i have multiple form, so i create a form with button submit outside form, this form
我有多个表单,所以我创建了一个表单,在表单外提交按钮
$("#buttonSubmit").click(function (event) {
event.preventDefault();
if (confirm("Anda yakin akan Checkout ?")) {
var formData = new FormData("form#formData");
$(".loader").show();
$.ajax({
url: 'belanja/belanja_crud.php',
type: 'POST',
data: formData,
async: false,
cache: false,
contentType: false,
processData: false,
dataType: 'json',
success: function (data) {
//console.log(data);
}
});
}
return false;
});
<form id='formData'>
<!-- input bla bla bla -->
</form>
<button id='buttonSubmit' type='button'>Submit</button>
how to get all input from form on above and submit with ajax ?
如何从上面的表单中获取所有输入并使用ajax提交?
1 个解决方案
#1
2
To get all data in the form in one instruction you can use .serialize()
要在一条指令中获取表单中的所有数据,可以使用.serialize()
E.g
如
$.ajax({
type: "POST",
url: 'belanja/belanja_crud.php',
data: $("#formData").serialize()
})
.done(function (data) {
//do somthing
})
.fail(function (xhr, ajaxOptions, thrownError) {
//do something
});
Or you can see this topic if you want it to JSON Format
如果想要JSON格式,也可以看到这个主题
#1
2
To get all data in the form in one instruction you can use .serialize()
要在一条指令中获取表单中的所有数据,可以使用.serialize()
E.g
如
$.ajax({
type: "POST",
url: 'belanja/belanja_crud.php',
data: $("#formData").serialize()
})
.done(function (data) {
//do somthing
})
.fail(function (xhr, ajaxOptions, thrownError) {
//do something
});
Or you can see this topic if you want it to JSON Format
如果想要JSON格式,也可以看到这个主题