如何使用group-concat mysql创建json格式?
(我使用MySQL)
例1:
表格1:
email | name | phone
-------------------------------------
[email protected] | Ben | 6555333
[email protected] | Tom | 2322452
[email protected] | Dan | 8768768
[email protected] | Joi | 3434356
像语法代码,不给我格式:
select email, group-concat(name,phone) as list from table1 group by email
我需要的输出:
email | list
------------------------------------------------
[email protected] | {name:"Ben",phone:"6555333"},{name:"Joi",phone:"3434356"}
[email protected] | {name:"Tom",phone:"2322452"},{name:"Dan",phone:"8768768"}
谢谢
试试这个查询 -
SELECT
email,
GROUP_CONCAT(CONCAT('{name:"', name, '", phone:"',phone,'"}')) list
FROM
table1
GROUP BY
email;
JSON格式结果 -
+---------------+-------------------------------------------------------------+
| email | list |
+---------------+-------------------------------------------------------------+
| [email protected] | {name:"Ben", phone:"6555333"},{name:"Joi", phone:"3434356"} |
| [email protected] | {name:"Tom", phone:"2322452"},{name:"Dan", phone:"8768768"} |
+---------------+-------------------------------------------------------------+
使用较新版本的MySQL,您可以使用JSON_OBJECT函数来实现所需的结果,如下所示:
GROUP_CONCAT(
JSON_OBJECT(
'name', name,
'phone', phone
)
) AS list
要准备好将SQL响应解析为数组:
CONCAT(
'[',
GROUP_CONCAT(
JSON_OBJECT(
'name', name,
'phone', phone
)
),
']'
) AS list
这将为您提供一个字符串:[{name: 'ABC', phone: '111'}, {name: 'DEF', phone: '222'}]
,可以对JSON进行解析。希望这可以帮助。
Devart的答案很棒,但K2xL的问题是有效的。我找到的答案是使用HEX()对名称列进行十六进制编码,这确保它将创建有效的JSON。然后在应用程序中,将十六进制转换回字符串。
(对不起自我推销,但是)我写了一篇关于这个的小博客帖子,更详细一点:http://www.alexkorn.com/blog/2015/05/hand-rolling-valid-json-in-mysql-using-group_concat/
[编辑Oriol]这是一个例子:
SELECT email,
CONCAT(
'[',
COALESCE(
GROUP_CONCAT(
CONCAT(
'{',
'\"name\": \"', HEX(name), '\", ',
'\"phone\": \"', HEX(phone), '\"',
'}')
ORDER BY name ASC
SEPARATOR ','),
''),
']') AS bData
FROM table
GROUP BY email
另请注意,我已添加COALESCE,以防该电子邮件没有任何项目。
我希望这能找到合适的眼睛。
您可以使用:
对于数组(documentation):
JSON_ARRAYAGG(col_or_expr) as ...
对象(documentation):
JSON_OBJECTAGG(key, value) as ...
关于@Devart的答案......如果该字段包含换行符或双引号,则结果将不是有效的JSON。
因此,如果我们知道“手机”字段偶尔包含双引号和换行符,我们的SQL将如下所示:
SELECT email, CONCAT( '[', GROUP_CONCAT(CONCAT( '{name:"', name, '", phone:"', REPLACE(REPLACE(phone, '"', '\\\\"'),'\n','\\\\n'), '"}' )), ']' ) AS list FROM table1 GROUP BY email;
如果Ben手机中间有一个引号,并且Joi有一个换行符,则SQL会给出(有效的JSON)结果,如:
[{name:"Ben", phone:"655\"5333"},{name:"Joi", phone:"343\n4356"}]
像这样使用
SELECT email,concat('{name:"',ur_name_column,'",phone:"',ur_phone_column,'"}') as list FROM table1 GROUP BY email;
干杯