Question

从代码中，我希望这两个表（个人和信息）来比较它是否具有相同的位置。如果匹配，我想显示评论中所述的个人表中的数据。

$selectall = "SELECT * FROM information";
$stmt = mysqli_query($connection, $selectall);

$compare = "SELECT * FROM personal INNER JOIN information ON personal.location = information.location";
$comparing = mysqli_query($connection, $compare);

while($row = mysqli_fetch_array($stmt)) {
   $output .= '<tr>
               <td>' .$row['name']. '</td>
               <td>' .$row['location']. '</td>
               <td>' .$row['postal']. '</td>
               </tr>
               <tr>
               <td>//This is where I want to show the matched data from personal table</td>
               </tr>';
}

Answer 1

我认为您需要的是$comparing查询的结果，而不是$selectall。

如果要返回特定列，可以操作SQL以仅返回所需的列。例如：

SELECT column_from_information, other_column_you_want, personal.location FROM personal INNER JOIN information ON personal.location = information.location

您的想法是可以使用前置表名定义列，即personal.name，information.location等。

while($row = mysqli_fetch_array($comparing)) {
   $output .= '<tr>
           <td>' .$row['name']. '</td>
           <td>' .$row['location']. '</td>
           <td>' .$row['postal']. '</td>
           </tr>
           <tr>
           <td><!--Matched Data--></td>
           </tr>';
}

Answer 2

如果您希望获取information中的所有内容，即使它在personal中没有匹配项，也请使用外部联接。对于不匹配的行，personal中的列将包含null。

您应该明确选择所需的列。如果您使用SELECT *并且两个表中都有相同名称的列，则$row['columnname']会从第二个表中获取值，如果没有匹配则为null。由于您需要第一个表中的值，请专门选择它。

$compare = "SELECT i.name, i.location, i.postal, p.somecolumn, p.anothercolumn FROM information AS i 
            LEFT JOIN personal AS p
            ON p.location = i.location";
$comparing = mysqli_query($compare);
while ($row = mysqli_fetch_assoc($comparing)) {
    $output .= '<tr>
           <td>' .$row['name']. '</td>
           <td>' .$row['location']. '</td>
           <td>' .$row['postal']. '</td>
           </tr>
           <tr>
           <td>' . $row['somecolumn'] . '</td>
           <td>' . $row['anothercolumn'] . '</td>
           </tr>';
}

如何使用php匹配来自2个不同表的列

2 个答案: