提交后保留查询生成的下拉框的值

Question

我使用查询结果填充了下拉框的选项。用户提交后如何保留选定的值？

下面的代码：

$query="SELECT trainingName,trainingID FROM training ORDER BY trainingName"; 

$result = mysql_query ($query); 
echo "<select name='training' value=selected>Training Name</option>"; 
$training = strip_tags(@$_POST['training']); 

echo "<option>---------------------Select---------------------</option>"; 
while($nt=mysql_fetch_array($result)){ 
echo "<option value=$nt[trainingID]>$nt[trainingName]</option>"; 
} 

谢谢！

Answer 1

试试这个：

$query="SELECT trainingName,trainingID FROM training ORDER BY trainingName"; 

$result = mysql_query ($query); 
echo "<select name='training'>"; 
echo "<option>---------------------Select---------------------</option>"; 
while($nt=mysql_fetch_array($result)){ 
    $selected = false; 
    // check if the current value equals the value submited 
    if($_POST['training'] == $nt['trainingID']){ 
     $selected = true; 
    } 

    // show selected attribute only if $selected is true 
    echo "<option value='{$nt['trainingID']}' ". ($selected ? "selected" : "") .">{$nt['trainingName']}</option>"; 
} 
echo '</select>';