SQLite - INSERT INTO SELECT - 如何插入"将3个现有表连接到新表中"的数据？

Question

所以这里的场景是,我在数据库中有4个表:

1. "question_info":
   CREATE TABLE question_info ( q_id mediumint(9) NOT NULL, q_type_id int(11) NOT NULL, q_options_id mediumint(9) NOT NULL, q_category_id int(11) NOT NULL, q_text varchar(2048) NOT NULL, status tinyint(4) NOT NULL DEFAULT '0', q_date_added date NOT NULL DEFAULT '2013-01-01', q_difficulty_level tinyint(4) NOT NULL DEFAULT '0', PRIMARY KEY(q_id) );
   

2. "question_options_info":
   CREATE TABLE question_options_info ( q_options_id mediumint(9) NOT NULL, q_options_1 varchar(255) NOT NULL, q_options_2 varchar(255) NOT NULL, q_options_3 varchar(255) NOT NULL, q_options_4 varchar(255) NOT NULL, q_options_ex_1 varchar(1024) DEFAULT NULL, q_options_ex_2 varchar(1024) DEFAULT NULL, q_options_ex_3 varchar(1024) DEFAULT NULL, q_options_ex_4 varchar(1024) DEFAULT NULL, PRIMARY KEY(q_options_id) );

3. "question_answer_info":
   CREATE TABLE question_answer_info ( q_id mediumint(9) NOT NULL, q_options mediumint(9) NOT NULL );

4. "trivia_data":
   CREATE TABLE trivia_data ( q_id mediumint(9) NOT NULL, q_text varchar(2048) NOT NULL, q_options_1 varchar(255) NOT NULL, q_options_2 varchar(255) NOT NULL, q_options_3 varchar(255) NOT NULL, q_options_4 varchar(255) NOT NULL, q_options mediumint(9) NOT NULL, q_difficulty_level tinyint(4) NOT NULL DEFAULT '0', q_date_added date NOT NULL DEFAULT '2015-04-8', PRIMARY KEY(q_id) );

所以我需要的是,将数据插入trivia_data表中.此查询返回数据:

SELECT question_info.q_id, question_info.q_text, question_options_info.q_options_1, question_options_info.q_options_2, question_options_info.q_options_3, question_options_info.q_options_4, question_answer_info.q_options, question_info.q_difficulty_level, question_info.q_date_added
FROM question_info JOIN question_options_info ON question_info.q_options_id = question_options_info.q_options_id JOIN question_answer_info ON question_info.q_id = question_answer_info.q_id;

此查询将返回如下数据:

我已经尝试过这个特定的查询来插入数据:
INSERT INTO trivia_data VALUES(q_id, q_text, q_options_1, q_options_2, q_options_3, q_options_4, q_options, q_difficulty_level, q_date_added) SELECT question_info.q_id, question_info.q_text, question_options_info.q_options_1, question_options_info.q_options_2, question_options_info.q_options_3, question_options_info.q_options_4, question_answer_info.q_options, question_info.q_difficulty_level, question_info.q_date_added FROM question_info JOIN question_options_info on question_info.q_options_id = question_options_info.q_options_id JOIN question_answer_info on question_info.q_id = question_answer_info.q_id;

但它总是返回此错误:
near "SELECT": syntax error:

老实说,我是SQL的新手.所以请尽可能简单地解释一下.任何帮助,将不胜感激.谢谢.

Answer 1

您不需要VALUES关键字,因为您从查询中进行选择:

INSERT INTO trivia_data (
    q_id, 
    q_text, 
    q_options_1, 
    q_options_2, 
    q_options_3, 
    q_options_4, 
    q_options, 
    q_difficulty_level, 
    q_date_added)  
SELECT 
    question_info.q_id, 
    question_info.q_text, 
    question_options_info.q_options_1, 
    question_options_info.q_options_2, 
    question_options_info.q_options_3, 
    question_options_info.q_options_4, 
    question_answer_info.q_options, 
    question_info.q_difficulty_level, 
    question_info.q_date_added 
FROM question_info 
    JOIN question_options_info on question_info.q_options_id = question_options_info.q_options_id 
    JOIN question_answer_info on question_info.q_id = question_answer_info.q_id;

通常,如果要插入记录,则语法为

INSERT INTO <tablename> (<column1>, <column2>, ..., <columnN>)
VALUES (<value1>, <value2>, ..., <valueN>)

如果要插入结果,语法如下:

INSERT INTO <tablename> (<column1>, <column2>, ..., <columnN>)
SELECT <value1>, <value2>, ..., <valueN> FROM ...

如您所见,VALUES在这种情况下没有关键字