Question

我无法正确使用mysql语法。我想在同一个数据库和今天的日期CURDATE（）中从另一个表中插入2个数据post_id和meta_value。

INSERT into wp_top_voted

（post_id，number_votes，todays_date）

以下内容：

SELECT `post_id`  
FROM  `wp_postmeta` WHERE  `meta_key` =  'votes' ORDER BY  `meta_value` DESC   
LIMIT 10
SELECT `meta_value`  
FROM  `wp_postmeta` WHERE  `meta_key` =  'votes' ORDER BY  `meta_value` DESC   
LIMIT 10
CURDATE())

如何将上述内容合并为一个参数？我试过了

INSERT into wp_top_voted  
(`post_id`, `number_votes`, `todays_date`)  
values (SELECT `post_id`  
FROM  `wp_postmeta` WHERE  `meta_key` =  'votes' ORDER BY  `meta_value` DESC   
LIMIT 10, SELECT `meta_value`  
FROM  `wp_postmeta` WHERE  `meta_key` =  'votes' ORDER BY  `meta_value` DESC   
LIMIT 10,   CURDATE())

但我收到错误。

Answer 1

你可以在这里找到一个很好的解释http://www.w3schools.com/sql/sql_insert_into_select.asp

您需要INSERT INTO SELECT语句

INSERT INTO Customers (CustomerName, Country)
SELECT SupplierName, Country FROM Suppliers;

您的具体情况应该是：

INSERT into wp_top_voted ('post_id','meta_value','todays_date' ) SELECT 'post_id' ,'meta_value' ,CURDATE()
FROM  'wp_postmeta' WHERE  'meta_key' =  'votes' ORDER BY 'meta_value' DESC   
LIMIT 10

Answer 2

由于INSERT INTO TABLE() SELECT的格式不正确，您收到错误。试试这个：

INSERT into wp_top_voted  
(`post_id`, `number_votes`, `todays_date`)  
SELECT `post_id`, `meta_value`, CURDATE() 
FROM  `wp_postmeta` WHERE  `meta_key` =  'votes' ORDER BY  `meta_value` DESC   
LIMIT 10;

在mysql语句中INSERT INTO和SELECT多个值

2 个答案: