发表新帖
发表新帖

XMLHttpRequest to Post HTML Form

后端 未结
关注
6 1999
遇见更好的自我
遇见更好的自我 2020-12-01 04:50

Current Setup

I have an HTML form like so.
相关标签:
6条回答
  • 滥情空心
    2020-12-01 05:12

    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.getAttribute("action"));
  xhr.send(new FormData(oFormElement));
  return false;
}
</script>
</head>

<body>
<form method="POST"
      action="post-handler.php"
      onsubmit="return submitForm(this);" >
   <input type="text"   value="previousValue" name="name"/>
   <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.

    0 讨论(0)
  • 生来不讨喜
    2020-12-01 05:12

    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");

    0 讨论(0)
  • 忘掉有多难
    2020-12-01 05:14

    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).

    0 讨论(0)
  • 滥情空心
    2020-12-01 05:21 
    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

    0 讨论(0)
  • 有刺的猬
    2020-12-01 05:29

    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);
    0 讨论(0)
  • 迷失自我
    2020-12-01 05:30

    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();
});
    0 讨论(0)
 看不清?
提交回复
热议问题