[{"subscription_id":"2","service_id":"2","service_name":"product1","service_category":"sms","rate_type":"high","rate_amount":101.0}]
我想提取＆＃34; product1＆＃34;从这个字符串，无法为这个需要帮助写出正确的表达式。
答案 0 :(得分：0)
你去。
Regex101.com演示here。
-- create usage table
create table `usage` ( job varchar(100),
start_time date,
end_time date,
`session` varchar(100)
);
-- create `sessions` table
create table `sessions` ( session_label varchar(100) ,
start_time date,
end_time date );
-- create procedure prc_insert_into_usage
delimiter $$
Create Procedure prc_insert_into_usage
(
IN p_job varchar(100),
IN p_start_time date,
IN p_end_time date
)
Begin
Declare v_session varchar(100);
-- check if record is there in the session table
-- it is assumed session table will have only one record for composite key (start_time, end_time)
Select session_label into v_session
from `sessions`
where start_time = p_start_time and end_time = p_end_time;
IF (v_session IS NULL ) THEN
-- if session_label is generated using auto increment key in sessions table then
-- last_insert_id() can be used to fetch that value after insert into sessions table
-- which can be used while inserting into usage table
-- example below
-- Insert into `sessions` (start_time,end_time) values (p_start_time,p_end_time);
-- set @v_session = last_insert_id();
-- Insert into `usage` values (p_job,p_start_time,p_end_time,@v_session);
-- dummy logic to create the session_label
-- dummy logic is used you can replace it with whatever logic you need
set @v_session = concat('sess_',left(uuid(),8));
-- insert record in both table (first in session table and then in usage table)
Insert into `sessions` values (@v_session,p_start_time,p_end_time);
Insert into `usage` values (p_job,p_start_time,p_end_time,@v_session);
else
-- record already present in sessions table for this session label
-- so insert record only in usage table
Insert into `usage` values (p_job,p_start_time,p_end_time,@v_session);
End If;
End;
$$
-- call this procedure to insert data into usage table
-- sample execution below when both tables have no rows
-- below call will insert 1 rows in both table having same session label
mysql> call prc_insert_into_usage('job_a','2015-01-01','2015-01-02');
Query OK, 1 row affected (0.00 sec)
mysql> select * from `usage`;
+-------+------------+------------+---------------+
| job | start_time | end_time | session |
+-------+------------+------------+---------------+
| job_a | 2015-01-01 | 2015-01-02 | sess_abc376bf |
+-------+------------+------------+---------------+
1 row in set (0.00 sec)
mysql> select * from `sessions`;
+---------------+------------+------------+
| session_label | start_time | end_time |
+---------------+------------+------------+
| sess_abc376bf | 2015-01-01 | 2015-01-02 |
+---------------+------------+------------+
1 row in set (0.01 sec)
-- below call will insert only in usage table as row already present in sessions table
mysql> call prc_insert_into_usage('job_b','2015-01-01','2015-01-02');
Query OK, 1 row affected (0.00 sec)
mysql> select * from `usage`;
+-------+------------+------------+---------------+
| job | start_time | end_time | session |
+-------+------------+------------+---------------+
| job_a | 2015-01-01 | 2015-01-02 | sess_abc376bf |
| job_b | 2015-01-01 | 2015-01-02 | sess_abc376bf |
+-------+------------+------------+---------------+
2 rows in set (0.00 sec)
mysql> select * from `sessions`;
+---------------+------------+------------+
| session_label | start_time | end_time |
+---------------+------------+------------+
| sess_abc376bf | 2015-01-01 | 2015-01-02 |
+---------------+------------+------------+
1 row in set (0.00 sec)
-- below call will again insert rows in both table
mysql> call prc_insert_into_usage('job_c','2015-01-02','2015-01-04');
Query OK, 1 row affected (0.01 sec)
mysql> select * from `usage`;
+-------+------------+------------+---------------+
| job | start_time | end_time | session |
+-------+------------+------------+---------------+
| job_a | 2015-01-01 | 2015-01-02 | sess_abc376bf |
| job_b | 2015-01-01 | 2015-01-02 | sess_abc376bf |
| job_c | 2015-01-02 | 2015-01-04 | sess_dcfa6853 |
+-------+------------+------------+---------------+
3 rows in set (0.00 sec)
mysql> select * from `sessions`;
+---------------+------------+------------+
| session_label | start_time | end_time |
+---------------+------------+------------+
| sess_abc376bf | 2015-01-01 | 2015-01-02 |
| sess_dcfa6853 | 2015-01-02 | 2015-01-04 |
+---------------+------------+------------+
2 rows in set (0.00 sec)
mysql>
- 首先检查文字“服务名称”，冒号和开头引号，然后捕获内部到下一个结束引号的内容。
但是，由于这看起来像是一个正确的JSON响应，因此您应该使用JSON序列化程序而不是RegEx。
编辑：根据@ neuronaut的评论，编辑出多余的"service_name":"([\w]+)"，因为'\ w'已经覆盖了它。可以找到旧的RegEx101演示here，以获得相关性。
答案 1 :(得分：0)
我已经弄清楚了，要查看数组中的各个元素，我们可以使用以下内容： 在hive中的arrayName.ElementName [0]它将给出所有元素。要在画面中使用它，需要将其传递给直通函数，如：RAWSQL_STR（“％1.service_name [0]”，[服务详细信息]）。