mysql 查询语法有问题吗？尝试从 INNER JOIN 中插入

发布于 2025-01-07 23:15:51 字数 338 浏览 0 评论 0原文

下面的代码给了我一个模糊且无用的错误：

INSERT INTO database VALUES * FROM dbtemp WHERE dbtemp.number NOT IN (SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price);

我已经成功地在这些数据库上运行了其他查询，所以我知道这只是这个语句的语法问题。但就我的一生而言，我看不出有什么问题。任何建议/意见将不胜感激。

谢谢！

原文

The following code is giving me a vague and unhelpful error:

INSERT INTO database VALUES * FROM dbtemp WHERE dbtemp.number NOT IN (SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price);

I've successfully run other queries on these databases, so I know its just a problem with the syntax of this statement. But for the life of me I can't see what's wrong. Any suggestions/advice would be greatly appreciated.

Thanks!

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我还不会笑 2025-01-14 23:15:51

这应该是一个 INSERT INTO SELECT 语句吗？您的 INSERT INTO database VALUES * 语法值得怀疑...

INSERT INTO database
  SELECT * FROM dbtemp 
  WHERE dbtemp.number NOT IN (
    SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price
  );

在执行这样的 INSERT SELECT 操作时，建议使用显式列名称，而不是 SELECT *，以确保它们以正确的顺序返回：

INSERT INTO database
  SELECT
    col1,
    col2,
    col3
  FROM dbtemp 
  WHERE dbtemp.number NOT IN (
    SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price
  );

Should this have been an INSERT INTO SELECT statement? Your syntax of INSERT INTO database VALUES * is suspect...

INSERT INTO database
  SELECT * FROM dbtemp 
  WHERE dbtemp.number NOT IN (
    SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price
  );

When doing an INSERT SELECT like this, it is advisable to use explicit column names, rather than SELECT *, to be sure they are returned in the correct order:

INSERT INTO database
  SELECT
    col1,
    col2,
    col3
  FROM dbtemp 
  WHERE dbtemp.number NOT IN (
    SELECT dbtemp.number FROM dbtemp INNER JOIN database ON dbtemp.number = database.number AND dbtemp.price = database.price
  );

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