2012-02-22 52 views
2

我使用查詢結果填充了下拉框的選項。用戶提交後如何保留選定的值?提交後保留查詢生成的下拉框的值

下面的代碼:

$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>"; 
} 

謝謝!

+0

http://stackoverflow.com/questions/9339225/how-do-retain-value-of-dropdown-box-in-php – CodeZombie 2012-02-22 09:40:11

回答

1

試試這個:

$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>'; 
+0

感謝的可能的複製!爲我工作很好。 – Programmer 2012-02-22 09:49:34

+0

太好了。我很高興它工作:) – 2012-02-22 09:51:54