php while循环输出为每个项目生成select

php while循环输出为每个项目生成select,php,Php,我知道我的编码不正确,但似乎找不到纠正的方法。其目的是显示一个下拉列表,其中填充来自mysql数据库的结果。它当前显示每个地址的下拉列表。我知道这是为什么,但似乎无法纠正它。回音应该在while循环之外吗?或者位置正确,但在其他地方是错误的?如果有人能检查一下并告诉我哪里出了问题,那就太好了。非常感谢 <?php $customer = mysql_real_escape_string( $_GET["customer"] ); $con = mysql_

我知道我的编码不正确,但似乎找不到纠正的方法。其目的是显示一个下拉列表,其中填充来自mysql数据库的结果。它当前显示每个地址的下拉列表。我知道这是为什么,但似乎无法纠正它。回音应该在while循环之外吗?或者位置正确,但在其他地方是错误的?如果有人能检查一下并告诉我哪里出了问题,那就太好了。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>';

            }

?>
while循环每次都生成一个select元素

你应该把select放在while之外

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

如果您的目标是使用一个下拉列表来显示所有选项,那么您需要将select的open和close标记的回显放置在while之外,并将选项标记保留在while中:

// 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>';
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>';

把你的时间换成这个

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>';

试着这样做:

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>';

您正在为表中的每一行设置不同的地址,希望这能有所帮助:。

选择送货地址选项也应该在表外:-