我有一个来自以下命令的表：
CREATE TABLE treatment_costs AS SELECT * FROM
(SELECT r.patient_id, r.transaction_date, r.paid_transaction_amount, o.dob, o.department_name, o.reason_of_visit FROM ReceiptTransactions r
LEFT OUTER JOIN OpdPatientQ o ON (r.patient_id = o.patient_id)
);我现在想把今天（给定日期）插入的所有记录插入上表。为此，我写了：
INSERT INTO TABLE treatment_costs SELECT * FROM
(SELECT r.patient_id, r.transaction_date, r.paid_transaction_amount, o.dob, o.department_name, o.reason_of_visit FROM ReceiptTransactions r WHERE timestamp_column = today_date
LEFT OUTER JOIN OpdPatientQ o ON (r.patient_id = o.patient_id)
);这是向表中插入多个查询的正确方法吗？
编辑1：例如我的表 treatment_costs 的内容包括以下行：
patient_id, transaction_date, paid_transaction_amount, dob, department_name, reason_of_visit
001 01/01/2014 30000 01/01/1985 Cardiology reason_1
002 01/01/2014 35000 01/01/1975 Cardiology reason_2
003 02/01/2014 40000 01/01/1965 Oncology reason_3
004 02/01/2014 30000 01/01/1985 Cardiology reason_4
005 02/01/2014 20000 01/01/1975 Gynecology reason_5我现在怀疑的是insert查询中的select语句，它是：
SELECT * FROM
(SELECT r.patient_id, r.transaction_date, r.paid_transaction_amount, o.dob, o.department_name, o.reason_of_visit FROM ReceiptTransactions r WHERE timestamp_column = today_date
LEFT OUTER JOIN OpdPatientQ o ON (r.patient_id = o.patient_id)
);例如，给出以下结果：
patient_id, transaction_date, paid_transaction_amount, dob, department_name, reason_of_visit
011 01/01/2015 30000 01/01/1986 Cardiology reason_11
012 01/01/2015 35000 01/01/1976 Cardiology reason_21
013 02/01/2015 40000 01/01/1966 Oncology reason_31
014 02/01/2015 30000 01/01/1986 Cardiology reason_41
015 02/01/2015 20000 01/01/1976 Gynecology reason_51在执行insert查询之后，我的表的内容是否如下所示？
patient_id, transaction_date, paid_transaction_amount, dob, department_name, reason_of_visit
001 01/01/2014 30000 01/01/1985 Cardiology reason_1
002 01/01/2014 35000 01/01/1975 Cardiology reason_2
003 02/01/2014 40000 01/01/1965 Oncology reason_3
004 02/01/2014 30000 01/01/1985 Cardiology reason_4
005 02/01/2014 20000 01/01/1975 Gynecology reason_5
011 01/01/2015 30000 01/01/1986 Cardiology reason_11
012 01/01/2015 35000 01/01/1976 Cardiology reason_21
013 02/01/2015 40000 01/01/1966 Oncology reason_31
014 02/01/2015 30000 01/01/1986 Cardiology reason_41
015 02/01/2015 20000 01/01/1976 Gynecology reason_51
1条答案
按热度按时间6rvt4ljy1#
摘自Hive语言手册，
insert-into将附加到表或分区，保持现有数据完整。
不会覆盖表中已存在的任何数据。这个
INSERT查询将运行mapreduce（基于引擎类型）作业，该作业将新生成的文件写入表位置，而不删除现有文件。