Javascript Jquery延迟停止代码,就像警报框一样
如果我使用警报停止运行ddl“Suburb1”的代码,如果我使用“delay”,则使用正确的值填充该值,但未设置该值。我需要某种方法来停止在“change”之后运行的代码,以便$(“#Suburb1').val(SuburbVC);从DB填充ddl Suburb1时,不会立即触发Javascript Jquery延迟停止代码,就像警报框一样,javascript,jquery,Javascript,Jquery,如果我使用警报停止运行ddl“Suburb1”的代码,如果我使用“delay”,则使用正确的值填充该值,但未设置该值。我需要某种方法来停止在“change”之后运行的代码,以便$(“#Suburb1').val(SuburbVC);从DB填充ddl Suburb1时,不会立即触发 if ($(this).attr("checked") == true) { var PostCode = $('#PostCode').val(); var SuburbVC = $('#SuburbVC').val
if ($(this).attr("checked") == true) {
var PostCode = $('#PostCode').val();
var SuburbVC = $('#SuburbVC').val();
$('#PostCode1').val(PostCode);
// Another function is called which populates Dropdown list from DB
// If I use delay Suburb1 is not populated
$('#PostCode1').change().delay(5000);
//If I use an alert Suburb1 is populated
// alert('delay');
$('#Suburb1').val(SuburbVC);
} else {
$('#PostCode1').val("");
}
谢谢你解决这个问题的方法不对;在从DB填充值之后,应该添加回调函数来执行代码的其余部分
if ($(this).attr("checked") == true) {
var PostCode = $('#PostCode').val();
var SuburbVC = $('#SuburbVC').val();
$('#PostCode1').val(PostCode);
// Another function is called which populates Dropdown list from DB
// If I use delay Suburb1 is not populated
$('#PostCode1').change().delay(5000);
//If I use an alert Suburb1 is populated
// alert('delay');
$('#Suburb1').val(SuburbVC);
} else {
$('#PostCode1').val("");
}
如何填充数据库中的值?AJAX如果是这样,请在数据准备就绪后向success
处理程序添加一个函数调用,其中包含要执行的代码