为什么我的ifnull()函数不能工作?

8iwquhpp  于 2021-06-20  发布在  Mysql
关注(0)|答案(4)|浏览(496)
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;

这是我的查询sql,full\ u province是一个包含中国所有省份的表,project\ u name\ u data是指主数据。现在我要计算每个省有多少记录,没有记录的省给0。但是我的sql只返回包含记录的省份。里面有错误吗?
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

预期输出:

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
k2fxgqgv

k2fxgqgv1#

我会代替 sum(case when f.province is not null then 1 else 0 end)

2lpgd968

2lpgd9682#

IFNULL 可以移除,因为如果 d.province 价值be
NULL COUNT 不会累积。会回来的 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样本数据。
您可以尝试此查询。

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公司
[结果]:

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

bkhjykvo3#

因为count永远不会返回null值。

huwehgph

huwehgph4#

替换:

IFNULL(COUNT(f.province), 0) AS count

使用:

sum(case when d.location is not null then 1 else 0 end)

可能效果更好—然后计算主表中链接到省表的非空位置的数量,如果(由于正确的联接)位置值为空,则得到零。

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