iam*_*esy 6 javascript php forms postback
这基本上就是标题所说的..我有一个带有选择控件的表单,我想强制表单在更改时回发给自己.
$bmsclientlist = $clientobj->getBMSClientList();
echo '<form name="changebmsid" method="post" action="' . $_SERVER['PHP_SELF'] . '"><select name="bmsid">';
foreach($bmsclientlist as $bmsclient) {
$var = '';
if($client['bmsid'] == $bmsclient['id']) {
$var = ' selected="selected"';
}
echo '<option value="' . $bmsclient['id'] .'"'. $var .'>' .$bmsclient['clientname'] . '</option>';
}
echo '</select></form>';
$backupobj = new AdminBackup();
if(isset($_POST['bmsid']){
$statusarray = $backupobj->getStatusTotalsbyId($_POST['bmsid']);
}else{
$statusarray = $backupobj->getStatusTotals();
}
我知道它会涉及一些javascript,但我不太清楚如何实现这一点.
任何帮助最感谢!
谢谢,
Jonesy
acm*_*acm 12
这<select>将提交父表单
<form method="post" action="#" name="myform">
<select name="x" onchange="myform.submit();">
<option value="y">y</option>
<option value="z">z</option>
</select>
</form>
您所要做的就是为您的名字命名<form>,并将onchange事件添加到您的<select>...
亚当是对的.虽然上面的例子很完美,但我会这样做:
使用jQuery但还有许多其他选项可用...
<head>
<script type="text/javascript" src="jquery.js"></script>
<script>
$(document).ready(function(){
$('#mySelect').change(function(){
myform.submit();
});
});
</script>
</head>
和形式
<body>
<form method="post" action="" name="myform">
<select name="x" id="mySelect">
<option value="y">y</option>
<option value="z">z</option>
</select>
</form>
</body>
小智 7
快速解决这个问题:
将onchange属性添加到选择列表中
<select name="bmsid" onchange="javascript: form.submit();">