3

在 PostgreSQL 9.5 表中,我有一integersocial.

当我尝试在存储过程中更新它时,给定in_users类型变量中的以下 JSON 数据(一个包含 2 个对象的数组,每个对象都有一个“社交”键) jsonb

'[{"sid":"12345284239407942","auth":"ddddc1808197a1161bc22dc307accccc",**"social":3**,"given":"Alexander1","family":"Farber","photo":"https:\/\/graph.facebook.com\/1015428423940942\/picture?type=large","place":"Bochum,
Germany","female":0,"stamp":1450102770},
  {"sid":"54321284239407942","auth":"ddddc1808197a1161bc22dc307abbbbb",**"social":4**,"given":"Alxander2","family":"Farber","photo":null,"place":"Bochum,
Germany","female":0,"stamp":1450102800}]'::jsonb

然后以下代码失败:

    FOR t IN SELECT * FROM JSONB_ARRAY_ELEMENTS(in_users)
    LOOP
            UPDATE words_social SET
                    social = t->'social',
            WHERE sid = t->>'sid';
    END LOOP;

带有错误消息:

ERROR:  column "social" is of type integer but expression is of type jsonb
LINE 3:                         social = t->'social',
                                         ^
HINT:  You will need to rewrite or cast the expression.

我尝试将该行更改为:

social = t->'social'::int,

但后来我得到了错误:

ERROR:  invalid input syntax for integer: "social"
LINE 3:                         social = t->'social'::int,
                                            ^

为什么 PostgreSQL 不能识别数据是integer

JSON-TYPE-MAPPING-TABLE我的印象是 JSON 数字会自动转换为 PostgreSQL 数字类型。

4

2 回答 2

6

单个基于集合的 SQL 命令比循环更有效:

UPDATE words_social w
SET    social = (iu->>'social')::int
FROM   JSONB_ARRAY_ELEMENTS(in_users) iu  -- in_user = function variable
WHERE  w.sid = iu->>'sid';                -- type of sid?

要回答您的原始问题:

为什么 PostgreSQL 不能识别数据是整数?

因为您试图将jsonb值转换为integer. 在您的解决方案中,您已经发现需要->>操作符而不是->提取text,可以将其转换为integer.

您的第二次尝试添加了第二个错误:

t->'social'::int

除上述内容外:运算符优先级。cast 运算符的::绑定比 json 运算符强->。就像你已经发现自己一样,你真的想要:

(t->>'social')::int

dba.SE 上非常相似的案例:

于 2016-02-13T09:08:39.117 回答
1

我最终使用了:

FOR t IN SELECT * FROM JSONB_ARRAY_ELEMENTS(in_users)
LOOP
        UPDATE words_social SET
                social = (t->>'social')::int
        WHERE sid = t->>'sid';

        IF NOT FOUND THEN
                INSERT INTO words_social (social)
                VALUES ((t->>'social')::int);
        END IF;
END LOOP;
于 2016-02-05T18:35:48.753 回答