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How to populate this form data and post it with jquery


How to populate this form data and post it with jquery

By : user2956560
Date : November 22 2020, 10:56 AM
Does that help I have a html form as follows:- , Change
code :
<button type="submit" class="btn btn-warning pull-right" .... >
<button type="button" class="btn btn-warning pull-right" .....>
event.preventDefault();
console.log( $( '.form-horizontal' ).serialize() );


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Post form in JQuery and populate DIV - broken in IE

Post form in JQuery and populate DIV - broken in IE


By : Ramanamurty
Date : March 29 2020, 07:55 AM
Any of those help I would say that you don't have to use the form tag at all in this scenario.
code :
<div id="add_value_form">
  <!-- input fields -->
  <input type="button" value="Add" onclick='post_form("/add_value");' />
</div>
<div id="add_value_form">
  <!-- input fields -->
  <input type="button" value="Add" class="formSubmitButton" />
</div>
$(document).ready(function() {
  $('.formSubmitButton').click(function() {
      $.post('/add_value', {'test':'test'}, function(data) {
          $("#add_value_form").empty().append($(data));
      }, "text");
  });
});
$.ajax({
  url: '/add_value',
  type: 'POST',
  cache: false,
  data: 'test=testValue',
  dataType: 'text',
  complete: function(xhr, textStatus) {
    alert('completed');
  },
  success: function(data) {
    alert(data);
  },
  error: function(xhr, textStatus, errorThrown) {
    alert('woops');
  }
});
Populate html drop down from POST data in php form

Populate html drop down from POST data in php form


By : ifateli
Date : March 29 2020, 07:55 AM
like below fixes the issue I would suggest naming your variable P[0], P[1], P[2], etc. on the form generating your list of variables. If you do, then in the form processing page you can do:
code :
if(is_array($_POST['P']))
{
    $arr = $_POST['P'];
    foreach($arr as $key => $val)
    {
        $buffer += '<option value="' . $key . '">' . $val . '</option>';
    }
}
$select = '<select name="somename">' . $buffer . '</select>';
Jquery Datatables populate after form post

Jquery Datatables populate after form post


By : Snir Morlevi
Date : March 29 2020, 07:55 AM
like below fixes the issue Ok, this is the full answer for you question
You need to make three events, the first load the database information in your datatable, the second event inserts the new information on the database, and the third refresh the datatable content.
code :
<html>
<head>
<script type="text/javascript" language="javascript" src="/global/js/jquery-1.9.1.min.js"></script>
<script type="text/javascript" language="javascript" src="/global/js/jquery.dataTables.min.js"></script>
<script type="text/javascript">
//Global variables 
var otable;
var dataTab;
$(document).ready(function () {
    chargeData();
    $('#btnSubmit').click(function () {
       insertData();
    });   
}); 

// 1. charge all data 
function chargeData() {
    $.ajax({
        type: "POST",
        //create a method for search the data and show in datatable
        url: "/servlets/service/competitions/",
        contentType: "application/json; charset=utf-8",
        data: '{ }',
        dataType: "json",
        success: AjaxGetFieldDataSucceeded,
        error: AjaxGetFieldDataFailed
    });
}

function AjaxGetFieldDataSucceeded(result) {
    if (result != "[]") {

        dataTab = $.parseJSON(result);
        //instance of datatable
        oTable = $('#example').dataTable({
            "bProcessing": true,
            "aaData": dataTab,
            //important  -- headers of the json
            "aoColumns": [{ "mDataProp": "season" }, { "mDataProp": "type" }],
            "sPaginationType": "full_numbers",
            "aaSorting": [[0, "asc"]],
            "bJQueryUI": true,

        });

    }
}

function AjaxGetFieldDataFailed(result) {
    alert(result.status + ' ' + result.statusText);
}

// 2. this function only insert the data in your database
function insertData() {
    var email = $("#season").val();
    var evento = $("#type").val();
    $.ajax({
        type: "POST",
        //in this method insert the data in your database
        url: "/servlets/service/competitions/",
        contentType: "application/json; charset=utf-8",
        data: '{ season : "' + season + '", type : "' + type + '"}',
        dataType: "json",
        success: AjaxUpdateDataSucceeded,
        error: AjaxUpdateDataFailed
    });
}

function AjaxUpdateDataSucceeded(result) {
    if (result != "[]") {
        alert("update ok");
        refreshDatatable();
    }
}

function AjaxUpdateDataFailed(result) {
    alert(result.status + ' ' + result.statusText);
}

// 3. This function refresh only the datatable not all page  in varius events you can call like INSERT,UPDATE,DELETE ;D
function refreshDatatable() {
    $.ajax({
        type: "POST",
        //same event used in chargeData function
        url: "/servlets/service/competitions/",
        contentType: "application/json; charset=utf-8",
        data: '{ }',
        dataType: "json",
        success: AjaxRefreshDataSucceeded,
        error: AjaxRefreshDataFailed
    });
}

function AjaxRefreshDataSucceeded(result) {
    if (result.d != "[]") {
        var jposts = result;
        dataTab = $.parseJSON(jposts);
        //when the instance of datatable exists, only pass the data :D
        oTable.fnClearTable();
        oTable.fnAddData(dataTab);
    }
}

function AjaxRefreshDataFailed(result) {
    alert(result.status + ' ' + result.statusText);
}

<script>
</head>
<body>
<form name="myform" id="myform" action="">  
  <label for="season">Season:</label>  
  <input type="text" name="season" id="season" value=""/> <br />
  <label for="type">Type:</label>  
  <input type="text" name="type" id="type" value=""/> <br/>
  <input type="button" id="btnSubmit" name="btnSubmit" value="Search"> 
</form>

<table class="display" id="example">
  <thead>
    <tr>
      <th>SEASON</th>
      <th>TYPE</th>
    </tr>
  </thead>
  <tbody>
    <!-- data goes here -->
  </tbody>
</table>
</body>
</html>
Keep or re-populate form data after ajax post using sessionStorage

Keep or re-populate form data after ajax post using sessionStorage


By : Tanjung Gading Muar
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further I found a way to make this work. I placed the following code onto the Create page itself. I have two modals whereby the user clicks to create new data. Once the modal has been submitted and becomes hidden, the data is reinserted into the form.
I've tested it and it is working!
code :
@section Scripts{

<script type="text/javascript">

$(document).ready(function () {

   $(function () {

       let customerdata = sessionStorage.getItem('customer');
       $('#Customer_Name').val(customerdata);
   });
});
</script>
}
How to populate a django form in a POST request with data from an ajax call?

How to populate a django form in a POST request with data from an ajax call?


By : Tran Ly Vu
Date : March 29 2020, 07:55 AM
will be helpful for those in need I'm sending form data via an ajax call after user hits submit button and then handling this request in my views, but the form is empty. How can I populate the form? , You can write as below:-
code :
$form_data = $("#idofform").serialize();
data : $form_data+'&key=value',
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