I wonder whether someone may be able to help me please.
Firstly, my apologies, I'm relatively new to JavaScript and jQuery, so perhaps this is a really stupid question.
Using these tutorials here and here I've put together this page to allow users to add records to a MySQL database but I'm having a little difficulty with the form 'validation' and jQuery 'submission' message.
If use select the above link, then once the page has loaded, select 'Save', you'll see that the correct field validation is activated, but despite being validation errors, the 'Location saved' message appears at the bottom of the page, and the page refreshes saving the record to the database.
Obviously this is not supposed to happen, but I'm having great difficulty in joining the 'validation' and 'submission' message. Independently they work fine, but as you can see, once together they don't.
The code below deals with the 'Save Record' and refresh of the page
UPDATE - Working Solution Below
<script>
jQuery(document).ready(function(){
jQuery("#addlocation").validationEngine();
$("#addlocation").bind("jqv.field.result", function(event, field, errorFound, prompText){ console.log(errorFound) })
});
</script>
<script type="text/javascript">
$(document).ready(function(){
$('#addlocation').submit(function(){
//check the form is not currently submitting
if($(this).data('formstatus') !== 'submitting'){
//setup variables
var form = $(this),
formData = form.serialize(),
formUrl = form.attr('action'),
formMethod = form.attr('method'),
responseMsg = $('#saverecordresponse');
//add status data to form
form.data('formstatus','submitting');
//show response message - waiting
responseMsg.hide()
.addClass('response-waiting')
.text('Please Wait...')
.fadeIn(200);
//send data to server for validation
$.ajax({
url: formUrl,
type: formMethod,
data: formData,
success:function(data){
//setup variables
var responseData = jQuery.parseJSON(data),
klass = '';
//response conditional
switch(responseData.status){
case 'error':
klass = 'response-error';
break;
case 'success':
klass = 'response-success';
break;
}
//show reponse message
responseMsg.fadeOut(200,function(){
$(this).removeClass('response-waiting')
.addClass(klass)
.text(responseData.message)
.fadeIn(200,function(){
//set timeout to hide response message
setTimeout(function(){
responseMsg.fadeOut(200,function(){
$(this).removeClass(klass);
form.data('formstatus','idle');
});
},3000)
});
});
}
});
}
//prevent form from submitting
return false;
});
});
</script>
and this is the 'saverecord.php' script which is called upon selecting the 'Save' button.
<?php
//sanitize data
$userid = mysql_real_escape_string($_POST['userid']);
$locationname = mysql_real_escape_string($_POST['locationname']);
$returnedaddress = mysql_real_escape_string($_POST['returnedaddress']);
//validate email address - check if input was empty
if(empty($locationname)){
$status = "error";
$message = "You didn't enter a name for this location!";
}
else if(!preg_match('/^$|^[A-Za-z0-9 _.,]{5,35}$/', $locationname)){ //validate email address - check if is a valid email address
$status = "error";
$message = "You have entered an invalid Location Name!";
}
else{
$query = mysql_query("INSERT INTO `table` (userid, locationname, returnedaddress) VALUES ('$userid', '$locationname', '$returnedaddress')");
if($query){ //if insert is successful
$status = "success";
$message = "Location Saved!";
}
else { //if insert fails
$status = "error";
$message = "I'm sorry, there has been a technical error! Please try again. If problems persist please contact Map My Finds support.";
}
}
//return json response
$data = array(
'status' => $status,
'message' => $message
);
echo json_encode($data);
exit;
?>
I just wondered whether someone could possibly take a look at this please and let me know where I'm going wrong.
Many thanks and kind regards
I believe you need:
if($.validationEngine.submitForm(this,settings) == true) {return false;}
somewhere before your $.ajax line
IRHM, check that the form is validate before submit in your event i.e.
$('#addlocation').submit(function(){
if($(this).validate()){
// put your all existing content here.
}
});
To prevent submitting the form after ajax put return false at the end of above script in if block i.e.
if($(this).validate()){
// put your all existing content here.
return false;
}
I guess the problem is occurring due to validation engine, so in that case to prevent form to submit try to use as follows:
$('#addlocation').submit(function(evt){
if($(this).validate()){
evt.preventDefault();
// put your all existing content here.
}
});
If the above code doesn't work then include onValidationComplete event with validationEngine and put you all existing stuff of if($(this).validate()) block in that i.e.
jQuery(document).ready(function(){
// binds form submission and fields to the validation engine
jQuery("#addlocation").validationEngine({ onValidationComplete: function(){
//setup variables
//add status data to form
//show response message - waiting
//send data to server for validation
return false;
}
});
});
Good Luck
After several days of working on this, I now have a working solution by using the example here which I've added to my original post.
Related
This is the code which sends data to the database. It works correctly when I filled form field and click the 'Add Record' button. It inserts the data successfully in the database.
