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问题:
Current Setup
I have an HTML form like so.
<form id="demo-form" action="POST" method="post-handler.php">
<input type="text" name="name" value="previousValue"/>
<button type="submit" name="action" value="dosomething">Update</button>
</form>
I may have many of these forms on a page.
My Question
How do I submit this form asynchronously and not get redirected or refresh the page? I know how to use XMLHttpRequest
. The issue I have is retrieving the data from the HTML in javascript to then put into a post request string. Here is the method I'm currently using for my zXMLHttpRequest`'s.
function getHttpRequest() {
var xmlhttp;
if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
} else {// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
return xmlhttp;
}
function demoRequest() {
var request = getHttpRequest();
request.onreadystatechange=function() {
if (request.readyState == 4 && request.status == 200) {
console.log("Response Received");
}
}
request.open("POST","post-handler.php",true);
request.setRequestHeader("Content-type","application/x-www-form-urlencoded");
request.send("action=dosomething");
}
So for example, say the javascript method demoRequest()
was called when the form's submit button was clicked, how do I access the form's values from this method to then add it to the XMLHttpRequest
?
EDIT
Trying to implement a solution from an answer below I have modified my form like so.
<form id="demo-form">
<input type="text" name="name" value="previousValue"/>
<button type="submit" name="action" value="dosomething" onClick="demoRequest()">Update</button>
</form>
However, on clicking the button, it's still trying to redirect me (to where I'm unsure) and my method isn't called?
Button Event Listener
document.getElementById('updateBtn').addEventListener('click', function (evt) {
evt.preventDefault();
// Do something
updateProperties();
return false;
});
回答1:
The POST string format is the following:
name=value&name2=value2&name3=value3
So you have to grab all names, their values and put them into that format.
You can either iterate all input elements or get specific ones by calling document.getElementById()
.
Warning: You have to use encodeURIComponent()
for all names and especially for the values so that possible &
contained in the strings do not break the format.
Example:
var input = document.getElementById("my-input-id");
var inputData = encodeURIComponent(input.value);
request.send("action=dosomething&" + input.name + "=" + inputData);
Another far simpler option would be to use FormData
objects. Such an object can hold name and value pairs.
Luckily, we can construct a FormData
object from an existing form and we can send it it directly to XMLHttpRequest
's method send():
var formData = new FormData( document.getElementById("my-form-id") );
xhr.send(formData);
回答2:
The ComFreek's answer is correct but a complete example is missing.
Therefore I have wrote an extremely simplified working snippet:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="X-UA-Compatible" content="IE=Edge, chrome=1"/>
<script>
"use strict";
function submitForm(oFormElement)
{
var xhr = new XMLHttpRequest();
xhr.onload = function(){ alert (xhr.responseText); }
xhr.open (oFormElement.method, oFormElement.action, true);
xhr.send (new FormData (oFormElement));
return false;
}
</script>
</head>
<body>
<form method="POST"
action="post-handler.php"
onsubmit="return submitForm(this);" >
<input type="text" name="name" value="previousValue"/>
<input type="submit" value="Update"/>
</form>
</body>
</html>
This snippet is basic and cannot use GET
. I have been inspired from the excellent Mozilla Documentation. Have a deeper read of this MDN documentation to do more. See also this answer using formAction
.
回答3:
By the way I have used the following code to submit form in ajax request.
$('form[id=demo-form]').submit(function (event) {
if (request) {
request.abort();
}
// setup some local variables
var $form = $(this);
// let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// serialize the data in the form
var serializedData = $form.serialize();
// fire off the request to specific url
var request = $.ajax({
url : "URL TO POST FORM",
type: "post",
data: serializedData
});
// callback handler that will be called on success
request.done(function (response, textStatus, jqXHR){
});
// callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
});
// callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
// reenable the inputs
});
// prevent default posting of form
event.preventDefault();
});
回答4:
With pure Javascript, you just want something like:
var val = document.getElementById("inputFieldID").value;
You want to compose a data object that has key-value pairs, kind of like
name=John&lastName=Smith&age=3
Then send it with request.send("name=John&lastName=Smith&age=3");
回答5:
I have had this problem too, I think.
I have a input element with a button. The onclick
method of the button uses XMLHTTPRequest
to POST a request to the server, all coded in the JavaScript.
When I wrapped the input and the button in a form the form's action property was used. The button was not type=submit
which form my reading of HTML standard (https://html.spec.whatwg.org/#attributes-for-form-submission) it should be.
But I solved it by overriding the form.onsubmit
method like so:
form.onsubmit = function(E){return false;}
I was using FireFox developer edition and chromium 38.0.2125.111 Ubuntu 14.04 (290379) (64-bit).
回答6:
function postt(){
var http = new XMLHttpRequest();
var y = document.getElementById("user").value;
var z = document.getElementById("pass").value;
var postdata= "username=y&password=z"; //Probably need the escape method for values here, like you did
http.open("POST", "chat.php", true);
//Send the proper header information along with the request
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.setRequestHeader("Content-length", postdata.length);
http.onreadystatechange = function() {//Call a function when the state changes.
if(http.readyState == 4 && http.status == 200) {
alert(http.responseText);
}
}
http.send(postdata);
}
how can I post the values of y and z here from the form
回答7:
var temp_data = $('#form_id').serialize();
data = new FormData();
$.each(temp_data.split('&'), function (i, o) {
data.append(o.split('=')[0], o.split('=')[1]);
});
var xhr = new XMLHttpRequest();
xhr.open('POST', '/pos_url/', true);
xhr.responseType = 'blob';
xhr.send(data);