在弹出框中显示消息

Question

I am trying to display popup box using jquery showing a success message after updating the table row.

This is code that I am trying :

$.ajax({
    type: "POST",
    url: "process.php", 
    dataType: 'html',
    data: { 
        name: $('#name').val(), 
        address: $('#address').val(), 
        city: $('#city').val() 
    },
    beforeSend: function(){$('#loading').show();},
    success:function(data){
        $('#manage_user table > tbody:last').find('tr:first').before(data);

        $('#success').dialog({  
            autoOpen: false,
            height: 'auto',
            width: 350,
            modal: true
        });

        setTimeout("$('#success').hide(); ", 3000);
    },
    error:function (xhr, ajaxOptions, thrownError){
        alert(thrownError);
    }, 
    complete: function(){
        //alert('update success'); 
    }
});

But my problem is after update finished this message is not display as a popup.

This is my HTML -

<div id="success" title="Hurray,">
    <p>User table is updated.</p>
</div>

Can anybody tell me where I am going wrong? Thank you.

Answer 1

Try removing this line

autoOpen: false,

From official documentation

autoOpen

Default: true

If set to true, the dialog will automatically open upon initialization. If false, the dialog will stay hidden until the open() method is called.

EDIT

Chage your setTimeout to this

setTimeout("$('#success').dialog('close');", 3000);