Question

我正在尝试为计划任务创建一些php和sql，该任务将在一个值上连接2个表，并根据table2中的unix时间戳值从table1中删除所有记录。到目前为止，我已经提出了以下代码，但是不想运行它，因为它可能会造成更大的损害。有人可以帮忙吗？预先感谢

$sql = "DELETE FROM tbl_code
LEFT JOIN ( SELECT * FROM `tbl_ctoon` WHERE tbl_code.ctoon_id=tbl_ctoon.ctoon_id AND tbl_code.code_for='C')
WHERE DATEDIFF( date(NOW( )) , FROM_UNIXTIME(expire_date)) > 3 ";
$del = new database();
$del->myquery($sql, 1);

因此最终它将合并两个表，并从{{1}中的tbl_code超过3天的expire_date中删除所有记录。

Answer 1

您应该在yle子查询中加入表以获得有效结果

DELETE tbl_code 
FROM tbl_code
INNER JOIN (
    SELECT tbl_ctoon.expire_date , tbl_ctoon.ctoon_id 
    FROM `tbl_ctoon` 
    LEFT JOIN tbl_code ON tbl_code.ctoon_id=tbl_ctoon.ctoon_id 
               AND tbl_code.code_for='C'
    ) t2 ON t2.ctoon_id =tbl_code.ctoon_id
            AND DATEDIFF( date(NOW( )) , FROM_UNIXTIME(t2. expire_date)) > 3 
WHERE tbl_code.code_for='C'

要测试涉及的行，您可以

SELECT *  
FROM tbl_code
INNER JOIN (
    SELECT tbl_ctoon.expire_date , tbl_ctoon.ctoon_id 
    FROM `tbl_ctoon` 
    LEFT JOIN tbl_code ON tbl_code.ctoon_id=tbl_ctoon.ctoon_id 
               AND tbl_code.code_for='C'
    ) t2 ON t2.ctoon_id =tbl_code.ctoon_id
            AND DATEDIFF( date(NOW( )) , FROM_UNIXTIME(t2. expire_date)) > 3 
WHERE tbl_code.code_for='C'

Answer 2

在删除之前，我会选择一个查询。您的查询看起来更简单：

$sql = "SELECT * FROM tbl_code 
LEFT JOIN tbl_ctoon 
ON tbl_code.ctoon_id = tbl_ctoon.ctoon_id AND tbl_code.code_for = 'C' 
WHERE DATEDIFF(DATE(NOW()), FROM_UNIXTIME(expire_date)) > 3 AND tbl_ctoon.ctoon_id IS NOT NULL";

然后，如果您的查询返回了您希望删除的条目，请执行删除操作：

$sql = "DELETE FROM tbl_code 
LEFT JOIN tbl_ctoon 
ON tbl_code.ctoon_id = tbl_ctoon.ctoon_id AND tbl_code.code_for = 'C' 
WHERE DATEDIFF(DATE(NOW()), FROM_UNIXTIME(expire_date)) > 3 AND tbl_ctoon.ctoon_id IS NOT NULL";

Answer 3

由于JOIN无效，所以我走了一条不同的路线。我运行了它，并在phpmyadmin中为我的数据库很好地执行了操作。

rdfFolder = rdsSession.GetDefaultFolder(rdoDefaultFolders.olFolderInbox)

如何根据另一个表中的时间戳值从一个表中删除记录？

3 个答案: