所以我有3个无线电按钮 具有不同的值,当我选择其中一个它总是给我的价值“3”或他们的最后一个值。我尝试提交没有AJAX它工作正常!AJAX发送错误信息
这里是我的.js代码
$(document).ready(function(){
$('form').on('submit', function() {
var that = $(this),
url = that.attr('action'),
type = that.attr('method') ,
data = {};
that.find('[name]').each(function(index, value){
var that = $(this),
name = that.attr('name'),
value = that.val();
data[name] = value;
});
$.ajax({
url: url,
type: type,
data: data,
success : function(rdata) {
$("#st").html(rdata);
}
});
return false;
});
});
,这里是我的html:
<form action="test.php" id="send" method="POST" accept-charset="UTF-8" autocomplete="off">
<table>
<tr><td><input type="radio" name="val" value="1">1</td></tr>
<tr><td><input type="radio" name="val" value="2">2</td></tr>
<tr><td><input type="radio" name="val" value="3">3</td></tr>
<tr><td><input type="submit"></td></tr><tr><td><div id='st'></div>/td></tr>
</table>
test.php的:
if (isset($_POST['val'])) {
$val = $_POST['val'];
echo $val;
}
谢谢先生,曾担任魅力:D – maxhill901 2014-08-27 11:47:18