我有以下html:
<html>
<body>
<div id='refresh'>
<form id='form1' action='do.php' method='post'>
<input type='checkbox' name='thetickets[]' value='1'>
<input type='hidden' name='gender'>
<input type='submit' value='Go'>
</form>
</div>
</body>
</html>
以下JS:
<script type="text/javascript">
$(document).ready(function() {// REFRESH DIV EVERY 5 SECONDS
interval = setTimeout(refreshpage, 5000);
function refreshpage() {
$('#refresh').load('page.php?&timer='+new Date().getTime()+' #refresh');
interval = setTimeout(refreshpage, 5000);
}
});
</script>
<script type="text/javascript"> /SUBMIT FORM
$(document).ready(function() {
var options = {
target: '',
dataType: 'html',
beforeSubmit: showRequest_reassign,
success: showResponse_reassign
};
$("#refresh").delegate("#form1,"submit",function () {
clearTimeout(interval);
$(this).ajaxSubmit(options);
return false;
});
});
function showRequest_reassign(formData, jqForm, options){
return true;
}
function showResponse_reassign(responseText, statusText, xhr, $form){
alert(responseText)
}
</script>
如果我在5秒之前提交表单,则表单正确提交,但如果我等待5秒,页面刷新,我提交表单,但输入类型“thetickets []”不提交。我检查开发人员工具,表单数据,并且值“thetickets []”甚至不存在....我迷失了为什么..
谢谢,
答案 0 :(得分:1)
我想在这里发布我的编辑代码,这可能有助于您找到问题所在:
<html>
<body>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function() {// REFRESH DIV EVERY 5 SECONDS
interval = setTimeout(refreshpage, 5000);
function refreshpage() {
var thisurl = document.URL;
$('#refresh').load(thisurl + '?timer='+new Date().getTime()+' #refresh');
interval = setTimeout(refreshpage, 5000);
console.log("reloaded");
}
});
</script>
<script type="text/javascript"> //SUBMIT FORM
$(document).ready(function() {
var options = {
target: '',
dataType: 'html',
beforeSubmit: showRequest_reassign,
success: showResponse_reassign
};
$("#refresh").delegate("#form1","submit",function () {
clearTimeout(interval);
$(this).ajaxSubmit(options);
return false;
});
});
function showRequest_reassign(formData, jqForm, options){
return true;
}
function showResponse_reassign(responseText, statusText, xhr, $form){
alert(responseText)
}
</script>
<div id='refresh'>
<form id='form1' action='http://atlantis.cit.ie/displayvalues.php' method='post'>
<input type='checkbox' name='thetickets[]' value='1'>
<input type='hidden' name='gender'>
<input type='submit' value='Go'>
</form>
</div>
</body>
</html>
如果您能向我提供有关您所犯错误的更多信息,我非常乐意为您提供帮助!
答案 1 :(得分:0)
我发现这个代码唯一的东西(除了上面提到的拼写错误)是你的“thisurl”var必须是一个php页面,它必须已经在查询字符串中有其他vars。
注意:如果您实际选中了该框,则只会获得$thetickets
数组。
答案 2 :(得分:0)
发现问题,我的原始代码结构不正确。
<div>
<form>
</div>
</form>
删除了div,现在表单正常工作。