How do I generate nested json objects using mysql native json functions?

Question

Using only the native JSON fuctions (no PHP, etc) in MySQL version 5.7.12 (section 13.16 in the manual) I am trying to write a query to generate a JSON document from relational tables that contains a sub object. Given the following example:

CREATE TABLE `parent_table` (
   `id` int(11) NOT NULL,
   `desc` varchar(20) NOT NULL,
   PRIMARY KEY (`id`)
);
CREATE TABLE `child_table` (
   `id` int(11) NOT NULL,
   `parent_id` int(11) NOT NULL,
   `desc` varchar(20) NOT NULL,
   PRIMARY KEY (`id`,`parent_id`)
);
insert `parent_table` values (1,'parent row 1');
insert `child_table` values (1,1,'child row 1');
insert `child_table` values (2,1,'child row 2');

I am trying to generate a JSON document that looks like this:

[{
    "id" : 1,
    "desc" : "parent row 1",
    "child_objects" : [{
            "id" : 1,
            "parent_id" : 1,
            "desc" : "child row 1"
        }, {
            "id" : 2,
            "parent_id" : 1,
            "desc" : "child row 2"
        }
    ]
}]

I am new to MySQL and suspect there is a SQL pattern for generating nested JSON objects from one to many relationships but I'm having trouble finding it.

In Microsoft SQL (which I'm more familiar with) the following works:

select 
 [p].[id]
,[p].[desc]
,(select * from [dbo].[child_table] where [parent_id] = [p].[id] for json auto) AS [child_objects]
from [dbo].[parent_table] [p]
for json path

I attempted to write the equivalent in MySQL as follows:

select json_object(
 'id',p.id 
,'desc',p.`desc`
,'child_objects',(select json_object('id',id,'parent_id',parent_id,'desc',`desc`) 
                  from child_table where parent_id = p.id)
)
from parent_table p;

select json_object(
  'id',p.id 
 ,'desc',p.`desc`
 ,'child_objects',json_array((select json_object('id',id,'parent_id',parent_id,'desc',`desc`) 
                              from child_table where parent_id = p.id))
 )
 from parent_table p

Both attempts fail with the following error:

Error Code: 1242. Subquery returns more than 1 row

Greg Faulk · Accepted Answer · 2016-05-27 20:06:28Z

37

The reason you are getting these errors is that the parent json object is not expecting a result set as one of its inputs, you need to have simple object pairs like {name, string} etc bug report - may be available in future functionality... this just means that you need to convert your multi row results into a concatination of results separated by commas and then converted into a json array.

You almost had it with your second example.

You can achieve what you are after with the GROUP_CONCAT function

select json_object(
  'id',p.id 
 ,'desc',p.`desc`
 ,'child_objects',json_array(
                     (select GROUP_CONCAT(
                                 json_object('id',id,'parent_id',parent_id,'desc',`desc`)
                             )   
                      from child_table 
                      where parent_id = p.id))
                   )
 from parent_table p;

This almost works, it ends up treating the subquery as a string which leaves the escape characters in there.

'{\"id\": 1, 
\"desc\": \"parent row 1\", 
\"child_objects\": 
    [\"
    {\\\"id\\\": 1,
     \\\"desc\\\": \\\"child row 1\\\", 
    \\\"parent_id\\\": 1
    },
    {\\\"id\\\": 2, 
    \\\"desc\\\": \\\"child row 2\\\", 
    \\\"parent_id\\\": 1}\"
    ]
}'

In order to get this working in an appropriate format, you need to change the way you create the JSON output as follows:

select json_object(
  'id',p.id 
 ,'desc',p.`desc`
 ,'child_objects',(select CAST(CONCAT('[',
                GROUP_CONCAT(
                  JSON_OBJECT(
                    'id',id,'parent_id',parent_id,'desc',`desc`)),
                ']')
         AS JSON) from child_table where parent_id = p.id)

 ) from parent_table p;

This will give you the exact result you require:

