mysql自定义函数,获取Json列中json数组某个属性的值

 1 DELIMITER $$
 2 DROP function IF EXISTS `func_fetch_spec_quantity` $$
 3 CREATE FUNCTION `func_fetch_spec_quantity`
 4 (service_spec varchar(1000),prop_value varchar(1000),prop_name varchar(1000)) RETURNS varchar(10)
 5 BEGIN
 6     declare prop_value_path varchar(50) default '';
 7     declare dest_prop_value varchar(10) default '';
 8     declare prop_name_deep varchar(10) default '';
 9     declare prop_name_path varchar(50) default '';
10         set prop_value_path = JSON_SEARCH(service_spec,'one',prop_value);
11             set prop_name_deep = substring_index(JSON_UNQUOTE(prop_value_path),'.',1);
12             set prop_name_path = concat_ws('.',prop_name_deep,prop_name);
13             set dest_prop_value = JSON_EXTRACT(service_spec,prop_name_path);
14     return JSON_UNQUOTE(dest_prop_value);
15 END$$

eg:

某json列的值为:

[{"code":"123","name":"张三"},{"code":"456","name":"李四"},{"code":"789","name":"王五"}]
要获取json字符串中code="789"的name值

1、通过JSON_SEARCH返回值为789的位置

select JSON_SEARCH('[{"code":"123","name":"张三"},{"code":"456","name":"李四"},{"code":"789","name":"王五"}]','one','789')

2、将结果分割,得到code=789在json字符串中的位置,注意:在上一个语句查询处理的结果中带有双引号,所以需要用JSON_UNQUOTE去掉

select substring_index(JSON_UNQUOTE("$[2].code"),'.',1);

 

 

 

3、将位置和name属性做拼接

select  concat_ws('.','$[2]','name');

 

 

 

4、获取指定路径属性的值

select JSON_EXTRACT('[{"code":"123","name":"张三"},{"code":"456","name":"李四"},{"code":"789","name":"王五"}]', '$[2].name')

 

 

 

5、结果去掉引号

select JSON_UNQUOTE("王五")

 用函数执行的结果:

 

 

 

posted on 2022-12-15 10:09  harry034  阅读(1657)  评论(1)    收藏  举报