However the main problem is that if the form field is empty and then I click the button, it sends empty data to the database.
function addRecord() {
var formData = new FormData($("#form1")[0]); //It automatically collects all fields from form
$.ajax({
url: "ajax/EditDeleteLecture.php",
type: "post",
data: formData,
async: false,
cache: false,
contentType: false,
processData: false,
success: function(output) {
alertify.set('notifier', 'delay', 3);
alertify.set('notifier', 'position', 'top-right');
alertify.success('Data Inserted Successfully');
readRecords();
$('#form1').trigger("reset");
}
});
}
you have 3 places where you can tackle empty data issue before saving to database
1- Put required attribute in your input elements so that user can not submit empty fields.
2- validate your form data in java-script function addRecord() before making ajax request. if the validation is complete send ajax call else show message to user to fill the data.
3- validate your data that you received in $_POST variable and if fields are empty, send error message back in ajax response and show error to user.
When You get form data value while you should the form field value is empty or not... or use print_r($_POST) data..
/* on submitting my form */
$( '#myform' ).submit(function() {
var errors = [];
/* else if Textarea is empty display message error */
if($('#myform textarea').val()=="") {
errors[errors.length] = 'enter your message';
}
/* if radio button is not being selected display message error */
if (!$(":radio:checked").attr('checked')) {
errors[errors.length] = 'select a radio';
}
if(errors.length > 0){
var errMsg = "Please "
for(e = 0; e < errors.length-1; e++){
errMsg += errors[e]+", ";
}
errMsg = errMsg.substr(0, errMsg.length -2)+" and "+errors[errors.length-1];
$('#errors').empty().text(errMsg);
return false;
}
/* Everthing is good display success message and add SENDING class to submit button */
else {
$('#myform :submit').addClass('sending').html('Sending message...');
$('#errors').empty().text('All Good :D');
}
return false;
});
});
Make sure to set the contentType property to "application/x-www-form-urlencoded" in order to parse this in the backend server.
I'm a really new coder and struggling with a task I'm now working on and trying out for days.
I searched Google and Stack Overflow but can't find a (for me understandable) solution to my problem:
I created a Twitter Bootstrap landing page and there a modal shows up when clicked. In this modal I have a form with a newsletter subscription:
<form id="newsletter" method="post">
<label for="email">Email:</label><br/>
<input type="text" name="email" id="email"/><br/>
<button type="submit" id="sub">Save</button>
</form>
<span id="result"></span>
Now I want to insert the data into a mySQL DB and do some basic validation that returns errors or a success message. The script works fine without ajax, but probably needs alterations on what it returns for ajax?
include("connection.php");
if ($_POST['email']) {
if(!empty($_POST['my_url'])) die('Have a nice day elsewhere.');
if (!$_POST['email']) {
$error.=" please enter your email address.";
} else if (!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) {
$error.=" please enter a valid email address.";
}
if($error) {
$error= "There was an error in your signup,".$error;
} else {
$query="SELECT * FROM `email_list` WHERE email='".mysqli_real_escape_string($link, $_POST['email'])."'";
$result = mysqli_query($link, $query);
$results = mysqli_num_rows($result);
if ($results) {
$error.=" this email address is already registered.";
} else {
$query = "INSERT INTO `email_list` (`email`) VALUES('".mysqli_real_escape_string($link, $_POST['email'])."')";
mysqli_query($link, $query);
$message = "Thanks for subscribing!";
}
}
}
After a lot of reading ajax seems to be the way to do it without the bootstrap modal closing after submit (to suppress default event).
The insertion into the DB works fine, also the validation.
But I can't manage to get the different error messages displayed (stored in the $error variable of the php file) or alternatively the $message in case of success.
This is the jquery script:
$("#sub").submit(function (){
event.preventDefault();
$.ajax( {
url: "newsletter2.php",
type: "POST",
data: {email: $("#email").val()},
success: function(message) {
$("#result").html(message);
},
error: function(error) {
$("#result").html(error);
}
});
I try to display the value of the error and message variable in the php script within the #result span.
Any help is appreciated. Please formulate it very straight forward since I'm really new to this field.
Thank you a lot in advance.
Edit:
Added some to the php file to create an array and store the messages within:
$response = array();
$response['success'] = $success;
$response['error']= $errors;
exit(json_encode($response));
But have still some trouble to get the ajax to work. Tried the shorthand $.post instead of $.ajax but can't them now even to get to work posting data...
$("#sub").submit(function (){
event.preventDefault();
$.post("newsletter.php", {email: $("#email").val() });
});
Quick time is much appreciated. I'm stuck after hours of testing and can't find the error. If I submit the form regularly it works fine, so the php/mysql part isn't the problem.
I also realized that when I click the "#sub" button, it still tries to submit the form via get (URL gets values passed). So I'm not sure if the event.preventDefault(); isn't working? jQuery is installed and working.
The $.ajax error function gets called when there is a connection error or the requested page cannot be found
You have to print some text out with the php and the ajax success function gets this output. Then you parse this output to see how it went.
