从AJAX呼叫处理成功

Question

我想处理来自下面提交的成功呼叫并重新加载当前页面。 我怎样才能最好地实现这一目标？

$('#modal-confirm').click(function() { 
    $(form).ajaxSubmit(); 
    $(form).resetForm(); 
    $('#myModal').modal('toggle'); 
}); 

Answer 1

您可以通过选项来ajaxSubmit()

$(form).ajaxSubmit({ 
    success:function(){ 
    alert("complete"); 
    // do other stuff 
    } 
}); 

Answer 2

如果你想让它重新加载当前页面可以或许试试这个了..

$('#modal-confirm').on("click", function(e){ 
    e.preventDefault(); 
    $.ajax({ 
    url  : '...' 
    }).done(function(data) { 
    //handle your conditions here once the request is successful. if conditions are met, then... 
    location.reload(true); 
    }); 
}); 

Answer 3

$(document).ajaxSuccess(function() { // http://api.jquery.com/ajaxSuccess/ 
    location.realod(); 
    }) 
    $('#modal-confirm').click(function() { 
     $(form).ajaxSubmit(); 
     $(form).resetForm(); 
     $('#myModalOne').modal('toggle'); 
    });