Php 根据$\u从动态选择框中获取值回显所选内容

我想使选择框成为GET变量中的选定值。 这是我的密码：

<select name="suraNo" id="suraNo" required="required">
  <option value="">-Select-</option>
    <?php
      do {  
        ?>
          <option value="<?php echo $row_SuraList['suraNo']?>">
            <?php echo $row_SuraList['suraNo']?>. <?php echo $row_SuraList['suraNameBN']?>
          </option>

          <?php
         } while ($row_SuraList = mysql_fetch_assoc($SuraList));

         $rows = mysql_num_rows($SuraList);
         if($rows > 0) {
           mysql_data_seek($SuraList, 0);
           $row_SuraList = mysql_fetch_assoc($SuraList);
         }
    ?>
</select> 

-挑选-
. 

如果要从列表中选择与GET参数对应的值，只需使用以下命令：

<option <?php echo ($_GET['yourval'] == $row_SuraList['suraNo'] ? "selected" : null); ?> value="<?php echo $row_SuraList['suraNo']?>"><?php echo $row_SuraList['suraNo']?>. <?php echo $row_SuraList['suraNameBN']?></option>

。

此外，我建议您不要在do…while中关闭php应答器，并这样编写“选项”：

echo "<option " . ($_GET['yourval'] == $row_SuraList['suraNo'] ? "selected" : null) . " value=\"" . $row_SuraList['suraNo'] . "\">" . $row_SuraList['suraNo'] . "." . $row_SuraList['suraNameBN'] ."</option>";

echo”“$第二行的SuraList['suraNo']。"." . $第四行“suraNameBN”]；

感谢您的善意帮助，效果非常好。