Question

我有三张my_location1＆amp;表my_location2＆amp; my_location3。 my_ is Prefix。 location1 = countrylist。 location2 = statslist。 location3 =城镇名单。在其他表中我有用户列表。现在我想列出所有位置（国家 - 统计 - 城镇）使用树木样式+水族用户在每个国家统计城镇（计数）。（如何生成这个？）

数据库注：pid＆amp; pid1 =国家/地区的ID，pid2 =统计信息的ID

my_location1：

id ++++++ name ++++++ active

my_location2：

id ++++++ pid +++++++ name ++++++ active

my_location3：

id ++++++ pid1 +++++++ pid2 ++++++ name ++++++ active

my_user

id ++++++ name ++++++ location1 ++++++ location2 ++++++ location3 ++++++ active

e.x

Country (print count Of users in this country )
+++++++stats (print count Of users in this stats )
++++++++++++town (print count Of users in this town )
++++++++++++town ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
+++++++stats ( ... )
++++++++++++town ( ... )
+++++++stats ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
Country1 ( ... )
+++++++stats ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
Country2 ( ... )
+++++++stats ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
++++++++++++town ( ... )
...more

感谢您的帮助。

Answer 1

我没有检查代码，但它应该是这样的：

function getCountries()
{
   return mysql_query('SELECT c.*, (COUNT(*) from u WHERE u.location1 = c.id) AS user_count
                           from countrylist c
                           LEFT JOIN users u
                           ON u.location1 = c.id
                           WHERE c.active = 1'
                        );
}

function getStats($countryId)
{
   return mysql_query('SELECT s.*, (COUNT(*) from u WHERE s.location2 = s.id) AS user_count
                           from statslist s
                           LEFT JOIN users u
                           ON u.location2 = s.id
                           WHERE s.pid = ' . $countryId
                        );   
}

function getTowns($stateId)
{
   return mysql_query('SELECT t.*, (COUNT(*) from u WHERE t.location3 = t.id) AS user_count
                           from townlist
                           LEFT JOIN users u
                           ON u.location3 = t.id
                           WHERE t.pid2 = ' . $stateId
                        );   
}

echo "<ul>\n";
while ($country = mysql_fetch_assoc(getCountries())) {
    echo "<li>" . $country["name"] . "(" . $country["user_count"] . ")\n";
    echo "<ul>\n";

    while ($state = mysql_fetch_assoc(getStats($country["id"]))) {
      echo "<li>" . $state["name"] . "(" . $state["user_count"] . ")\n<ul>\n";
      while ($town = mysql_fetch_assoc(getTowns($state["id"]))) {
         echo "<li>" . $town["name"] . "(" . $town["user_count"] . ")</li>\n";
      }
      echo "</ul>\n</li>\n";
    }

    echo "<ul>\n</li>\n";
}
echo "</ul>\n";

它可以而且应该进一步优化，但希望你能得到这个想法。还请注意你的mysql结构是多余的。

国家/统计/城镇+计数列表

1 个答案: