recipe_status, 
        IFNULL(COUNT(projectrecipes.fk_projectrecipes_projectid),0) AS CntRows
FROM projectrecipes
RIGHT JOIN recipes
ON projectrecipes.fk_projectrecipes_recipeid = recipes.id
WHERE projectrecipes.fk_projectrecipes_projectid = 213
GROUP BY recipes.fk_recipe_status

Answer 3

SELECT  recipes.fk_recipe_status,
        COALESCE(COUNT(projectrecipes.fk_projectrecipes_projectid),0) AS CntRows
FROM projectrecipes
RIGHT JOIN recipes
ON projectrecipes.fk_projectrecipes_recipeid = recipes.id
WHERE projectrecipes.fk_projectrecipes_projectid = 213
GROUP BY recipes.fk_recipe_status

Answer 4

我认为，加入和应该改变的地方。所以我接受了保罗的答案并修改：

SELECT  recipes.fk_recipe_status, COUNT(projectrecipes.fk_projectrecipes_projectid)

转动JOIN中的表格

FROM recipes
LEFT JOIN projectrecipes
ON projectrecipes.fk_projectrecipes_recipeid = recipes.id

GROUP BY recipes.fk_recipe_status

添加一个而不是在哪里

HAVING projectrecipes.fk_projectrecipes_projectid = 213
   OR COUNT(projectrecipes.fk_projectrecipes_projectid) = 0

对于某些配方（示例中为1和2），找不到projectrecipes中的数据。因此，如果您首先从projectrecipes中选择，则不会从配方中获取所有数据。

由于projectrecipes可能不包含结果中的数据，projectrecipes.fk_projectrecipes_projectid = 213将始终为false。如果没有找到数据，请检查是否找不到数据，例如count = 0。

Answer 5

您希望每个状态有一行，因此请从状态中选择。然后，您希望从每个状态的食谱和项目配方中计算记录，因此外部连接它们。将您的条件放在ON子句中，而不是在WHERE子句中，因为当没有匹配的记录时，fk_projectrecipes_projectid将为NULL。 （通过在where子句中使用外部联接列，将外部联接转换为内部联接。）

select s.statusses_namenl, count(*)
from statusses
left join recipes r on r.fk_recipe_status = s.id
left join projectrecipes pr on pr.fk_projectrecipes_recipeid = r.id and pr.fk_projectrecipes_projectid = 213
group by s.statusses_namenl;