Question

基本上，我想做的是找到具有最接近值的记录，从中提取一个值，然后添加一个创建包含该值的记录。我不知道有没有可能。

所以我的记录看起来像这样

Wire_tag|lenght|place|
----------------------
W1      |2     |closet
W2      |2.5   |NULL
W3      |3     |NULL
W4      |4     |conveyor
W5      |5     |NULL

我想实现的目标是将wire_tag分配给列位置不是空的列，而最接近的长度值是电线到列位置不同于空的地方。所以我想要达到的结果就是这样

Wire_tag|lenght|place   |wire_assign
------------------------------------
W1      |2.5   |closet  |W2
W4      |3     |conveyor|W3
W4      |5     |conveyor|W5

Answer 1

这里有两种方法，第一种（不错的）方法依赖于使用公共表表达式，我不确定它们是否在您使用的MySQL版本中可用，因此第二种版本不使用任何“花哨的” SQL。

首先，我创建了一个临时表，您将不需要此步骤（而且它也不是MySQL语法）：

CREATE TABLE #wires (
    wire_tag VARCHAR(50),
    lenght NUMERIC(9,1),
    place VARCHAR(50));
INSERT INTO #wires SELECT 'W1', 2, 'closet';
INSERT INTO #wires SELECT 'W2', 2.5, NULL;
INSERT INTO #wires SELECT 'W3', 3, NULL;
INSERT INTO #wires SELECT 'W4', 4, 'conveyor';
INSERT INTO #wires SELECT 'W5', 5, NULL;

因此，在以下查询中，将#wires替换为您的实际表名：

--Nice way
WITH distances AS (
    SELECT
        w.wire_tag,
        w2.lenght,
        w.place,
        w2.wire_tag AS wire_assign,
        ABS(w2.lenght - w.lenght) AS distance
    FROM
        #wires w
        LEFT JOIN #wires w2 ON w2.place IS NULL
    WHERE
        w.place IS NOT NULL),
min_distances AS (
    SELECT
        wire_tag,
        MIN(distance) AS min_distance
    FROM
        distances
    GROUP BY
        wire_tag)
SELECT
    d.wire_tag,
    d.lenght,
    d.place,
    d.wire_assign
FROM
    distances d
    INNER JOIN min_distances m ON m.wire_tag = d.wire_tag AND m.min_distance = d.distance;

--Nasty way 
SELECT
    y.wire_tag,
    y.lenght,
    y.place,
    y.wire_assign
FROM
    (SELECT
        w.wire_tag,
        w2.lenght,
        w.place,
        w2.wire_tag AS wire_assign,
        ABS(w2.lenght - w.lenght) AS distance
    FROM
        #wires w
        LEFT JOIN #wires w2 ON w2.place IS NULL
    WHERE
        w.place IS NOT NULL) y
    INNER JOIN (
        SELECT
            wire_tag,
            MIN(distance) AS min_distance
        FROM
            (SELECT
                w1.wire_tag,
                ABS(w2.lenght - w1.lenght) AS distance
            FROM
                #wires w1
                LEFT JOIN #wires w2 ON w2.place IS NULL
            WHERE
                w1.place IS NOT NULL) x
        GROUP BY
            wire_tag) m ON m.wire_tag = y.wire_tag AND m.min_distance = y.distance;

两种方法的结果相同：

wire_tag    lenght  place   wire_assign
W1          2.5     closet      W2
W4          3.0     conveyor    W3
W4          5.0     conveyor    W5

Answer 2

我在想：

select t.*,
       (select t2.wire_tag
        from t t2
        where t2.place is not null
        order by abs(t2.length - t.length)
        limit 1
       ) closest_wire_tag
from t
where t.place is null;

如何找到最接近的值并将其与记录合并？

2 个答案: