php while循环输出为每个项目产生选择

Question

我知道我的编码有误，但似乎找不到更正它的方法。目的是显示一个下拉列表，其中填充了来自 mysql 数据库的结果。它当前显示每个地址的下拉列表。我知道为什么会这样，但似乎无法纠正它。 echo 应该在 while 循环之外吗？还是位置正确但其他地方错误？如果有人可以检查它并告诉我哪里出了问题，我将不胜感激。非常感谢。

<?php
    $customer = mysql_real_escape_string( $_GET["customer"] );         
    $con = mysql_connect("localhost","root","");
    $db = "sample";
      if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }

        mysql_select_db($db, $con);
        $query_rs_select_address2 = sprintf("SELECT * FROM company_com where idcode_com = '$customer'");
        $rs_select_address2 = mysql_query($query_rs_select_address2, $con) or die(mysql_error());
        $row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2);
        $totalRows_rs_select_address2 = mysql_num_rows($rs_select_address2);

        while ($row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2))
            {  
        $address=$row_rs_select_address2['address1_com']. " ". $row_rs_select_address2['address2_com']. " ". $row_rs_select_address2['address3_com']. " ". $row_rs_select_address2['town_com']. " ". $row_rs_select_address2['postcode_com'];
        echo '<select name="customer">'.'<option value="">Select delivery address</option>'.'<option value="address">'.$address.'</option>'.'</select>';

            }

?>

Answer 1

如果您的目标是让一个下拉菜单包含所有选项，那么您需要将 echo 的 select 的打开和关闭标记放置在 outside 的 while 中，并保留选项标签在期间:

// also put first option on the outside as you *don't* want to repeat that.
echo '<select name="customer"><option value="">Select delivery address</option>';
while ($row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2))
{  
    // only place the code in here you want repeated for every value in the query
    /* 
        have you considered concatenating in the query?
        Basically:
        select concat( address1_com, ' ', address2_com, ' ', address3_com, ' '
               town_com, ' ', postcode_com ) as full_address From...

        Sometimes the greater number of records means slower execution.
        Of course, you should benchmark to be sure.
    */
    $address=  $row_rs_select_address2['address1_com']. " ". 
               $row_rs_select_address2['address2_com']. " ". 
               $row_rs_select_address2['address3_com']. " ". 
               $row_rs_select_address2['town_com']. " ". 
               $row_rs_select_address2['postcode_com'];
    // notice I swapped out $uid for address. You want to have each option 
    // reflect a different value so that the POST gives a uid to the server
    // You can probably get a uid from the primary key of the company_com 
    // table.
    echo '<option value="$uid">'.$address.'</option>';

}
echo '</select>';