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

Question

所以这里的情况是，我在数据库中有4个表，即：

1. ＆＃34; question_info＆＃34;：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. ＆＃34; question_options_info＆＃34 ;:
   question_info

3. ＆＃34; question_answer_info＆＃34 ;:
   q_id

4. ＆＃34; trivia_data＆＃34 ;:
   q_type_id

所以我需要的是，将数据插入q_options_id表。 此查询返回数据：

q_category_id

此查询将返回如下数据：

我已尝试使用此特定查询来插入数据：
q_text

但它始终会返回此错误：
status

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，此案例中没有VALUES个关键字

Answer 2

从SQL中删除VALUES，因为在这种情况下值来自SELECT。

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;

Answer 3

删除VALUES关键字。试试这个：

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;