如何在PL/SQL中解析JSON字符串

Question

我想解析表Tests_1中CLOB列中的JSON字符串,并将其插入另一个表(Test_2).

如何在不使用任何JSON库的情况下在PL/SQL中执行此操作？

create table Tests_1
(
  value   CLOB
)
create table Test_2 (a date,b date,c number,d number, e number)

INSERT INTO Tests_1
  (value)
VALUES
  ('{ 
"a":"01/01/2015",
"b":"31/12/2015",
"c":"11111111111",
"d":"1111111111",
"e":"1234567890"
}');

Answer 1

对于11.0.4版本(当然没有11.0.4版本),您至少有两个选择(除了自己编写解析器):

根据您使用的RDBMS版本,以下是几个选项:

第一个:对于Oracle 11.1.0.7及以上版本,安装Apex 5并使用apex_json包:

-- here I have 12.1.0.1 version with version 5 of apex installed

column ora_version format a21;
column apex_version format a21;


select (select version from v$instance) as ora_version
     , (select version_no from apex_release) as apex_version
  from dual;

--drop table test_2;
/* our test table */  
create table test_2(
  c_a date,
  c_b date,
  c_c number,
  c_d number,
  c_e number
);

select * from test_2;

declare
  l_json_doc clob; 
begin
  dbms_output.put_line('Parsing json...');
  l_json_doc := '{"a":"01/01/2015","b":"31/12/2015",
                  "c":"11111111111","d":"1111111111",
                  "e":"1234567890"}';
  apex_json.parse(l_json_doc);
  insert into test_2(c_a, c_b, c_c, c_d, c_e)
    values(apex_json.get_date(p_path=>'a', p_format=>'dd/mm/yyyy'),
           apex_json.get_date(p_path=>'b', p_format=>'dd/mm/yyyy'),
           to_number(apex_json.get_varchar2(p_path=>'c')),
           to_number(apex_json.get_varchar2(p_path=>'d')),
           to_number(apex_json.get_varchar2(p_path=>'e')));
  commit;
  dbms_output.put_line('Done!');
end;
/

column c_c format 99999999999;
select to_char(c_a, 'dd/mm/yyyy') as c_a
     , to_char(c_b, 'dd/mm/yyyy') as c_b
     , c_c
     , c_d
     , c_e 
  from test_2;

结果:

ORA_VERSION           APEX_VERSION         
--------------------- ---------------------
12.1.0.1.0            5.0.2.00.07          

1 row selected.

Table created.

no rows selected.

Parsing json...
Done!
PL/SQL procedure successfully completed.


C_A        C_B                 C_C        C_D        C_E
---------- ---------- ------------ ---------- ----------
01/01/2015 31/12/2015  11111111111 1111111111 1234567890

1 row selected.

第二个:使用opensource PL/JSON.从来没有使用它,所以我借此机会尝试一下.它非常相似apex_json.

declare
  l_json      json;  --json object
  l_json_doc  clob;
begin
  dbms_output.put_line('Parsing json...');

  -- parsing is done upon object instantiation

  l_json_doc := '{"a":"01/01/2015","b":"31/12/2015",
                  "c":"11111111111","d":"1111111111",
                  "e":"1234567890"}';
  l_json := json(l_json_doc);


  insert into test_2(c_a, c_b, c_c, c_d, c_e)
    values(to_date(l_json.get('a').get_string, 'dd-mm-yyyy'),
           to_date(l_json.get('b').get_string, 'dd-mm-yyyy'),
           to_number(l_json.get('c').get_string),
           to_number(l_json.get('d').get_string),
           to_number(l_json.get('e').get_string));
  commit;
  dbms_output.put_line('Done!');
end;

column c_c format 99999999999;
select to_char(c_a, 'dd/mm/yyyy') as c_a
     , to_char(c_b, 'dd/mm/yyyy') as c_b
     , c_c
     , c_d
     , c_e 
  from test_2;

结果:

C_A        C_B                 C_C        C_D        C_E
---------- ---------- ------------ ---------- ----------
01/01/2015 31/12/2015  11111111111 1111111111 1234567890
01/01/2015 31/12/2015  11111111111 1111111111 1234567890

2 rows selected.

json_table()在12.1.0.2版本中的介绍使JSON解析它更简单(仅为了演示):

insert into test_2
  select to_date(c_a, 'dd-mm-yyyy')
       , to_date(c_b, 'dd-mm-yyyy')
       , c_c
       , c_d
       , c_e
    from json_table('{"a":"01/01/2015",
                      "b":"31/12/2015",
                      "c":"11111111111",
                      "d":"1111111111",
                      "e":"1234567890"}'
                    , '$' 
                    columns ( 
                       c_a varchar2(21) path '$.a',
                       c_b varchar2(21) path '$.b',
                       c_c varchar2(21) path '$.c',
                       c_d varchar2(21) path '$.d',
                       c_e varchar2(21) path '$.e'
                    )) ;

结果:

 select *
   from test_2;


C_A         C_B                C_C        C_D        C_E
----------- ----------- ---------- ---------- ----------
1/1/2015    12/31/2015  1111111111 1111111111 1234567890

