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我有两个表，第一个是“用户”表，其中有一个名为“商店”的列，我还有一个名为“商店”的表，其中包含“商店编号”“商店位置”列。

用户表中的“商店”列是“商店编号”。

我想要做的是创建一个类似于

样本数据：

店铺编号：34 店铺地点：伦敦用户：34

店铺编号 | 店铺位置 | 本店用户数|

所以这就像从商店中选择 * 并为每个创建新行。

并且对于用户的数量，类似于 sum * from users where 'store' = 'store number' from stores 表。

我希望这是有道理的，

杰克。

更新：

这是对的：

$result = mysql_query("SELECT * FROM stores", $con) or die(mysql_error());

echo "<table border='1'>";
echo "<tr> <th>Store Number</th> <th>Store Location</th> <th>Number of Users</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
    // Print out the contents of each row into a table
    echo "<tr><td>"; 
    echo $row['storenumber'];
    echo "</td><td>"; 
    echo $row['location'];
    echo "</td><td>"; 
    echo 'Amount of users here';
    echo "</td></tr>"; 
} 

echo "</table>";

表：

CREATE TABLE IF NOT EXISTS `stores` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `storenumber` int(11) NOT NULL,
  `location` varchar(40) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=4 ;

如果不存在则创建表users（

  `id` int(50) NOT NULL AUTO_INCREMENT,
  `email` varchar(50) NOT NULL,
  `store` int(11) NOT NULL,
  `lastvisit` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=28 ;
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3 回答 3

1

试试这个 SQL：

SELECT storenumber, location, COUNT(users.store) as nbr_users FROM stores
LEFT JOIN users ON stores.storenumber = users.store
GROUP BY store.id

然后做

echo $row['nbr_users'];

打印用户数。

于 2013-01-05T15:56:56.820 回答
0

试试这个：

$result = mysql_query("SELECT * FROM stores", $con) or die(mysql_error());
echo "<table border='1'>";
echo "<tr> <th>Store Number</th> <th>Store Location</th> <th>Number of Users</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
    // gets the total amount of users
    $query = mysql_query("SELECT COUNT(*) AS total FROM users WHERE `store`='".$row['storenumber']."'") or die( mysql_error() );
    $r = mysql_fetch_array( $query );
    $total = $r['total'];
    unset($query, $r);

    // Print out the contents of each row into a table
    echo "<tr><td>"; 
    echo $row['storenumber'];
    echo "</td><td>"; 
    echo $row['location'];
    echo "</td><td>"; 
    echo $total;
    echo "</td></tr>"; 
} 

echo "</table>";
于 2013-01-05T16:03:31.700 回答
0

你可以试试这个-

$result = mysql_query("Select count(user_id) as CNT,storenumber,location from store Left join user ON user.store_number = store.store_number group by store.store_number", $con) or die(mysql_error());
$row = mysql_fetch_assoc($result );

echo "<table border='1'>";
echo "<tr> <th>Store Number</th> <th>Store Location</th> <th>Number of Users</th> </tr>";
// keeps getting the next row until there are no more to get
while($row = mysql_fetch_array( $result )) {
    // Print out the contents of each row into a table
    echo "<tr><td>"; 
    echo $row['storenumber'];
    echo "</td><td>"; 
    echo $row['location'];
    echo "</td><td>"; 
    echo $row['CNT'];
    echo "</td></tr>"; 
} 

echo "</table>";

如果您可以提供精确的表格结构，我可以更准确地构建它。

于 2013-01-05T16:09:19.983 回答