按钮点击后保持最后从下拉列表中选择的值

Question

就像标题所说，我在实现它时遇到困难。

这里是我的下拉列表：

<?php 
     $query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data"; 
     $result = mysql_query ($query); 
     echo "<select name=data value=''>Data</option>"; 
     while($nt=mysql_fetch_array($result)){ 
     echo "<option value=$nt[data] name=\"blabla\">$nt[data]</option>"; 
     } 
     echo "</select>"; 
?> 

这里的buttonclick：

<?php 
     if(isset($_POST['Submit'])) 
     { 
     $query = "SELECT SUM(suma), paskirtis FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' AND data ='".$_POST['data']."' group by paskirtis"; 
     $result = mysql_query ($query); 
     echo "<tr><td>Paskirtis:</td><td>Biudzetas:</td><td>Isleista:</td><td>Likutis:</td></tr>"; 
     while($nt=mysql_fetch_array($result)){ 
      if($nt['SUM(suma)'] != null){ 
      $suma = $nt['SUM(suma)']; 
     } 
      echo "<tr><td>$nt[paskirtis]</td> 
      <td><input type=\"text\" name=\"isleista[]\" value=\"Skiriamų pinigų kiekis...\" method=\"post\"></td><td>".$suma." Lt</td><td>--</td></tr> <br>"; 
     } 
     } 
?> 

后，我按它，它检索我从我从下拉列表中选择日期所需的数据还重置整个下拉列表，显示来自sql数据库的日期的第一个值，而不是我选择的那个。如果有人知道如何将选定的值保存在列表中，那么任何帮助都会非常有用！

Answer 1

试试这个，你需要将select="selected"放在while循环中。请参见下面的代码，我怎么放在$selected

<?php 
     $query = "SELECT data, rel_id FROM $tbl_rel_balansas INNER JOIN $tbl_balansas ON $tbl_rel_balansas.rel_id = $tbl_balansas.id WHERE $tbl_rel_balansas.member_id = '$_SESSION[id]' group by data"; 
     $result = mysql_query ($query); 
     echo "<select name=data value=''>Data</option>"; 
     while($nt=mysql_fetch_array($result)){ 
     $selected = ($_POST['blabla'] == $nt[data])?'selected="selected"':NULL; 
      echo "<option value=$nt[data] name=\"blabla\" $selected >$nt[data]</option>"; 
     } 
     echo "</select>"; 
?>