SQL将多行中的字段的空值填充为先前的非空值

Question

我在 netezza 中有一个表（基于 postgresql），如下所示。我需要创建一个新表，其中 name 的 NULL 值替换为前一个非空行的 name 值。

table1

id name time    value
---------------------
1 john  11:00   324
2 NULL  12:00   645
3 NULL  13:00   324
4 bane  11:00   132
5 NULL  12:00   30
6 NULL  13:00   NULL
7 NULL  14:00   -1
8 zane  11:00   152
9 NULL  12:00   60
10 NULL 13:00   NULL

输出表

name    time    value
---------------------
john    11:00   324
john    12:00   645
john    13:00   324
bane    11:00   132
bane    12:00   30
bane    13:00   NULL
bane    14:00   -1
zane    11:00   152
zane    12:00   60
zane    13:00   NULL

笔记：

1. 由于权限限制，无法更改 table1，因此创建一个新表 是方式。

2. 需要在Netezza（最好）或MS Access中运行。

用于在 Netezza 中创建测试数据的代码如下。

create temp table test (
    id int
    ,name varchar(10)
    ,time time
    ,value int
)distribute on random;

insert into test (id, name, time, value) values(1, 'joe', '10:00', 324);
insert into test (id, name, time, value) values(2, null, '11:00', 645);
insert into test (id, name, time, value) values(3, null, '12:00', 324);

insert into test (id, name, time, value) values(4, 'bane', '10:00', 132);
insert into test (id, name, time, value) values(5, null, '11:00', 30);
insert into test (id, name, time, value) values(6, null, '12:00', null);
insert into test (id, name, time, value) values(7, null, '13:00', -1);

insert into test (id, name, time, value) values(8, 'zane', '10:00', 152);
insert into test (id, name, time, value) values(9, null, '11:00', 60);
insert into test (id, name, time, value) values(10, null, '12:00', null);

Answer 1

试试这个递归 PostgreSQL 查询：

WITH RECURSIVE t(id, name, time, value) AS (
  SELECT id, name, time, value FROM test WHERE id = (
    SELECT MIN(id) FROM test
  )
  UNION
  SELECT test.id, COALESCE(test.name, t.name), test.time, test.value
  FROM test, t WHERE test.id = (
    SELECT id FROM test WHERE id > t.id ORDER BY id LIMIT 1
  )
) SELECT * FROM t ORDER BY id;

请注意，这可能会在每一行上发出SELECT。如果你不想这样，你可以使用视图解决方案：

CREATE VIEW test_view AS
SELECT id, LAG(id) OVER (ORDER BY id) lag_id, name, time, value FROM test;

WITH RECURSIVE t(id, name, time, value) AS (
  SELECT id, name, time, value FROM test_view WHERE lag_id IS NULL
  UNION ALL
  SELECT test_view.id, COALESCE(test_view.name, t.name),
    test_view.time, test_view.value
  FROM test_view, t WHERE test_view.lag_id = t.id
) SELECT * FROM t ORDER BY id;

这应该快得多。这个想法来自this article。 SQLFiddle：http://sqlfiddle.com/#!15/63f7b/1/1.

Answer 2

这在 Access 2010 中适用于我：

SELECT 
    t1.id,
    (
        SELECT TOP 1 t2.name
        FROM test t2
        WHERE t2.id<=t1.id AND t2.name IS NOT NULL
        ORDER BY t2.id DESC
    ) AS name,
    t1.time,
    t1.value
FROM test t1

它也应该适用于其他 SQL 方言，尽管它们的处理方式可能略有不同 TOP 1（例如，LIMIT 1 是一个常见的变体）。

Answer 3

您可以使用COALESCE 函数和子查询来实现这一点：

SELECT t.id, 
    COALESCE(t.name, (SELECT s.name FROM table s WHERE s.name IS NOT NULL AND s.id < t.id ORDER BY s.id LIMIT 1)) AS name,
    t.time,
    COALESCE(t.value, (SELECT s.value FROM table s WHERE s.value IS NOT NULL AND s.id < t.id ORDER BY s.id LIMIT 1)) AS value
FROM table t ORDER BY t.id

Answer 4

基于 MS Access 的解决方案

SELECT (SELECT last(name)
          FROM test AS temp
          WHERE test.id >= temp.id AND temp.name IS NOT NULL) AS new_name, *
FROM test;

Answer 5

尝试使用引导功能。不确定这是否适用于 postgre，对于 Oracle 则适用。我认为这会有所帮助。