MySQL 5.7使用新的json功能将行作为json返回 [英] mySQL 5.7 return row as json using new json features

问题描述

我正在研究一些新的 JSON功能并且想知道是否存在一种聪明(或明显)的方法来将行集作为JSON对象返回.理想情况下，无需命名键或使用任何类型的字符串操作.

I was going over some of the new JSON features and was wondering if there is a clever (or obvious) way to return a rowset as a JSON object. Ideally without having to name the keys or use any kind of string manipulation.

示例:

TABLE: people
id     name     age
1      bob      54
2      jay      32
3      john     10

SELECT * FROM people where id = 1

会返回

{"id":1,"name":"bob","age":54}

甚至更好

SELECT * FROM people

将返回所有3个对象的数组

would return an array of all 3 objects

如果您不熟悉新的JSON功能，则新功能之一是JSON_OBJECT

If you are not familiar with the new JSON features, one of the new functions is JSON_OBJECT

SELECT JSON_OBJECT('key1', 1, 'key2', 'abc')

将返回键值JSON对象.

would return a key value JSON object.

推荐答案

取决于您所说的给键命名或任何类型的字符串处理".如果您很乐意将上述键名和字符串操作的命名封装在存储的proc中，以使您在调用过程时无需命名键，那么可以，您可以:

Depends on what you mean by "name the keys or any kind of string manipulation". If you're happy to encapsulate said naming of keys and string manipulation inside a stored proc so that you don't need to name keys when you're calling the procedure, then yes, you can:

drop procedure if exists spGetJson;
delimiter $$
create procedure spGetJson(pTableName varchar(45), pId int)
begin

select  group_concat(concat("'", COLUMN_NAME, "', ", COLUMN_NAME) separator ',')
into    @cols
from    information_schema.columns
where   TABLE_NAME = pTableName and TABLE_SCHEMA = database();

set @q = concat('select json_object(', @cols, ') from ', pTableName);
if pId is not null then
    set @q = concat(@q, ' where id = ', pId);
end if;
set @q = concat(@q, ';');

prepare statement from @q;
execute statement;
deallocate prepare statement;

end $$
delimiter ;

然后您可以使用以下任一方法调用此proc:

You could then call this proc using either:

call spGetJson('people', 1);
call spGetJson('people', null);

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