当前位置: > 数据库 > MySQL >

MySQL如何复制表中的一条记录并插入

时间:2014-10-30 17:22来源:linux.it.net.cn 作者:it

先把需求说一下吧。从 MSSQL 中导出一个文章表,需要插入到 PHPCMS 中的内容表 phpcms_content 去,需要做到文章可以发布到不同的栏目中去。也就是说,需要复制一条记录,并修改其 catid,再插入到表尾的位置上。

MySQL 复制一条数据并插入的语句:

1 INSERT INTO phpcms_content (SELECT ".$r[$i]['aid']." + 520, ".$r[$i]['cateid'].", news_catid, catid, typeid, areaid, title, style, thumb, keywords, keywords, posids, url, listorder, status, userid, username, inputtime, updatetime, searchid, islink, prefix FROM phpcms_content WHERE contentid = '".$r[$i-1]['aid']."')

大致为:insert into a SELECT id+1, ...(其它字段) FROM a ; 

下面PHP具体程序:

01 $query  "SELECT * FROM articleincategory ORDER BY ArticleID  ";
02 $result $connector -> query($query);
03  
04 $i = 0;
05 while($myrow $connector -> fetch_array($result))
06 {
07     $r[$i]['aid'] = $myrow["ArticleID"];
08     $r[$i]['cateid'] = $myrow["CategoryID"];
09     $i++;
10 }
11  
12 for($i = 0; $i count($r); $i++)
13 {
14     if($i > 0)
15     {
16         if$r[$i]['aid'] == $r[$i-1]['aid'] )
17         {
18             echo '第 '$i' 条数据 '$r[$i]['aid'] .' 与前一条数据 '$r[$i-1]['aid'] .' 重复'.'<br/>';
19             $sql " INSERT INTO phpcms_content (SELECT ".$r[$i]['aid']." + 520, ".$r[$i]['cateid'].", news_catid, catid, typeid, areaid, title, style, thumb, keywords, keywords, posids, url, listorder, status, userid, username, inputtime, updatetime, searchid, islink, prefix FROM phpcms_content WHERE contentid = '".$r[$i-1]['aid']."') ";
20             //$sql = " INSERT INTO phpcms_c_news (SELECT ".$r[$i]['aid']." + 520, template, titleintact, content, groupids_view, readpoint, author, copyfrom, paginationtype, maxcharperpage, sub_title FROM phpcms_c_news WHERE contentid = '".$r[$i-1]['aid']."')  ";
21             echo $sql.'<br />';
22             //$result = $connector -> query($sql);
23             //INSERT INTO test (SELECT id + 10, name, class, score FROM test WHERE id = '1');
24         }
25         
26         else if$r[$i]['aid'] != $r[$i-1]['aid'] )
27         {
28             $sql " UPDATE phpcms_content SET origin_cateid = '".$r[$i]['cateid']."' WHERE contentid = '".$r[$i]['aid']."'  ";
29             echo $sql.'<br />';
30             //$result = $connector -> query($sql);
31         }
32         
33     }
34 }

如果不需要插入,则更简单:insert into mytable (select * from mytable where id=1) ON DUPLICATE KEY UPDATE id=2;

或者: update mytable set id=2 where id=1;

(责任编辑:IT)
------分隔线----------------------------
栏目列表
推荐内容