從兩個MYSql表中獲取數據並顯示為HTML表

Question

我有兩個表，第一個是“用戶”表，其中有一個名為“商店”的列，我也有一個名為“商店”的表，其中有“商店編號”，“商店位置”列。

用戶表中的“商店”列是“商店編號”。

我想做的是創建一個類似HTML表格

樣本數據：

店鋪編號：34店鋪位置：倫敦用戶：34

店鋪編號| 店鋪位置| 該商店的用戶數|

因此，就像從商店中選擇*並為每個創建新行。

對於用戶數量，可以使用諸如sum *來自用戶的數字，其中來自store表的'store'='store number'。

我希望這是有道理的，

插口。

更新：

這是對的：

$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>";

表格：

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users不存在，則創建表（

  `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 ;

Answer 1

試試這個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'];

打印用戶數。

Answer 2

嘗試這個：

$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>";

Answer 3

你可以試試這個

$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>";

如果您可以提供准確的表格結構，則可以更准確地進行構圖。