如何向查询中添加记录答案

【问题标题】：How to add records to query如何向查询中添加记录
【发布时间】：2014-12-11 18:13:10
【问题描述】：

我有问题。

SELECT * FROM '.PRFX.'sell 
WHERE draft = "0" '.$e_sql.' 
AND ID NOT IN (SELECT id_ FROM '.PRFX.'skipped WHERE uid = "'.$u.'") 
AND ID NOT IN (SELECT id_ FROM '.PRFX.'followed WHERE uid = "'.$u.'") 
ORDER BY raised DESC '.$sql_limit;

我想以最少的刷新次数添加 3 条记录；最佳第五名

它们必须是唯一的（所以如果你连接两个 UNION ALL...）

【问题讨论】：

不，我没喝醉。 SELECT DISTINCT * (SELECT * FROM '.PRFX.'sell WHERE draft = "0" '.$e_sql.' AND ID NOT IN (SELECT id_ FROM '.PRFX.'skipped WHERE uid = "'.$u.'") AND ID NOT IN (SELECT id_ FROM '.PRFX.'followed WHERE uid = "'.$u.'") ORDER BY raised DESC '.$sql_limit;) UNION ALL (SELECT * FROM '.PRFX.'sell WHERE draft = "0" '.$e_sql.' AND ID NOT IN (SELECT id_ FROM '.PRFX.'skipped WHERE uid = "'.$u.'") AND ID NOT IN (SELECT id_ FROM '.PRFX.'followed WHERE uid = "'.$u.'") ORDER BY refreshes ASC LIMIT 3) 但如何让它发挥作用？
我想出了：#$sql = '(SELECT * FROM '.PRFX.'sell WHERE draft = "0" '.$e_sql.' AND ID NOT IN (SELECT id_ FROM '.PRFX.'skipped WHERE uid = "'.$u.'") AND ID NOT IN (SELECT id_ FROM '.PRFX.'followed WHERE uid = "'.$u.'") ORDER BY raised DESC '.$sql_limit.') UNION (SELECT * FROM '.PRFX.'sell WHERE draft = "0" '.$e_sql.' AND ID NOT IN (SELECT id_ FROM '.PRFX.'skipped WHERE uid = "'.$u.'") AND ID NOT IN (SELECT id_ FROM '.PRFX.'followed WHERE uid = "'.$u.'") ORDER BY refreshes ASC LIMIT 3)';，但它会引发重复。
嗯。剥离你的代码。 “PRFX。”没有意义。这段 SQL 代码很难理解（对我来说）为什么要添加 SELECT 请求？

标签： mysql sql-limit

【解决方案1】：

首先，您需要使您的 SQL 更具可读性。像这样的

SELECT * FROM sell
WHERE draft = 0
AND ID NOT IN (SELECT id_ FROM skipped WHERE uid = '0')
AND ID NOT IN (SELECT id_ FROM followed WHERE uid = '0')
ORDER BY raised DESC LIMIT 15

那么，你想要什么？通过单个请求将数据添加到sell 表？这可以通过这样的请求来完成

INSERT INTO sell (key1, key2, keyN)
VALUES 
('aaa', 'bbb', 'ccc'),
('ddd', 'eee', 'fff');
-- and so forth.

【讨论】：