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

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所以这里的场景是，我在数据库中有4个表：

  • "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) );

  • "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) );

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

  • "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 . 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：
enter image description here

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 .

sql database sqlite sql-insert

3 回答

  • 0

    您不需要 VALUES 关键字，因为您从查询中选择：

    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，在这种情况下没有 VALUES 关键字

    回复于
  • 1

    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;
    回复于
  • 6

    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;
    回复于

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