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> \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?
_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
我不确定 - 我还在学习。这里有一个