php - php mysql 从2个表的列中选择加入

Question

我有两张桌子。visitor_details，带有 id、scanner_id、time 列，visitors_info 带有scanner_id、name、surname 列

我想在表格中取回 id、name、surname、time

我写了这个，但不工作

$result = mysql_query("SELECT visitors_details.id AS id, 
visitors_info.name AS name, visitors_info.surname AS surname, visitors_details.time 
AS time FROM visitors_details AS d LEFT JOIN visitors_info AS i ON 
d.scanner_id=i.scanner_id ");

echo "<table border='1'>
<tr>
<th>id</th>
<th>name</th>
<th>surname</th>
<th>Time</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
 echo "<tr>";
 echo "<td>" . $row['id'] . "</td>";
 echo "<td>" . $row['name'] . "</td>"; 
 echo "<td>" . $row['surname'] . "</td>";
 echo "<td>" . $row['time'] . "</td>";
  echo "</tr>";
  }
   echo "</table>";

有任何想法吗？？

Answer 1

试试这个查询

 $result = mysql_query("SELECT d.id , i.name , i.surname , d.time 
                       FROM visitors_details AS d LEFT JOIN visitors_info AS i 
                        ON d.scanner_id=i.scanner_id ");

Answer 2

最好为您的代码启用一些调试，如下所示：

<?php
error_reporting(E_ALL);
$sql = "
SELECT d.id AS id, i.name AS name, i.surname AS surname, d.time  AS time 
  FROM visitors_details AS d 
  LEFT JOIN visitors_info AS i ON d.scanner_id=i.scanner_id
";

$result = mysql_query($sql);
if (!$result) {
    die('Invalid query: ' . mysql_error());
}

?> 

Answer 3

添加它以捕获错误。节省大量时间：

if(!$result) {
    echo mysql_error();
}