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

在Postgresql 9.5中,可以执行以下命令-

UPDATE test
SET data = data - 'a' || '{"a":5}'
WHERE data->>'b' = '2';

OR

UPDATE test
SET data = jsonb_set(data, '{a}', '5'::jsonb);

有人问如何一次更新jsonb值中的多个字段。假设我们创建一个表:

CREATE TABLE testjsonb ( id SERIAL PRIMARY KEY, object JSONB );

然后我们插入一个实验行:

INSERT INTO testjsonb
VALUES (DEFAULT, '{"a":"one", "b":"two", "c":{"c1":"see1","c2":"see2","c3":"see3"}}');

然后我们更新行:

UPDATE testjsonb SET object = object - 'b' || '{"a":1,"d":4}';

它的作用如下:

更新a字段 删除b字段 添加d字段

选择数据:

SELECT jsonb_pretty(object) FROM testjsonb;

会导致:

      jsonb_pretty
-------------------------
 {                      +
     "a": 1,            +
     "c": {             +
         "c1": "see1",  +
         "c2": "see2",  +
         "c3": "see3",  +
     },                 +
     "d": 4             +
 }
(1 row)

要更新里面的字段,不要使用concat操作符||。请改用jsonb_set。这并不简单:

UPDATE testjsonb SET object =
jsonb_set(jsonb_set(object, '{c,c1}','"seeme"'),'{c,c2}','"seehim"');

使用concat操作符{c,c1},例如:

UPDATE testjsonb SET object = object || '{"c":{"c1":"seedoctor"}}';

将移除{c,c2}和{c,c3}。

要获得更强大的功能,请查看postgresql json函数文档。有人可能会对#-操作符、jsonb_set函数和jsonb_insert函数感兴趣。

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" + ")"

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"')

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'