Question

我知道我编码错误，但似乎找不到纠正它的方法。目的是显示一个下拉列表，其中填充了mysql数据库的结果。它目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。回声是否应该超出while循环？或者是正确的位置，但在其他地方是错误的？如果有人能检查它并告诉我我的错误在哪里，我会很感激的。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

如果你的目标是让所有选项都有一个下拉列表，那么你需要为while的外部的打开和关闭标签放置回声并离开选项标签 in while：

// 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>';

Answer 2

试试这个：

echo '<select name="customer">';
echo '<option value="">Select delivery address</option>';
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']. " ". $row_rs_select_address2['postcode_com'];
    echo '<option value="address">'.$address.'</option>';
}
echo '</select>';

Answer 3

while循环每次都会生成一个select元素。

你应该把select放在while

之外

echo "<select>";
while(/* while statement */){
    echo "<option></option>";
}
echo "</select>";

Answer 4

将您的时间改为此

echo '<select name="customer">';
echo '<option value="">Select delivery address</option>';
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']. " ". $row_rs_select_address2['postcode_com'];       
     echo '<option value="address">'.$address.'</option>'
}
echo '</select>';

Answer 5

尝试这样的事情：

echo '<select name="customer">';
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']. " ". $row_rs_select_address2['postcode_com'];
        echo '<option value="">Select delivery address</option>'.'<option value="address">'.$address.'</option>';
            }
echo '</select>';

你们每个人（表中的一行）都有所不同，希望这有助于：）。

php while循环输出为每个项目生成select

5 个答案: