我尝试获取内容ID,这里是查询...
$getIDs = mysql_query("SELECT content_id FROM playlist_content WHERE playlist_id=".$rrow['playlist_id']."") or die(mysql_error());
while($row = mysql_fetch_array($getIDs)) {
$array[] = $row;
}
这是结果
Array
(
[0] => Array
(
[0] => 33
[content_id] => 33
)
[1] => Array
(
[0] => 13
[content_id] => 13
)
[2] => Array
(
[0] => 8
[content_id] => 8
)
)
现在从第二个表中获取此ID必须获取值 注意:ID = content_id
$result = mysql_query("SELECT SUM( length ) as total FROM content WHERE ID ...
并且所有行的回声总和['length'在哪里是这个ID 需要支持 102 或 256 等结果......
或者有更好的方法来获得总和?
答案 0 :(得分:1)
您可以通过连接两个表
在单个查询中实际执行此操作SELECT a.content_id, SUM(b.length) TotalSum
FROM playlist_content a
INNER JOIN content b
ON a.content_id = b.ID
WHERE a.playlist_id = IDHERE
GROUP BY a.content_id
查询结果将为每个length
提供content_id
的总和。
要进一步了解联接,请访问以下链接:
答案 1 :(得分:1)
试试这个:
$query = "SELECT content_id, " .
" ( select sum (length) from content where id = content_id) as content_length " .
" FROM playlist_content " .
" WHERE playlist_id=".$rrow['playlist_id'];
$getIDs = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($getIDs)) {
$array[] = $row;
}