使用子查询更新postgres中的表行

使用postgres 8.4，我的目标是更新现有表：

CREATE TABLE public.dummy
(
  address_id SERIAL,
  addr1 character(40),
  addr2 character(40),
  city character(25),
  state character(2),
  zip character(5),
  customer boolean,
  supplier boolean,
  partner boolean

)
WITH (
  OIDS=FALSE
);

最初我使用插入语句测试了我的查询：

insert into address customer,supplier,partner
SELECT  
    case when cust.addr1 is not null then TRUE else FALSE end customer, 
    case when suppl.addr1 is not null then TRUE else FALSE end supplier,
    case when partn.addr1 is not null then TRUE else FALSE end partner
from (
    SELECT *
        from address) pa
    left outer join cust_original cust
        on (pa.addr1=cust.addr1 and pa.addr2=cust.addr2 and pa.city=cust.city 
            and pa.state=cust.state and substring(cust.zip,1,5) = pa.zip  )
    left outer join supp_original suppl 
        on (pa.addr1=suppl.addr1 and pa.addr2=suppl.addr2 and pa.city=suppl.city 
                and pa.state=suppl.state and pa.zip = substring(suppl.zip,1,5))
    left outer join partner_original partn
        on (pa.addr1=partn.addr1 and pa.addr2=partn.addr2 and pa.city=partn.city
                  and pa.state=partn.state and pa.zip = substring(partn.zip,1,5) )
where pa.address_id = address_id

作为新手，我无法转换为更新语句，即用select语句返回的值更新现有行。非常感谢您的帮助。

Postgres允许：

UPDATE dummy
SET customer=subquery.customer,
    address=subquery.address,
    partn=subquery.partn
FROM (SELECT address_id, customer, address, partn
      FROM  /* big hairy SQL */ ...) AS subquery
WHERE dummy.address_id=subquery.address_id;

此语法不是标准SQL，但是对于这种类型的查询，它比标准SQL更加方便。我相信Oracle（至少）接受类似的东西。

您正在使用UPDATE FROM语法。

UPDATE 
  table T1  
SET 
  column1 = T2.column1 
FROM 
  table T2 
  INNER JOIN table T3 USING (column2) 
WHERE 
  T1.column2 = T2.column2;

参考文献

• 此处的代码示例：UPDATE FROM子句中的GROUP BY
• 和这里
• 正式语法规范

如果使用联接没有性能提升，那么出于可读性考虑，我更喜欢使用通用表表达式（CTE）：

WITH subquery AS (
    SELECT address_id, customer, address, partn
    FROM  /* big hairy SQL */ ...
)
UPDATE dummy
SET customer = subquery.customer,
    address  = subquery.address,
    partn    = subquery.partn
FROM subquery
WHERE dummy.address_id = subquery.address_id;

恕我直言有点现代。

有许多更新行的方法。

对于UPDATE使用子查询的行，可以使用这些方法中的任何一种。

1. 方法1 [使用直接表参考]

UPDATE
  <table1>
SET
  customer=<table2>.customer,
  address=<table2>.address,
  partn=<table2>.partn
FROM
  <table2>
WHERE
  <table1>.address_id=<table2>.address_i;

说明：table1是要更新table2 的表，是要从中获取要替换/更新的值的表。我们正在使用FROM子句，以获取table2的数据。WHERE 子句将有助于设置正确的数据映射。

2. 方法2 [使用子查询]

UPDATE
  <table1>
SET
  customer=subquery.customer,
  address=subquery.address,
  partn=subquery.partn
FROM
  (
    SELECT
      address_id, customer, address, partn
    FROM  /* big hairy SQL */ ...
  ) AS subquery
WHERE
  dummy.address_id=subquery.address_id;

说明：这里我们在FROM子句中使用子查询，并为其赋予别名。这样它就可以像桌子一样工作。

3. 方法3 [使用多个联接表]

UPDATE
  <table1>
SET
  customer=<table2>.customer,
  address=<table2>.address,
  partn=<table2>.partn
FROM
  <table2> as t2
  JOIN <table3> as t3
  ON
    t2.id = t3.id
WHERE
  <table1>.address_id=<table2>.address_i;

说明：有时我们遇到这样的情况，表联接对于获取正确的更新数据非常重要。为此，Postgres允许我们在FROM子句中联接多个表。

4. 方式4 [使用WITH语句]

   • 4.1 [使用简单查询]

WITH subquery AS (
    SELECT
      address_id,
      customer,
      address,
      partn
    FROM
      <table1>;
)
UPDATE <table-X>
SET customer = subquery.customer,
    address  = subquery.address,
    partn    = subquery.partn
FROM subquery
WHERE <table-X>.address_id = subquery.address_id;

• 4.2 [使用带有复杂JOIN的查询]

WITH subquery AS (
    SELECT address_id, customer, address, partn
    FROM
      <table1> as t1
    JOIN
      <table2> as t2
    ON
      t1.id = t2.id;
    -- You can build as COMPLEX as this query as per your need.
)
UPDATE <table-X>
SET customer = subquery.customer,
    address  = subquery.address,
    partn    = subquery.partn
FROM subquery
WHERE <table-X>.address_id = subquery.address_id;

说明：从Postgres 9.1开始，WITH引入了this（）概念。使用它，我们可以进行任何复杂的查询并生成期望的结果。在这里，我们使用这种方法来更新表。

我希望这会有所帮助。😊

update json_source_tabcol as d
set isnullable = a.is_Nullable
from information_schema.columns as a 
where a.table_name =d.table_name 
and a.table_schema = d.table_schema 
and a.column_name = d.column_name;

@Mayur “ 4.2 [使用具有复杂JOIN的查询]”和“ 公用表表达式（CTE） ”对我来说很有效。

WITH cte AS (
SELECT e.id, e.postcode
FROM employees e
LEFT JOIN locations lc ON lc.postcode=cte.postcode
WHERE e.id=1
)
UPDATE employee_location SET lat=lc.lat, longitude=lc.longi
FROM cte
WHERE employee_location.id=cte.id;

希望这可以帮助...：D