Answer 2

由于您指定不想使用任何 JSON 库，如果格式是固定的，您可以将其强制转换为可以解析为 XML 的内容，首先是去除花括号，用等号替换冒号，并删除双-从每个名称/值对的第一部分引用：

select regexp_replace(regexp_replace(value, '(^{|}$)'),
  '^"(.*)":(".*")($|,)', '\1=\2', 1, 0, 'm')
from tests_1;

REGEXP_REPLACE(REGEXP_REPLACE(VALUE,'(^{|}$)'),'^"(.*)":(".*")($|,)','\1=\2',1,0
--------------------------------------------------------------------------------

a="01/01/2015"
b="31/12/2015"
c="11111111111"
d="1111111111"
e="1234567890"

您可以将其用作虚拟 XML 节点的属性；将其转换为 XMLType，您可以使用 XMLTable 来提取属性：

select x.a, x.b, x.c, x.d, x.e
from tests_1 t
cross join xmltable('/tmp'
  passing xmltype('<tmp ' ||regexp_replace(regexp_replace(value, '(^{|}$)'),
    '^"(.*)":(".*")($|,)', '\1=\2', 1, 0, 'm') || ' />')
  columns a varchar2(10) path '@a',
    b varchar2(10) path '@b',
    c number path '@c',
    d number path '@d',
    e number path '@e'
) x;

A          B                      C             D             E
---------- ---------- ------------- ------------- -------------
01/01/2015 31/12/2015   11111111111    1111111111    1234567890

然后您可以在插入期间将字符串转换为日期：

insert into test_2 (a, b, c, d, e)
select to_date(x.a, 'DD/MM/YYYY'), to_date(x.b, 'DD/MM/YYYY'), x.c, x.d, x.e
from tests_1 t
cross join xmltable('/tmp'
  passing xmltype('<tmp ' || regexp_replace(regexp_replace(value, '(^{|}$)'),
    '^"(.*)":(".*")($|,)', '\1=\2', 1, 0, 'm') || ' />')
  columns a varchar2(10) path '@a',
    b varchar2(10) path '@b',
    c number path '@c',
    d number path '@d',
    e number path '@e'
) x;

select * from test_2;

A          B                      C             D             E
---------- ---------- ------------- ------------- -------------
2015-01-01 2015-12-31   11111111111    1111111111    1234567890

这将处理一些不存在的名称/值对，如果发生这种情况，您将获得空值。

如果所有对都在那里，你可以标记字符串并拉出相关部分：

select to_date(regexp_substr(value, '[^"]+', 1, 4), 'DD/MM/YYYY') as a,
  to_date(regexp_substr(value, '[^"]+', 1, 8), 'DD/MM/YYYY') as b,
  to_number(regexp_substr(value, '[^"]+', 1, 12)) as c,
  to_number(regexp_substr(value, '[^"]+', 1, 16)) as d,
  to_number(regexp_substr(value, '[^"]+', 1, 20)) as e
from tests_1;

A          B                      C             D             E
---------- ---------- ------------- ------------- -------------
2015-01-01 2015-12-31   11111111111    1111111111    1234567890

Answer 3

Oracle 12c 支持 JSON

如果你有一个现有的表就做

ALTER TABLE table1 ADD CONSTRAINT constraint_name CHECK (your_column IS json);
SELECT t.your_column.id FROM table1 t;

请注意，出于某种原因，t那里需要昵称

或完整示例：

CREATE TABLE json_documents (
  id    RAW(16) NOT NULL,
  data  CLOB,
  CONSTRAINT json_documents_pk PRIMARY KEY (id),
  CONSTRAINT json_documents_json_chk CHECK (data IS JSON)
);

INSERT INTO json_documents (id, data)
VALUES (SYS_GUID(),
        '{
          "FirstName"      : "John",
          "LastName"       : "Doe",
          "Job"            : "Clerk",
          "Address"        : {
                              "Street"   : "99 My Street",
                              "City"     : "My City",
                              "Country"  : "UK",
                              "Postcode" : "A12 34B"
                             },
          "ContactDetails" : {
                              "Email"    : "john.doe@example.com",
                              "Phone"    : "44 123 123456",
                              "Twitter"  : "@johndoe"
                             },
          "DateOfBirth"    : "01-JAN-1980",
          "Active"         : true
         }');




SELECT a.data.FirstName,
       a.data.LastName,
       a.data.Address.Postcode AS Postcode,
       a.data.ContactDetails.Email AS Email
FROM   json_documents a;


FIRSTNAME       LASTNAME        POSTCODE   EMAIL
--------------- --------------- ---------- -------------------------
Jayne           Doe             A12 34B    jayne.doe@example.com
John            Doe             A12 34B    john.doe@example.com

2 rows selected.

更多信息

• https://oracle-base.com/articles/12c/json-support-in-oracle-database-12cr1

• https://docs.oracle.com/database/122/ADJSN/using-PLSQL-object-types-for-JSON.htm#ADJSN-GUID-F0561593-D0B9-44EA-9C8C-ACB6AA9474EE