2014-09-05 58 views
0

我有以下代码:如何让从选定字段中选择的值不再消失?

<form method="post" enctype="multipart/form-data"> 
<table id="box-table-a" summary="PAGE HEADER" style='width:10%'> 
<tr> 
<th> 
<?php 
include('./db.php'); 
$PM = mysqli_query($con, "SELECT DISTINCT Name FROM report ORDER BY Name ASC"); 
echo "<b>Sort by Name:</b>&nbsp;\n"; 
echo " <select name='Name' onChange='submit(this.form)'>\n"; 
while($row = mysqli_fetch_row($PM)) { 
$sel = ($table === $row[0]) ? "id='sel' selected" : ""; 
printf(" <option %s value='%s'>%s</option>\n", $sel, $row[0], $row[0]); 
    } 
    echo " </select>\n"; 
    ?> 
</th> 
</tr> 
</form> 

一旦做出选择,选择框默认回任何值是在列表的顶部(从SELECT SQL查询...)。如何阻止这种情况的发生并保持在选择框中选定的值,直到做出其他选择?

+0

_once作出了选择,选择框默认回来......你正在离开一步。选择完成之后和选择变回默认之前会发生什么?一个页面刷新?表单帖子? JavaScript操作? – 2014-09-05 18:28:24

+0

我不确定 - 我还在学习。这里有一个