Postgresql 8.4更新plpgsql函数查询语法错误

Question

我正在使用 PostgreSQL 8.4 并创建一个 plpgsql 函数。在这个函数的主体中，我有一个更新记录的查询。

...
  UPDATE device_syncfiles SET 
  state_code = 1, updated_at = NOW() at time zone 'UTC'
  WHERE 
  ((state_code = 2 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > 3600) OR 
  (state_code = 3 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > 3600));
...

当我将此函数加载到数据库中时，出现语法错误

ERROR:  syntax error at or near ""
LINE 1: UPDATE device_syncfiles SET    = 1,    = NOW() at time z...
                                     ^
QUERY:  UPDATE device_syncfiles SET    = 1,    = NOW() at time zone 'UTC' WHERE ((   = 2 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' -   ::timestamp without time zone)) >   ) OR (   = 3 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' -   ::timestamp without time zone)) >   ))
CONTEXT:  SQL statement in PL/PgSQL function "syncfile_get" near line 19

我找不到这个查询有任何问题。这里有什么问题？

更新：（缺少信息）

Table: device_syncfiles

  id PK integer auto inc
  user_id integer FK
  file_name character varying(255) NOT NULL,
  state_code integer NOT NULL FK,
  md5 character varying(255) NOT NULL,
  msg character varying(255),
  created_at timestamp without time zone,
  updated_at timestamp without time zone

函数：syncfile_get()

CREATE OR REPLACE FUNCTION syncfile_get()
  RETURNS TABLE(id integer, user_id integer, file_name character varying, state_code integer, md5 character varying, created_at timestamp without time zone, updated_at timestamp without time zone) AS
$BODY$
      DECLARE
        _device_syncfile_id integer;
        _download_timeout integer;
        _processing_timeout integer;
      BEGIN
        -- GET all timeout info
        SELECT state_timeout INTO _download_timeout FROM device_syncfile_states 
        WHERE state_name = 'downloading';
        SELECT state_timeout INTO _processing_timeout FROM device_syncfile_states 
        WHERE state_name = 'processing';
        -- GET syncfile id
        _device_syncfile_id = NULL;

        -- Reset timed out file to idel state
        UPDATE device_syncfiles SET 
        state_code = 1, updated_at = NOW() at time zone 'UTC'
        WHERE 
        ((state_code = 2 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > _download_timeout) OR 
        (state_code = 3 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > _processing_timeout));

        -- GET the id of one idel/timed out file => result could be a integer or NULL
        SELECT device_syncfiles.id INTO _device_syncfile_id FROM device_syncfiles 
        WHERE 
        device_syncfiles.state_code = 1 OR 
        (device_syncfiles.state_code = 2 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - device_syncfiles.updated_at::timestamp without time zone)) > _download_timeout) OR
        (device_syncfiles.state_code = 3 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - device_syncfiles.updated_at::timestamp without time zone)) > _processing_timeout) 
        LIMIT 1;

        -- WHEN NULL skip state update and return empty set of record
        -- Otherwise return the set of record with the id found in last step
        IF _device_syncfile_id IS NOT NULL THEN
          PERFORM syncfile_update(_device_syncfile_id, 2, NULL);
        END IF;

        RETURN QUERY SELECT 
        device_syncfiles.id,
        device_syncfiles.user_id ,
        device_syncfiles.file_name ,
        device_syncfiles.state_code ,
        device_syncfiles.md5 ,
        device_syncfiles.created_at ,
        device_syncfiles.updated_at 
        FROM device_syncfiles WHERE device_syncfiles.id = _device_syncfile_id;
      END;
      $BODY$
  LANGUAGE plpgsql VOLATILE

Answer 1

很多个问题。

0.

I am using Postgresql 8.4

Postgres 8.4 于 2014 年 7 月停产。Consider upgrading to a current version. 紧急。

1.

你的问题没有透露完整的功能（至少是页眉和页脚），也没有任何 table 定义和一些示例数据来帮助我们帮助你。

我必须做出假设，我有根据的猜测是您有一个名为 state_code 的函数参数，它与相同的列名冲突。在三个处。基础知识：

How to return result of a SELECT inside a function in PostgreSQL?

您必须知道 RETURNS TABLE 子句中声明的所有字段也是有效的 OUT 参数。（如第一句第一句所述link。）所以你的Q更新证实了我的假设。

2.

您的错误消息在此处报告了 第一个 实例：

UPDATE device_syncfiles SET 
  <b>state_code</b> = 1 ...

这是 0. 的结果。您正被早已死去和遗忘的 Postgres 版本绊倒，在该版本中，函数创建时的表面语法检查用于检测 UPDATE 语句的目标列与函数参数之间的命名冲突。这很愚蠢，后来被删除了：那些目标列不能与主体上的函数参数冲突。

您的错误在 Postgres 8.4 中重现：dbfiddle here

在 Postgres 9.4 中不会发生同样的事情：dbfiddle here

要修复，最好重命名函数参数，避免冲突。相关：

3.

还有两个实例：

WHERE 
  ((<b>state_code</b> = 2 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > 3600) OR 
   (<b>state_code</b> = 3 AND EXTRACT(EPOCH FROM (NOW() at time zone 'UTC' - updated_at::timestamp without time zone)) > 3600));

任何版本都需要修复。 Postgres 无法判断是解析为函数参数还是 table 列。最好 table-限定所有列以避免与参数名称先验的任何可能的冲突。（UPDATE 目标列除外，它们不需要也不允许 table 限定。）

4.

这仍然是猪口红。像这样改进查询：

UPDATE device_syncfiles d
SET    state_code = 1
     , updated_at = NOW() AT TIME ZONE 'UTC'
WHERE  d.state_code IN (2, 3)
AND    d.updated_at < (now() - interval '1 hour') AT TIME ZONE 'UTC';

更短，更快，可以在 updated_at.

上使用索引

5.

最后考虑使用 timestamp with time zone 而不是 timestamp without time zone 开头：

Ignoring timezones altogether in Rails and PostgreSQL

Postgresql 8.4更新plpgsql函数查询语法错误

Postgresql 8.4 update query syntax error in plpgsql function

postgresql

plpgsql

postgresql-8.4

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