SQL 选择 - 多个孩子

Question

干杯，

我有4张桌子：

 DIRECTORY - iddir, idcust, idfile
 CUSTOMERS - id_cust, fname, lname
 FILES - idf, id_dir, id_type
 TYPEF - idtype, type_name 

基本上，我想显示这样的所有值：

ID DIR、CUST FNAME、CUST LNAME、文件名

我试过这个 SQL 查询：

  SELECT * FROM directory
        JOIN customers ON idcust = id_cust
        JOIN files ON idfile = idf AND 
        JOIN typef ON id_type = idtype

一切正常，直到最后一次 JOIN。 没有它，我没有任何错误，但是我无法打印与文件 id 对应的文件类型的名称。

我搜索过，我尝试了一些选择，但没有任何效果。 如何将“孙子表”中的一条信息带到父表？ 谢谢

PHP代码

$conn = mysqli_connect("localhost", "root", "", "demodb");
if ($conn -> connect_error){
  die("Connection failed:". $conn-> connect_error);
}
$sql = "SELECT * FROM directory
    JOIN customers ON idcust = id_cust
    JOIN files ON idfile = idf AND 
    JOIN typef ON id_type = idtype";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
  while ($row = $result->fetch_assoc()) {
    echo "<tr><td>". $row["iddir"] ." </td>
    <td>". $row["fname"] ." </td>
        <td>". $row["lname"] ." </td>
        <td>". $row["id_type"] ." </td>
        <td> ". $row["type_name"] ." </td>
        </tr>";
  }
  echo " </table>";
} else {
  echo "0 result";
}
$conn->close();

Answer 1

我希望这对你有用

$conn = mysqli_connect("localhost", "root", "", "demodb");
if ($conn -> connect_error){
  die("Connection failed:". $conn-> connect_error);
}
$sql = "SELECT d.iddir, c.fname, c.lname, f.id_type, t.type_name 
    FROM directory AS d
    LEFT JOIN customers AS c ON d.idcust = c.id_cust
    LEFT JOIN files AS f ON d.idfile = f.idf 
    LEFT JOIN typef AS t ON f.id_type = d.idtype";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
  while ($row = $result->fetch_assoc()) {
    echo "<tr><td>". $row["iddir"] ." </td>
    <td>". $row["fname"] ." </td>
        <td>". $row["lname"] ." </td>
        <td>". $row["id_type"] ." </td>
        <td> ". $row["type_name"] ." </td>
        </tr>";
  }
  echo " </table>";
} else {
  echo "0 result";
}
$conn->close();