如何在SQL中聯接2個表

Question

$results = mysqli_query($con,"SELECT * FROM dayalpha WHERE d_id= '".$_POST['dtb']."'");

echo "<table border='0'>
<tr>
<td>Day Name</td>
<td>Type</td>
<td>Alphabet</td>
</tr>";

while($row = mysqli_fetch_array($results))
 {
  echo "<tr>";
  echo "<td>" . $row['dayname'] . "</td>";
  echo "<td>" . $row['type'] . "</td>";
  echo "<td>" . $row['alpha'] ."</td>";
  echo "<td>" . $row['alpha1'] ."</td>";
  echo "<td>" . $row['alpha2'] ."</td>";
  echo "<td>" . $row['alpha3'] ."</td>";
  echo "<td>" . $row['alpha4'] ."</td>";
  echo "<td>" . $row['alpha5'] ."</td>";
  echo "<td>" . $row['alpha6'] ."</td>";
  echo "</tr>";
  }
echo "</table>";

在這里，我顯示了來自dayalpha表的字母。 每個字母都應鏈接到babyname表中的多個bname，而無論alpha == iname（即名字最初存儲在babyname表中）是什么。

    -----------------
    My Babyname Table
    -----------------
    iname    bname    gender    mean

    K        Komal    Female    Tender
    K        Kiran    Male      Ray
    K        Kamlesh  Male      God
    N        Nityesh  Male      Yash

    -----------------
    My dayalpha table 
    -----------------
    dayname   type     alpha  alpha1  alpha2....
    Monday    vyainjan K      G       D
    Wednesday vyainjan T      D       N

如何將dayalpha的值鏈接到babyname的多個值？

Answer 1

如果按原樣保留這些表，則此SQL可能有效。

SELECT b.*
FROM babyname a
   INNER JOIN dayalpha b ON (a.iname = b.alpha OR a.iname = b.alpha2 OR a.iname = b.alpha3 ...)
WHERE b.dayname = 'Monday'

據我了解您的問題，這是正確的答案。