使用postgresql 9.3我可以选择JSON数据类型的特定字段,但如何使用UPDATE修改它们?我在postgresql文档或网上任何地方都找不到这样的例子。我尝试了一些显而易见的方法:

postgres=# create table test (data json);
CREATE TABLE
postgres=# insert into test (data) values ('{"a":1,"b":2}');
INSERT 0 1
postgres=# select data->'a' from test where data->>'b' = '2';
 ?column?
----------
 1
(1 row)
postgres=# update test set data->'a' = to_json(5) where data->>'b' = '2';
ERROR:  syntax error at or near "->"
LINE 1: update test set data->'a' = to_json(5) where data->>'b' = '2...

当前回答

在PostgreSQL 9.4中,我们实现了以下python函数。它也可以用于PostgreSQL 9.3。

create language plpython2u;

create or replace function json_set(jdata jsonb, jpaths jsonb, jvalue jsonb) returns jsonb as $$
import json

a = json.loads(jdata)
b = json.loads(jpaths)

if a.__class__.__name__ != 'dict' and a.__class__.__name__ != 'list':
  raise plpy.Error("The json data must be an object or a string.")

if b.__class__.__name__ != 'list':
   raise plpy.Error("The json path must be an array of paths to traverse.")

c = a
for i in range(0, len(b)):
  p = b[i]
  plpy.notice('p == ' + str(p))

  if i == len(b) - 1:
    c[p] = json.loads(jvalue)

  else:
    if p.__class__.__name__ == 'unicode':
      plpy.notice("Traversing '" + p + "'")
      if c.__class__.__name__ != 'dict':
        raise plpy.Error("  The value here is not a dictionary.")
      else:
        c = c[p]

    if p.__class__.__name__ == 'int':
      plpy.notice("Traversing " + str(p))
      if c.__class__.__name__ != 'list':
        raise plpy.Error("  The value here is not a list.")
      else:
        c = c[p]

    if c is None:
      break    

return json.dumps(a)
$$ language plpython2u ;

使用示例:

create table jsonb_table (jsonb_column jsonb);
insert into jsonb_table values
('{"cars":["Jaguar", {"type":"Unknown","partsList":[12, 34, 56]}, "Atom"]}');

select jsonb_column->'cars'->1->'partsList'->2, jsonb_column from jsonb_table;

update jsonb_table
set jsonb_column = json_set(jsonb_column, '["cars",1,"partsList",2]', '99');

select jsonb_column->'cars'->1->'partsList'->2, jsonb_column from jsonb_table;

Note that for a previous employer, I have written a set of C functions for manipulating JSON data as text (not as a json or jsonb type) for PostgreSQL 7, 8 and 9. For example, extracting data with json_path('{"obj":[12, 34, {"num":-45.67}]}', '$.obj[2]['num']'), setting data with json_path_set('{"obj":[12, 34, {"num":-45.67}]}', '$.obj[2]['num']', '99.87') and so on. It took about 3 days work, so if you need it to run on legacy systems and have the time to spare, it may be worth the effort. I imagine the C version is much faster than the python version.

其他回答

对于那些使用mybatis的人,这里有一个更新语句示例:

<update id="saveAnswer">
    update quiz_execution set answer_data = jsonb_set(answer_data, concat('{', #{qid}, '}')::text[], #{value}::jsonb), updated_at = #{updatedAt}
    where id = #{id}
</update>

参数:

Qid, field的键。 是一个有效的json字符串,对于字段值, 通过jackson将对象转换为json字符串,

当我尝试更新字符串类型字段时,这对我有用。

UPDATE table_name 
SET body = jsonb_set(body, '{some_key}', to_json('value'::text)::jsonb);

希望它能帮助其他人!

假设表table_name有一个名为body的jsonb列,并且您想要更改body。Some_key = 'value'

从PostgreSQL 14开始,你可以使用jsonb下标直接访问JSON字段的元素,并最终更新它们。

UPDATE test SET data['a'] = '5' WHERE data['b'] = '2';

我为自己写了一个小函数,在Postgres 9.4中递归地工作。下面是函数(我希望它对你有用):

CREATE OR REPLACE FUNCTION jsonb_update(val1 JSONB,val2 JSONB)
RETURNS JSONB AS $$
DECLARE
    result JSONB;
    v RECORD;
BEGIN
    IF jsonb_typeof(val2) = 'null'
    THEN 
        RETURN val1;
    END IF;

    result = val1;

    FOR v IN SELECT key, value FROM jsonb_each(val2) LOOP

        IF jsonb_typeof(val2->v.key) = 'object'
            THEN
                result = result || jsonb_build_object(v.key, jsonb_update(val1->v.key, val2->v.key));
            ELSE
                result = result || jsonb_build_object(v.key, v.value);
        END IF;
    END LOOP;

    RETURN result;
END;
$$ LANGUAGE plpgsql;

下面是使用示例:

select jsonb_update('{"a":{"b":{"c":{"d":5,"dd":6},"cc":1}},"aaa":5}'::jsonb, '{"a":{"b":{"c":{"d":15}}},"aa":9}'::jsonb);
                            jsonb_update                             
---------------------------------------------------------------------
 {"a": {"b": {"c": {"d": 15, "dd": 6}, "cc": 1}}, "aa": 9, "aaa": 5}
(1 row)

正如你所看到的,它深入分析并在需要的地方更新/添加值。

如果您的字段类型是json,下面将为您工作。

UPDATE 
table_name
SET field_name = field_name::jsonb - 'key' || '{"key":new_val}' 
WHERE field_name->>'key' = 'old_value'.

操作符“-”从左操作数中删除键/值对或字符串元素。键/值对根据键值进行匹配。

运算符'||'将两个jsonb值连接成一个新的jsonb值。

因为这些都是jsonb操作符,你只需要将它们类型转换为::jsonb

更多信息:JSON函数和操作符

你可以在这里阅读我的笔记