从jsonb成员更新整数列失败,其中:column的类型为integer,但expression的类型为jsonb

时间:2016-02-05 15:40:50

标签: json postgresql plpgsql postgresql-9.5 postgresql-json

在PostgreSQL 9.5表中,我有一个psexec -i -s \\Remote-Pc -u USERNAME -p ****** "locationoftheexe\Initializator.exe" integer

当我尝试在存储过程中更新它时,给定以下JSON数据(一个包含2个对象的数组,每个对象都有一个" social"键)social变量类型{{ 1}}:

in_users

然后以下代码失败:

jsonb

错误消息:

'[{"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

我已尝试将该行更改为:

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

然后我收到错误:

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.

为什么PostgreSQL不识别数据是social = t->'social'::int,

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

2 个答案:

答案 0 :(得分:4)

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?

回答你原来的问题:

  

为什么PostgreSQL不识别数据是整数?

因为您尝试将jsonb值转换为integer。在您的解决方案中,您已经发现需要->>运算符而不是->来提取text,可以将其转换为integer

您的第二次尝试又添加了第二个错误:

<击> t->'social'::int

除上述内容外:operator precedence。转换运算符::比json运算符->绑定得更强。就像你已经发现的那样,你真的想要:

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

dba.SE非常相似:

答案 1 :(得分:0)

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;