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基于两个具有值替换的表创建视图

[英]Create a view based on two tables with value substitution

我在查询中挣扎,我不确定该怎么做。

我想基于表original_data_table创建一个视图my_view ,其中每当指定一个ID时,每个值都将由来自表replace_data_table的非空值替换

original_data_tablereplacement_data_table都具有相同的结构,但具有不同的值。

我尝试在查询中使用JOIN,但不确定是否要这样做。

SELECT * FROM original_data_table AS o
LEFT JOIN replacement_data_table AS r
ON o.id = r.id

original_data_table

id         name        value

1          David       10
2          John        20
3          Sarah       30
4          Amy         40

replace_data_table

id         name        value

1          NULL        50
2          Rick        NULL
4          Emma        60

我的观点

id         name        value

1          David       50
2          Rick        20
3          Sarah       30
4          Emma        60

您需要COALESCE()从正确的表中获得正确的值:

SELECT o.id,
       COALESCE(r.name, o.name) as name,
       COALESCE(r.value, o.value) as value
FROM original_data_table o LEFT JOIN
     replacement_data_table r
     ON o.id = r.id;

尝试这个:

SELECT o.id as id, IFNULL(r.name, o.name) as new_name, IFNULL(r.value, o.value) as new_value
FROM original_data_table AS o
LEFT JOIN replacement_data_table AS r
ON o.id = r.id 

尝试使用case when

SELECT 
  o.`id`,case when o.`name` <> r.`name` and r.`name` is not null then
      r.`name`
    else o.`name`
  end as name
  ,case when o.`value` <> r.`value` and r.`value` is not null then
      r.`value`
    else o.`value`
  end as value
FROM original_data_table AS o
LEFT JOIN replacement_data_table AS r
  ON o.id = r.id 
order by o.id

SQL小提琴演示链接

| id |  name | name |
|----|-------|------|
|  1 | David |   50 |
|  2 |  Rick |   20 |
|  3 | Sarah |   30 |
|  4 |  Emma |   60 |

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