Question

php下拉框点击跳转不了？请高手帮忙一下

<form action='ss1.php' method="GET">
<select name='select'>
<option value="title">标题</option>

<option value="content">内容</option>
</select>
<input type= "submit" value="提交">
</form>

ss1.php

<?php
$title=$_GET[title];
echo"<META HTTP-EQUIV=\"Refresh\" Content=\" 1;URL=http://localhost/delete.php\">";

?>
<?php
$content=$_GET[content];
echo"<META HTTP-EQUIV=\"Refresh\" Content=\" 1;URL=http://localhost/ff.php\">";

?>

Answer 1

支持IE、FF、chrome的标准语法用document.getElementById()
如下：

<form action="ss1.php" method="GET" id="form_jump">

<select name="select" onChange="javascript:document.getElementById('form_jump').submit();">

<option value="delete">标题</option>
<option value="ff">内容</option>

</select>
<input type= "submit" value="提交">
</form>

ss1.php

<?php
echo '<META HTTP-EQUIV="Refresh" Content=" 0;URL=/' . $_GET['select'] . '.php">';
exit();
?>

Answer 2

<select name='select' onchange="form.submit();">

Answer 3

<select name='select' onchange="javascript:location.href=xx">

Answer 4

<select name='select' onchange="top.location.href='ss1.php?action='+this.value">

Answer 5

我试过了，跳转没有问题，但是，你接的参数，或者说$_GET[]的键不应该是titile，应该是select的name的值，即select
不清楚你想要这个title、content做什么，呵呵