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;

在@pozs回答的基础上，这里有两个可能对一些人有用的PostgreSQL函数。(需要PostgreSQL 9.3+)

按键删除:按键从JSON结构中删除一个值。

CREATE OR REPLACE FUNCTION "json_object_del_key"(
  "json"          json,
  "key_to_del"    TEXT
)
  RETURNS json
  LANGUAGE sql
  IMMUTABLE
  STRICT
AS $function$
SELECT CASE
  WHEN ("json" -> "key_to_del") IS NULL THEN "json"
  ELSE (SELECT concat('{', string_agg(to_json("key") || ':' || "value", ','), '}')
          FROM (SELECT *
                  FROM json_each("json")
                 WHERE "key" <> "key_to_del"
               ) AS "fields")::json
END
$function$;

按键递归删除:按键路径从JSON结构中删除一个值。(需要@pozs的json_object_set_key函数)

CREATE OR REPLACE FUNCTION "json_object_del_path"(
  "json"          json,
  "key_path"      TEXT[]
)
  RETURNS json
  LANGUAGE sql
  IMMUTABLE
  STRICT
AS $function$
SELECT CASE
  WHEN ("json" -> "key_path"[l] ) IS NULL THEN "json"
  ELSE
     CASE COALESCE(array_length("key_path", 1), 0)
         WHEN 0 THEN "json"
         WHEN 1 THEN "json_object_del_key"("json", "key_path"[l])
         ELSE "json_object_set_key"(
           "json",
           "key_path"[l],
           "json_object_del_path"(
             COALESCE(NULLIF(("json" -> "key_path"[l])::text, 'null'), '{}')::json,
             "key_path"[l+1:u]
           )
         )
       END
    END
  FROM array_lower("key_path", 1) l,
       array_upper("key_path", 1) u
$function$;

使用例子:

s1=# SELECT json_object_del_key ('{"hello":[7,3,1],"foo":{"mofu":"fuwa", "moe":"kyun"}}',
                                 'foo'),
            json_object_del_path('{"hello":[7,3,1],"foo":{"mofu":"fuwa", "moe":"kyun"}}',
                                 '{"foo","moe"}');

 json_object_del_key |          json_object_del_path
---------------------+-----------------------------------------
 {"hello":[7,3,1]}   | {"hello":[7,3,1],"foo":{"mofu":"fuwa"}}

我发现以前的答案更适合有经验的PostgreSQL用户。这是给初学者的:

假设你有一个JSONB类型的表列，值如下:

{
    "key0": {
        "key01": "2018-05-06T12:36:11.916761+00:00",
        "key02": "DEFAULT_WEB_CONFIGURATION",
        
    "key1": {
        "key11": "Data System",
        "key12": "<p>Health,<p>my address<p>USA",
        "key13": "*Please refer to main screen labeling"
    }
}

让我们假设我们想在行中设置一个新值:

"key13": "*Please refer to main screen labeling"

取而代之的是放置值:

"key13": "See main screen labeling"

我们使用json_set()函数将一个新值赋给key13

jsonb_set()的参数

jsonb_set(target jsonb, path text[], new_value jsonb[, create_missing boolean])

在“target”-我将放置jsonb column-name(这是正在修改的表列)

"path"-是指向(包括)我们要覆盖的键的"json keys路径"

"new_value" -这是我们分配的新值

在我们的例子中，我们想要更新位于key1下的key13的值(key1 -> key13):

因此路径语法是:'{key1,key13}' (路径是最棘手的部分-因为教程是可怕的)

jsonb_set(jsonb_column,'{key1,key13}','"See main screen labeling"')

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.

你可以尝试更新如下:

格式:UPDATE table_name SET column_name = column_name::jsonb || '{"key":new_value}' WHERE column_name条件;

举个例子:

UPDATE test SET data = data::jsonb || '{"a":new_value}' WHERE data->>'b' = '2';

如果你想添加新的字段，你可以尝试:

typeorm代码

let keyName:string = '{key2}'
let valueName:string = '"new_value"'

emailLog: () => "jsonb_set(cast(email_log as jsonb), '" + keyNAme + "','" + valueName + "'," + "true" + ")"