使用外鍵將數據從一個表插入另一個表中從下拉菜單中插入數據

Question

好的，我想插入/更新一個新區域到我的數據庫的區域表中，一個區域應該鏈接到一個建築物（buildingID是一個fk in區表），目前我已經做了一個表格，以便我可以進入與建築物鏈接的新區域，但是我必須手動鍵入buildingID，我想要的是從建築物表格中的所有建築物名稱顯示在下拉框中，以便我可以選擇它，而不必在添加新區域時鍵入buildingID。

希望這是有道理的，非常感謝任何能夠幫助我的人！

?php 
$con = mysql_connect ("localhost","user","pass"); 
if (!$con){ 
die("can not connect: " . mysql_error()); 
} 
mysql_select_db("dbname",$con); 

if(isset($_POST['update'])){ 
$updatequery = "UPDATE zones SET Zonename='$_POST[zonename]', Zonenumber='$_POST[zonenumber]', Zonecapacity='$_POST[zonecapacity]', buildingID='$_POST[buildingid]'WHERE zoneID='$_POST[hidden]'"; 
mysql_query($updatequery, $con); 
}; 

if(isset($_POST['add'])){ 
$addquery = "INSERT INTO zones (Zonename, Zonenumber, Zonecapacity, buildingID) VALUES  ('$_POST[uname]', '$_POST[ucapacity]', '$_POST[unumber]', '$_POST[ubuilding]')"; 
mysql_query($addquery, $con); 
}; 


$sql = "SELECT * FROM zones"; 

$myData = mysql_query($sql,$con); 
echo "<table> 
<tr> 
<th> </th> 
<th>Zone Name</th> 
<th>Zone Number</th> 
<th>Zone Capacity</th> 
<th>Building ID</th> 
</tr>"; 
while ($record = mysql_fetch_array($myData)){ 
echo "<form action=addeditzone.php method=post>"; 
echo " <tr>"; 
echo "<td>" . $record['zoneID'] . " </td>"; 
echo "<td>" . "<input type=text name=zonename value='" . $record['Zonename'] . "' </td>"; 
echo "<td>" . "<input type=text name=zonenumber value='" . $record['Zonenumber'] . "' </td>"; 
echo "<td>" . "<input type=text name=zonecapacity value='" . $record['Zonecapacity'] . "' </td>"; 
echo "<td>" . "<input type=text name=buildingid value='" . $record['buildingID'] . "' </td>"; 
echo "<td>" . "<input type=hidden name=hidden value=" . $record['zoneID'] . " </td>"; 
echo "<td>" . "<input type=submit name=update value=update" . " </td>"; 
echo "</form>"; 
} 
echo "<form action=addeditzone.php method=post>"; 
echo "<tr>"; 
echo "<td><td><input type=text name=uname></td>"; 
echo "<td><input type=text name=unumber></td>"; 
echo "<td><input type=text name=ucapacity></td>"; 
echo "<td><input type=text name=ubuilding></td>"; 
echo "<td>" . "<input type=submit name=add value=add>" . " </td>"; 
echo "<tr>"; 
echo "</form>"; 
echo "</table>"; 
mysql_close($con); 

?> 

Answer 1

這是用數據庫表格數據填充保存箱的代碼。

<?php 
    $con = mysql_connect ("127.0.0.1","root"); 
    if (!$con){ 
     die("can not connect: " . mysql_error()); 
    } 
    mysql_select_db("examples",$con); 
if (isset($_POST['submit'])){ 




    $sql = "INSERT INTO zones (Zonename,Zonecapacity,buildingID) VALUES 
    ('$_POST[Zonename]','$_POST[Zonecapacity]','$_POST[BuildingID]')"; 
    $inserted=mysql_query($sql,$con); 
    if($inserted) { 
     echo "New Zone succesfully entered"; 
    } else { 
     echo "fail ". mysqli_error($con); 
    }; 
} else { 
    echo "<form action='insertzone1.php' method='post'> 
      Insert New Zone:<br><br> 
      Zone Name:<br><input type='text' name='Zonename'><br><br> 
      Zone Capacity:<br><input type='int' name='Zonecapacity'><br><br> 
      Building:<br>"; 
    $sql = "select * from building"; 
    $res = mysql_query($sql,$con); 
    echo mysql_error($con); 
    echo "<select name='BuildingID'>"; 
    while ($row = mysql_fetch_array($res)){ 

     echo "<option value=" . $row['buildingID'] . ">". $row['buildname'] . "</option>"; 
    } 
    echo "</select><br><input type='submit' name='submit'> 
      </form>"; 


    mysql_close($con); 
};