我的.load函数存在问题。加载的页面没有在加载的div中执行Jquery。我确保Jquery确实在标签中。我在想.load调用存在问题,但我是新手。有什么想法吗?
<script type="text/javascript">
var partyid = '<?php echo sanitize_text_field($_GET["Party_ID"]); ?>';
jQuery(document).ready( function($){
if (localStorage.user2==localStorage.username2){
var e = document.getElementById('display');
if(e.style.display == 'block')
e.style.display = 'none';
$('#Submission-2').load('http://example.com/play/?Party_ID=' + partyid + ' #Submission' );
refresh();
}
});
</script>
<script type="text/javascript">
function refresh()
{
setTimeout( function() {
$('#Submission-2').fadeOut('slow').load('http://example.com/play/?Party_ID=' + partyid + ' #Submission' , function() {
var username = '<?php echo $party_information[0]->Username; ?>';
if (document.getElementById('total').value!=username){
window.location.reload();
}
});
$('#Submission-2').fadeIn('slow').load('http://example.com/play/?Party_ID=' + partyid + ' #Submission' );
{
refresh();
}
}, 5000);
}
</script>
未执行的Jquery
<script type="text/javascript">
jQuery("#Answertoggle3 .lietruth").click(function(){
jQuery("#Answertoggle3 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
jQuery("#Answertoggle2 .lietruth").click(function(){
jQuery("#Answertoggle2 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
jQuery("#Answertoggle1 .lietruth").click(function(){
jQuery("#Answertoggle1 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
</script>
也代替执行window.location.reload();功能。我想结束包含函数refresh()的执行。每5秒后刷新一次。所以基本上停止刷新。
答案 0 :(得分:0)
根据我对你的问题的理解,试试这样的事情:
<script type="text/javascript">
$(document).ready(function(){
$("#Answertoggle3 .lietruth").click(function(){
$("#Answertoggle3 .lietruth").removeClass('active');
$(this).toggleClass('active');
});
$("#Answertoggle2 .lietruth").click(function(){
$("#Answertoggle2 .lietruth").removeClass('active');
$(this).toggleClass('active');
});
$("#Answertoggle1 .lietruth").click(function(){
$("#Answertoggle1 .lietruth").removeClass('active');
$(this).toggleClass('active');
});
})
</script>
答案 1 :(得分:0)
这是我添加的内容,它使切换工作暂时通过fadeOut.load和fadeIn。当这是真的时,我需要停止执行refresh()函数。
if (document.getElementById('total').value!=username)
而不是
window.location.reload();
的Javascript
<script type="text/javascript">
function refresh()
{
var quitfunction = 0;
setTimeout( function() {
$('#Submission-2').fadeOut('slow').load('http://example.com/play-with-friends/?Party_ID=' + partyid + ' #Submission' ,function() {
jQuery("#Answertoggle3 .lietruth").click(function(){
jQuery("#Answertoggle3 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
jQuery("#Answertoggle2 .lietruth").click(function(){
jQuery("#Answertoggle2 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
jQuery("#Answertoggle1 .lietruth").click(function(){
jQuery("#Answertoggle1 .lietruth").removeClass('active');
jQuery(this).toggleClass('active');
});
var username = '<?php echo $party_information[0]->Username; ?>';
if (document.getElementById('total').value!=username)
{
window.location.reload();
}
});
$('#Submission-2').fadeIn('slow').load('http://example.com/play-with-friends/?Party_ID=' + partyid + ' #Submission' );
{
if (quitfunction==1)
return;
refresh();
}
}, 5000);
}
</script>