查询:
$query=mysql_query(SELECT content_id, COUNT(*) FROM votingapi_vote WHERE value_type = option AND value = 1 GROUP BY content_id)
如果我将其分配给$result
。
while ($obj = db_fetch_object ($result)) {
$output .= $obj->content_id.'<br>' . $obj->count;
}
如何输出计数?在$obj->content_id
之后。没有打印数字。谢谢。
答案 0 :(得分:3)
您的列COUNT(*)
需要别名(即重命名该列):
SELECT content_id, COUNT(*) as count FROM [...]
然后使用变量:
$obj->count
答案 1 :(得分:3)
为计数添加别名,例如:
SELECT content_id, COUNT(*) as the_count FROM ...
然后你可以用这个名字来引用它:
$output .= $obj->content_id.'<br>' . $obj->the_count;
答案 2 :(得分:2)
$query=mysql_query(SELECT content_id, COUNT(*) as count FROM votingapi_vote WHERE value_type = option AND value = 1 GROUP BY content_id)
答案 3 :(得分:1)
更改
COUNT(*) FROM
到
COUNT(*) as count FROM
答案 4 :(得分:1)
尝试添加:$ query = mysql_query(SELECT content_id,COUNT(*) * as the_count * FROM votingapi_vote WHERE value_type = option AND value = 1 GROUP BY content_id)
然后计数应该是$ obj-&gt; the_count