Javascript 5秒后取下一个div
我有这个AJAX表单:Javascript 5秒后取下一个div,javascript,jquery,ajax,Javascript,Jquery,Ajax,我有这个AJAX表单: function form(){ var icao = document.getElementById('icao').value; var name = document.getElementById('name').value; var weightempty = document.getElementById('weightempty').value; var weightfull = documen
function form(){
var icao = document.getElementById('icao').value;
var name = document.getElementById('name').value;
var weightempty = document.getElementById('weightempty').value;
var weightfull = document.getElementById('weightfull').value;
var cargofull = document.getElementById('cargofull').value;
var cruisespeed = document.getElementById('cruisespeed').value;
var range = document.getElementById('range').value;
var price = document.getElementById('price').value;
var firstclassseats = document.getElementById('firstclassseats').value;
var businessclassseats = document.getElementById('businessclassseats').value;
var economyclassseats = document.getElementById('economyclassseats').value;
ajax.open("POST","new_aircraft_process.php",true);
ajax.onreadystatechange=function(){
if(ajax.readyState==4)
{
var respuesta=ajax.responseText;
document.getElementById('result').innerHTML=ajax.responseText;
$("#newaircraftdialog").dialog('close');
refreshTable(function(){$("#loadingdialog").dialog('close');});
}
}
ajax.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
ajax.send("icao="+icao+"&name="+name+"&weightempty="+weightempty+"&weightfull="+weightfull+"&cargofull="+cargofull+"&cruisespeed="+cruisespeed+"&range="+range+"&price="+price+"&firstclassseats="+firstclassseats+"&businessclassseats="+businessclassseats+"&economyclassseats="+economyclassseats);
$("#loadingdialog").dialog('open');
}
我想知道的是,当结果显示在result div中时(它是一个写着“Successful Form”的文本),它只显示5秒钟,然后消失。我已经找到了各种方法,但不想删除div。因为如果我想再次制作表单,其结果必须再次显示在div中,并在5秒后消失。您可以使用delay()
$('.myDiv').fadeIn(); // Show the div to start.
// All your other code here.
setTimeout(function() { $('.myDiv').fadeOut() }, 5000); // hide the div after we're done
您可以在重新提交表单时删除div并在稍后删除它
ajax.onreadystatechange=function(){
if(ajax.readyState==4)
{
var respuesta=ajax.responseText;
var $result = $('#result');
$result.html(ajax.responseText);
$result.show();
setTimeout(function() { $result.hide() }, 5000);
$("#newaircraftdialog").dialog('close');
refreshTable(function(){$("#loadingdialog").dialog('close');});
}
}
以后还可以使用它并将其重新附加到DOM 对不起,我没有把它放在匿名函数中。现在应该可以用了,我有点问题。当我想再做一次,没有刷新页面的对话框不工作!我不太明白你的意思。请进一步解释。如果OP希望稍后再次显示div,“remove”是错误的方法。他应该分离或者隐藏它。OP的标题是“5秒后移除一个div”,所以我在5秒后移除了一个div。我更喜欢
$('.myDiv).fadeOut()
尽管其他答案的问题相同:我做表单,一切正常,div被删除了。但是,当我想在其他时间执行表单而不重新加载页面时,AJAX脚本找不到结果div来放置结果。@user1972864,我更新了代码以显示和隐藏您的div。请尝试一下,并告诉我它是否有效。
ajax.onreadystatechange=function(){
if(ajax.readyState==4)
{
var respuesta=ajax.responseText;
var $result = $('#result');
$result.html(ajax.responseText);
$result.show();
setTimeout(function() { $result.hide() }, 5000);
$("#newaircraftdialog").dialog('close');
refreshTable(function(){$("#loadingdialog").dialog('close');});
}
}