Question

您好我有一个下拉列表，它将从表（data_builder）中检索值，并希望组合另一个表（data_suburb）的现有郊区。因此，当我点击下拉列表时，它将显示所选的郊区和data_suburb的所有郊区

示例I具有：

TABLE（data_builder）

<b>id | builder_name | builder_suburb | builder_state | builder_postcode</b><br>

   1  | Biz Shop     | Richmond       | VIC           | 3121<br>
   2  | Tony Sykes   | Melbourne      | VIC           | 3000<br>
   3  | HDI Group    | Abbotsford     | VIC           | 3607<br>

等...

TABLE（data_suburb）

<b>id | suburb       | suburb_state  | suburb_postcode</b><br>

   1  | Abbotsford   | VIC           | 3067<br>
   2  | Aberfeldie   | VIC           | 3040<br>
   3  | Aberfeldy    | VIC           | 3825<br>

等...

这是我尝试过的代码

include $_SERVER['DOCUMENT_ROOT']."/includes/conn.php";
$id =$_REQUEST['id'];
$result = mysql_query("SELECT * FROM data_builder WHERE id  = '$id'");
$retrieve = mysql_fetch_array($result);
$BuildSuburb=$retrieve['builder_suburb'] ;

echo "<select name='buildersuburb'>";
      $query="SELECT suburb FROM data_suburbs ORDER BY suburb ASC";
      $result = mysql_query($query);


            if(!$result){

                die ("Could not query the database: <br />" . mysql_error());

            }

while ($data=mysql_fetch_array($result)){
             echo '<option value="' . $data[suburb];
             if($data[suburb]=$BuildSuburb): echo $data[suburb] . '"' .     'selected="selected"';
             else: echo $data[suburb] . '"';
             endif;
             echo '" >' . $data[suburb] . '</option>'. "\n";

}




echo "</select>";

Answer 1

在循环$ data = mysql_fetch_array（$ result）中将以下行更改为

$ data = mysql_fetch_assoc（$ result）因为您正在使用$ data ['suburb']

同样在您的$查询中，不需要ASC

在下拉选项中组合两个mysql表结果并检索选中

1 个答案: