Question

我有8个不同产品类别的数据库供下载。

pic, app, ebo, tem, des, cod, mus, cat

我想计算客户下载的总产品数和每个产品类别的总下载量。类别产品的每日下载限额为3。用户登录时应查看剩余的下载量。这是工作代码

$query = "SELECT COUNT(*) as sum FROM service_downloads where client_id like '$client'";
$result = mysql_query($query) or die(mysql_error()); 
// Print out result
while($row = mysql_fetch_array($result))
{
echo "You have downloaded". $row['sum'] ." products.";
echo "<br />";
}
$query = "SELECT COUNT(*) as sum FROM service_downloads where client_id like '$client' and product like 'pic'";
$result = mysql_query($query) or die(mysql_error());

// Print out result
while($row = mysql_fetch_array($result))
{
echo "". $row['sum'] ." downloaded pictures";
$leftovers = 3 - $row['sum'];
echo " $leftovers pictures remain for download";
echo "<br />";
}
$query = "SELECT COUNT(*) as sum FROM service_downloads where client_id like '$client' and product like 'app'";
$result = mysql_query($query) or die(mysql_error());

// Print out result
while($row = mysql_fetch_array($result))
{
echo "". $row['sum'] ."downloaded applications";
$leftovers = 3 - $row['sum'];
echo " $leftovers applications remain for download.";
echo "<br />";
}

$query = "SELECT CO.... This procedure repeat eight times for different product category.

结果

You have downloaded 12 products.

3 downloaded pictures 0 pictures remain for download.

1 downloaded applications 2 applications remain for download.

3 downl.......

Answer 1

您可以使用GROUP BY语句对结果进行分组。

SELECT   COUNT(`Product`) AS `Sum`, `Product`
FROM     `service_downloads`
WHERE    `client_id` = '<client_id>'
GROUP BY `Product`

然后你可以使用一个while语句遍历每个产品：

// Print out result
while($row = mysql_fetch_array($result))
{
    echo "". $row['Sum'] ."downloaded " . $row['Product'];
    $leftovers = 3 - $row['Sum'];
    echo " $leftovers " . $row['Product'] . " remain for download.";
    echo "<br />";
}

Answer 2

您应该在一个查询中细分每个类别的下载量：

SELECT product,COUNT(*)
FROM service_downloads
WHERE client_id like '$client';

我也不认为你需要使用LIKE;你可能想用=

Answer 3

您可以使用此查询获得包含其中所有总和的单个结果集。

SELECT COUNT(*) as sum, product
  FROM service_downloads 
 WHERE client_id = '$client' 
   AND PRODUCT IN ('pic', 'app', 'abc', 'def', 'ghi') 
 GROUP BY product WITH ROLLUP
 ORDER BY product NULLS FIRST

这将为每个特定产品提供一行，并为产品列中的NULL值提供汇总（汇总）行。

如果此查询需要很长时间在(client, product上创建一个索引，那么它应该会非常快。

Answer 4

如果您经常显示此数据，这听起来像是这样，那么您应该有一个表示这些SUM的单独表，并且是CLIENT_ID索引。

每次添加新条目时，都可以递增/递减该值。

例如，当您向service_downloads添加一个新行时，如果CLIENT_ID为'pic'，则您还会增加此快捷方式表：

UPDATE service_counts SET pic=pic+1 WHERE client_id=1;

Answer 5

$query = "SELECT COUNT(*) as sum,product FROM service_downloads where client_id like '$client' GROUP BY product";
$result = mysql_query($query) or die(mysql_error()); 
// Print out result
while($row = mysql_fetch_array($result))
{
echo "You have downloaded". $row['sum'] ." ".$row['product'];
echo "<br />";
}

这应该有效

我怎样才能简化这个php mysql计数代码并减少查询？

5 个答案: