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MySQL如何复制表中的一条记录并插入

2012-09-09 21:00 浏览: 1800941 次 我要评论(0 条) 字号:


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:


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:


$query = "SELECT * FROM articleincategory ORDER BY ArticleID ";
$result = $connector -> query($query);

$i = 0;
while($myrow = $connector -> fetch_array($result))
{
$r[$i]['aid'] = $myrow["ArticleID"];
$r[$i]['cateid'] = $myrow["CategoryID"];
$i++;
}

for($i = 0; $i < count($r); $i++)
{
if($i > 0)
{
if( $r[$i]['aid'] == $r[$i-1]['aid'] )
{
echo '第 '. $i. ' 条数据 '. $r[$i]['aid'] .' 与前一条数据 '. $r[$i-1]['aid'] .' 重复'.'<br/>';
$sql = " INSERT INTO phpcms_content (SELECT ".$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[$i-1]['aid']."') ";
//$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']."') ";
echo $sql.'<br />';
//$result = $connector -> query($sql);
//INSERT INTO test (SELECT id + 10, name, class, score FROM test WHERE id = '1');
}

else if( $r[$i]['aid'] != $r[$i-1]['aid'] )
{
$sql = " UPDATE phpcms_content SET origin_cateid = '".$r[$i]['cateid']."' WHERE contentid = '".$r[$i]['aid']."' ";
echo $sql.'<br />';
//$result = $connector -> query($sql);
}

}
}

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;



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