<form action="index.php?page=checkin" method="post" name="regForm">
Fullständiga namn: <input name="full_name" type="text" id="full_name" class="required">
<br >
Email: <input name="usr_email" type="text" id="usr_email3" class="required email">
<br>
Sex: <select name="sex"><option value="male">Kille</option><option value="female">Tjej</option></select>
<input type="submit" id="t" value="CheckIn">
<script>
$("#t").click(function() {
$("#Next_2").toggle();
});
</script>
<br>
<div id="Next_2" style="display: none">
Lösenord: <input name="pwd" type="password" class="required password" id="pwd"> <br>
En gång till..
<input name="pwd2" id="pwd2" class="required password" type="password" >
<br>
<input name="doRegister" type="submit" id="doRegister" value="Register">
</div>
</form>
我有这种形式。但我的问题是,当我按下CheckIn来切换其他两个表单元素时,它认为它应该是action =“index.php?page = checkin”,我希望它在下一个按钮(register)上执行,应该如何我这样做对不对?jQuery:切换按钮问题与表单动作