查询多个表格，显示错误

Question

<?php 

DEFINE ('DB_USER', 'username'); 
DEFINE ('DB_PASSWORD', 'password'); 
DEFINE ('DB_HOST', 'xxxxxxxxxxxxxxxx'); 
DEFINE ('DB_NAME', 'bigbobsveterinarysurgery'); 


$dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME) OR die ('Could not connect, check the details ' . mysqli_connect_error()); 
mysqli_set_charset($dbc, 'utf8'); 
$q = "SELECT dog_name, owner_name, vet_date, Breed, cost, address, contact_number FROM owner, dogs ORDER BY dog_name ASC";  
$r = @mysqli_query ($dbc, $q); 


    echo '<table> 
    <tr> 
    <th scope="col">Dog</th> 
    <th scope="col">Owner</th> 
    <th scope="col">Entry Date</th> 
    <th scope="col">Breed</th> 
    <th scope="col">Cost</th> 
    <th scope="col">Address</th> 
    <th scope="col">Contact Number</th> 
    </tr>' 
    ; 


    while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
     echo '<tr><td>' . $row['Dog_name'] . '</td><td>' . $row['owner_name'] . '</td><td>' . $row['vet_Date'] . 
     '</td></tr>' . $row['Breed'] . '</td></tr> ' . $row['cost'] . '</td></tr> '. $row['address'] . '</td></tr> ' . $row['contact_number'] . '</td></tr> '; 
    } 

echo '</table>'; 
?> 

它将所有数据的品种，成本和联系号码显示在一个大块中。我想要的表格的格式可以在下面找到，但是每条狗都列在每个其他所有者列表中，而其他所有列都是空白的。发布日期心不是显示的，但我认为是有问题的，当我从一个Excel工作表导入数据和日期是错误的格式

编辑

我把它更新为：

while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
     echo '<tr><td>' . $row['Dog_name'] . '</td><td>' . $row['owner_name'] . '</td><td>' . $row['vet_Date'] . 
    '</td><td>' . $row['Breed'] . '</td><td> ' . $row['cost'] . '</td><td> '. $row['address'] . '</td><td> ' . $row['contact_number'] . '</td></tr> '; 
} 

这已经停止了开始时的整个混乱，所有数据现在都格式化在一张表中。然而，每只狗仍然与每个单独的主人匹配。这是在这篇文章或数据库中的代码中的问题？

Answer 1

我认为你需要连接两个表

$q = "SELECT dog_name, owner_name, vet_date, Breed, cost, address, contact_number 
     FROM owner o 
     LEFT JOIN dogs d ON o.id = d.owner_id 
     ORDER BY d.dog_name ASC";  
// make sure you join on valid columns. 
// also, if there is a column that appears in both tables, 
// to avoid ambiguity, prefix it with the corresponding table alias 
$r = @mysqli_query ($dbc, $q); 

echo "<table> 
    <tr> 
    // skipping this so it doesn't eat up too much space 
    </tr>"; 

// also, every `$row[something]` should be placed inside a `<td></td>`, like so: 
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) { 
    echo '<tr> 
      <td>' . $row['Dog_name'] . '</td> 
      <td>' . $row['owner_name'] . '</td> 
      <td>' . $row['vet_Date'] . '</td> 
      <td>' . $row['Breed'] . '</td> 
      <td>' . $row['cost'] . '</td> 
      <td>' . $row['address'] . '</td> 
      <td>' . $row['contact_number'] . '</td> 
     </tr> '; 
} 

两个表中的选择使他们之间的笛卡尔乘积。
使用join，您只能得到相应的行（根据ON子句）。

在你的特殊情况下，使用join将保持狗主的关系，而使用FROM owner, dogs，你会得到每个所有者的所有狗，这可能不是你所追求的。