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mysql - 为什么我的 IfNull() 函数不起作用?

转载 作者:搜寻专家 更新时间:2023-10-30 21:49:09 29 4
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SELECT
f.province AS loc,
IFNULL(COUNT(f.province), 0) AS count
FROM
project_name_data d
RIGHT JOIN full_province f ON (
TRIM(
REPLACE (
REPLACE (
SUBSTRING_INDEX(d.location, '|', 1),
'省',
''
),
'壮族',
''
)
) = f.province
)
GROUP BY
loc;

这是我查询的Sql,full_province是一张包含中国所有省份的表,project_name_data是指主要数据。现在我想计算每个省有多少记录,对于没有任何记录的省,给它们0。但是我的Sql只返回包含记录的省份。有没有错误?

举个不恰当的例子:

-- main table
id location comment other columns
1 Los Angeles aaa ...
2 New York bbb ...
3 Cambridge ccc ...
4 Philadelphia ddd ...
5 New York eee ...
6 Cambridge fff ...

--full_province table
id location
1 Los Angeles
2 New York
3 Cambridge
4 Philadelphia
5 Beijing
6 Tokyo
7 Barcelona
8 Paris
9 Toronto

期望输出:

      location      count
1 Los Angeles 1
2 New York 2
3 Cambridge 2
4 Philadelphia 1
5 Beijing 0
6 Tokyo 0
7 Barcelona 0
8 Paris 0
9 Toronto 0

最佳答案

IFNULL 可以删除,因为如果 d.province 值为 NULL COUNT 将不会累加。将返回 0

SELECT
f.province AS loc,
COUNT(d.province) AS count
FROM
project_name_data d
RIHGT JOIN full_province f ON (
TRIM(
REPLACE (
REPLACE (
SUBSTRING_INDEX(d.location, '|', 1),
'省',
''
),
'壮族',
''
)
) = f.province
)
GROUP BY f.province;

编辑

我看到你添加了一些示例数据。

你可以试试这个查询。

SELECT
f.location AS loc,
COUNT(d.location) AS CNT
FROM
project_name_data d
right JOIN full_province f
ON d.location= f.location
GROUP BY f.location
ORDER BY d.id desc

sqlfiddle

[结果]:

|          loc | CNT |
|--------------|-----|
| Philadelphia | 1 |
| Cambridge | 2 |
| New York | 2 |
| Los Angeles | 1 |
| Beijing | 0 |
| Barcelona | 0 |
| Toronto | 0 |
| Tokyo | 0 |
| Paris | 0 |

关于mysql - 为什么我的 IfNull() 函数不起作用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51472819/

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