Adding POST parameters before submit

Question

I\'ve this simple form:

Answer 1

You can do a form.serializeArray(), then add name-value pairs before posting:

var form = $(this).closest('form');

form = form.serializeArray();

form = form.concat([
    {name: "customer_id", value: window.username},
    {name: "post_action", value: "Update Information"}
]);

$.post('/change-user-details', form, function(d) {
    if (d.error) {
        alert("There was a problem updating your user details")
    } 
});

Answer 2

If you want to add parameters without modifying the form, you have to serialize the form, add your parameters and send it with AJAX:

var formData = $("#commentForm").serializeArray();
formData.push({name: "url", value: window.location.pathname});
formData.push({name: "time", value: new Date().getTime()});

$.post("api/comment", formData, function(data) {
  // request has finished, check for errors
  // and then for example redirect to another page
});

See .serializeArray() and $.post() documentation.

Answer 3

PURE JavaScript:

Creating the XMLHttpRequest:

function getHTTPObject() {
    /* Crea el objeto AJAX. Esta funcion es generica para cualquier utilidad de este tipo, 
       por lo que se puede copiar tal como esta aqui */
    var xmlhttp = false;
    /* No mas soporte para Internet Explorer
    try { // Creacion del objeto AJAX para navegadores no IE
        xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
    } catch(nIE) {
        try { // Creacion del objet AJAX para IE
            xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
        } catch(IE) {
            if (!xmlhttp && typeof XMLHttpRequest!='undefined') 
                xmlhttp = new XMLHttpRequest();
        }
    }
    */
    xmlhttp = new XMLHttpRequest();
    return xmlhttp; 
}

JavaScript function to Send the info via POST:

function sendInfo() {
    var URL = "somepage.html"; //depends on you
    var Params = encodeURI("var1="+val1+"var2="+val2+"var3="+val3);
    console.log(Params);
    var ajax = getHTTPObject();     
    ajax.open("POST", URL, true); //True:Sync - False:ASync
    ajax.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
    ajax.setRequestHeader("Content-length", Params.length);
    ajax.setRequestHeader("Connection", "close");
    ajax.onreadystatechange = function() { 
        if (ajax.readyState == 4 && ajax.status == 200) {
            alert(ajax.responseText);
        } 
    }
    ajax.send(Params);
}

Answer 4

To add that using Jquery:

$('#commentForm').submit(function(){ //listen for submit event
    $.each(params, function(i,param){
        $('<input />').attr('type', 'hidden')
            .attr('name', param.name)
            .attr('value', param.value)
            .appendTo('#commentForm');
    });

    return true;
}); 

Answer 5

you can do this without jQuery:

    var form=document.getElementById('form-id');//retrieve the form as a DOM element

    var input = document.createElement('input');//prepare a new input DOM element
    input.setAttribute('name', inputName);//set the param name
    input.setAttribute('value', inputValue);//set the value
    input.setAttribute('type', inputType)//set the type, like "hidden" or other

    form.appendChild(input);//append the input to the form

    form.submit();//send with added input

Answer 6

Previous answer can be shortened and be more readable.

$('#commentForm').submit(function () {
    $(this).append($.map(params, function (param) {
        return   $('<input>', {
            type: 'hidden',
            name: param.name,
            value: param.value
        })
    }))
});