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

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

你也可以像这样在jsonb中原子地增加键:

UPDATE users SET counters = counters || CONCAT('{"bar":', COALESCE(counters->>'bar','0')::int + 1, '}')::jsonb WHERE id = 1;

SELECT * FROM users;

 id |    counters
----+------------
  1 | {"bar": 1}

>假设起始值为0。

更详细的解释,请看我的回答:https://stackoverflow.com/a/39076637

遗憾的是,我没有在文档中找到任何东西,但您可以使用一些变通方法,例如您可以编写一些扩展函数。

例如,在Python中:

CREATE or REPLACE FUNCTION json_update(data json, key text, value json)
returns json
as $$
from json import loads, dumps
if key is None: return data
js = loads(data)
js[key] = value
return dumps(js)
$$ language plpython3u

然后

update test set data=json_update(data, 'a', to_json(5)) where data->>'b' = '2';

这个解决方案是jsonb_set的替代方案,即使JSONB的列中有NULL,它也能工作。只有当对象存在时,jsonb_set才有效。

在下面的示例中,settings是用户表上的JSONB列。

 UPDATE public."Users"
 SET settings = coalesce("settings", '{}')::jsonb || '{ "KeyToSet" : "ValueToSet" }'
 WHERE id=35877;

例如,我的字符串是这样的: {" a1 ":{“a11”:“x”,“a22”:“y”、“a33”:“z”}}

我使用临时表更新jsons,这对于相当小的数据量(<1.000.000)已经足够了。我找到了另一条路,但后来去度假了,就忘了……

所以。查询将是这样的:

with temp_table as (
select 
a.id,
a->'a1'->>'a11' as 'a11',
a->'a1'->>'a22' as 'a22',
a->'a1'->>'a33' as 'a33',
u1.a11updated
from foo a
join table_with_updates u1 on u1.id = a.id)
    update foo a
    set a = ('{"a1": {"a11": "'|| t.a11updated ||'",
        "a22":"'|| t.a22 ||'",
        "a33":"'|| t.a33 ||'"}}')::jsonb
    from temp_table t
    where t.id = a.id;

它与string的关系比json更大,但它是有效的。基本上,它将所有数据拉到临时表中,在用备份的数据插入concat漏洞时创建一个字符串,并将其转换为jsonb。

Json_set可能更有效,但我仍然掌握了它的窍门。当我第一次尝试使用它时,我完全把弦弄乱了……