'{\"id\": 1, 
\"desc\": \"parent row 1\", 
\"child_objects\": 
    [{\"id\": 1, 
    \"desc\": \"child row 1\", 
    \"parent_id\": 1
    }, 
    {\"id\": 2, 
    \"desc\": \"child row 2\", 
    \"parent_id\": 1
    }]  
}'

edited May 27, 2016 at 20:06

Greg Faulk

3621 gold badge2 silver badges9 bronze badges

answered May 27, 2016 at 3:07

Brad Baskin

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6 Comments

Sandeep Over a year ago

hello, i am not able to use json_object this function. #1305 - FUNCTION json_object does not exist. What to do???

Brad Baskin Over a year ago

Check which version of Mysql you are using. The older versions do not support json_object

Sandeep Over a year ago

current version is 5.6

toddsby Over a year ago

JSON aggregate functions JSON_ARRAYAGG() and JSON_OBJECTAGG() are now available in MySQL 8 mysqlserverteam.com/mysql-8-0-labs-json-aggregation-functions

Johny Bravo Over a year ago

This works fine but for two level only, can we get till 'n' level

|

JarmoP · Accepted Answer · 2019-11-15 21:28:22Z

7

For MariaDb, CAST AS JSON does not work. But JSON_EXTRACT may be used to convert a string to a JSON object:

select json_object(
  'id',p.id 
 ,'desc',p.`desc`
 ,'child_objects',JSON_EXTRACT(IFNULL((select
    CONCAT('[',GROUP_CONCAT(
      json_object('id',id,'parent_id',parent_id,'desc',`desc`)
    ),']')   
   from child_table where parent_id = p.id),'[]'),'$')
 ) from parent_table p;

answered Nov 15, 2019 at 21:28

JarmoP

3914 silver badges5 bronze badges

Comments

Ban Nguyen - VN · Accepted Answer · 2021-08-30 15:53:53Z

I tried group_concat solution but I found it's problems larger string because of group_concat limitations (group_concat_max_len). I wrote the new function resolve the problem about converting a string to JSON object as bellow and how to use it. Tested on MariaDB 10.5.12

Usage: https://i.sstatic.net/cWfd7.jpg

    CREATE FUNCTION `ut_tf_array`(input_json longtext) RETURNS longtext CHARSET utf8mb4 COLLATE utf8mb4_unicode_ci
    COMMENT 'Function for transform json array agg'
BEGIN

    DECLARE transformed_data_list longtext ;
    DECLARE record longtext ;
    DECLARE i_count int ;    
    DECLARE i_count_items int ;  
    

    SET i_count = 0;
    SET i_count_items = JSON_LENGTH(JSON_EXTRACT(input_json,'$'));
    
    SET transformed_data_list = '[]';
    
    -- return array with length = zero
    IF input_json is NULL THEN
        RETURN transformed_data_list;
    END IF;

    WHILE i_count < i_count_items DO
      -- fetch into record  
      SELECT JSON_EXTRACT( JSON_EXTRACT( input_json ,'$') , CONCAT('$[',i_count,']')) INTO record;
      -- append to transformed_data_list    
      SELECT JSON_ARRAY_APPEND(transformed_data_list, '$', JSON_EXTRACT(record, '$')) into transformed_data_list;

      SET i_count := i_count + 1;
    END WHILE;
    
    -- done
    RETURN transformed_data_list;


END

chirag chauhan · Accepted Answer · 2022-06-05 14:48:54Z

0

Below Query works for me.

SELECT JSON_ARRAYAGG(JSON_OBJECT('Id', p.id, 'desc', p.`desc`, 'child_objects', temp_json)) AS json_value
FROM (
SELECT p.id, p.`desc`,
JSON_ARRAYAGG(JSON_OBJECT('id', p.id, 'parent_id', p.parent_id, 'desc', p.`desc`)) AS temp_json
FROM parent_table p
GROUP BY p.id
) t;

edited Jun 5, 2022 at 14:48

answered Jun 5, 2022 at 14:45

chirag chauhan

11 bronze badge

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