The best practice is this:
php part:
$response = array();
$response['success'] = $success;
$response['general_message'] = $message;
$response['errors'] = $errors;
exit(json_encode($response));
js/html part:
$.post("yourpage.php", a , function (data) {
response = JSON.parse(data);
if(response['success']){
//handle success here
}else{
//handle errors here with response["errors"] as error messages
}
});
Good luck with your project
You need to echo your messages back to your AJAX. There is no place in you PHP code where the messages are going back to the message variable in your AJAX success.
include("connection.php");
if ($_POST['email']) {
if(!empty($_POST['my_url'])) die('Have a nice day elsewhere.');
if (!$_POST['email']) {
$error.=" please enter your email address.";
echo $error; die;
} else if (!filter_var($_POST['email'], FILTER_VALIDATE_EMAIL)) {
$error.=" please enter a valid email address.";
echo $error; die;
}
if($error) {
$error= "There was an error in your signup,".$error;
echo $error; die;
} else {
$query="SELECT * FROM `email_list` WHERE email='".mysqli_real_escape_string($link, $_POST['email'])."'";
$result = mysqli_query($link, $query);
$results = mysqli_num_rows($result);
if ($results) {
$error.=" this email address is already registered.";
echo $error; die;
} else {
$query = "INSERT INTO `email_list` (`email`) VALUES('".mysqli_real_escape_string($link, $_POST['email'])."')";
mysqli_query($link, $query);
$message = "Thanks for subscribing!";
echo $message; die;
}
}
}
I basicly just had the same case. I structured my code a little bit different but it works so...
$("#sub").submit(function (){
event.preventDefault();
$.ajax( {
url: "newsletter2.php",
type: "POST",
dataType: 'json',
data: {email: $("#email").val()},
})
.success(function(message) {
$("#result").html(message);
}),
.error(function(error) {
$("#result").html(error);
})
on server side I used C#(asp.net) and just returned a Json
return Json(new { Message = "Something...", Passed = true}, JsonRequestBehavior.AllowGet);
Oukay, finally I managed to solve the problem with the great inputs here. I did the following:
PHP:
$response = array();
$response['success'] = $success;
$response['error'] = $error;
exit(json_encode($response));
JS:
$("#newsletter").submit(function(event) {
event.preventDefault();
$.ajax({
url: 'newsletter3.php',
method: 'post',
data: {email: $('#email').val()},
success: function(data) {
var response = JSON.parse(data);
console.log(response);
if (response['success']) {
$("#error").hide();
$("#success").html(response['success']);
$("#success").toggleClass("alert alert-success");
} else {
$("#error").html(response['error']);
if(!$("#error").hasClass("alert alert-danger"))
$("#error").toggleClass("alert alert-danger");
}
}
});
});
The functionality is now that you click on a button and a modal pops-up, then you can enter your email and the php script validates if its valid and if it's already in the db. Error and success messages get JSON encoded and then are displayed in a span that changes color according to bootstrap classes danger or success.
Thank you very much for helping me, I'm very happy with my first coding problem solved :)
I use this on my ajax
request.done(function (response, data) {
$('#add--response').html(response);
});
and this on the PHP
die("Success! Whatever text you want here");
At this moment I am using laravel. In this context I am having a form which is successfully submitted by using ajax to a controller. and that controller make it to the database. But the problem is as the ajax is doing its job the whole page remain unmoved / unchanged after the submission even the database is updated.
Now what I want
I want to give feedback to the user that your post is successfully submitted there. or what I want to do in further, I want to refresh the section in which the post is collected from the database as this post can be retrieved from there. But by using ajax only.
So there is no need to collect the whole page or refresh.
here is my form structure
`
{{ Form::open(array('route' => array('questions.store'), 'class' => 'form-horizontal' )) }}
blah blah blaaa .......
<script type="text/javascript">
$(".form-horizontal").submit(function(e){
$(this).unbind("submit")
$("#ask").attr("disabled", "disabled")
var that = $(this),
url = that.attr('action'),
type = that.attr('method'),
data = {};
that.find('[name]').each(function(index, value){
var that = $(this),
name = that.attr('name'),
value = that.val();
data[name] = value;
});
$.ajax({
url: url,
type: type,
data: data,
success: function(response){
console.log(response);
}
});
return false;
});
</script>
{{ Form::close() }}
`
As it is very much visible that the post is updated through a route & controller I want to have another dive and a success message at this script to be displayed after the success of posting. I am looking for some professional structure using what there is minimal need to have interaction with the server side and give user a better page viewing experience.
Thanks a lot for helping me in this research.
I am not sure if I understand you well, but if you want to notify the user about the result of an ajax-called db update you need to have
a route for the ajax save db call - it should point to a method that does the db update.
the db update method should return some value indicating the success/failure of update (for example OK or FAIL)
the only result of calling the method will be just plain text page with OK or FAIL as body
fetch the result by ajax and inform user accordingly (after form submit button)
check out the below code for ajax call itself (inside the form submit handler) to see what I mean
var db_ajax_handler = "URL_TO_YOUR_SITE_AND_ROUTE";
var $id = 1; //some id of post to update
var $content = "blablabla" //the cotent to update
$.ajax({
cache: false,
timeout: 10000,
type: 'POST',
tryCount : 0,
retryLimit : 3,
url: db_ajax_handler,
data: { content: $content, id: $id }, /* best to give a CSRF security token here as well */
beforeSend:function(){
},
success:function(data, textStatus, xhr){
if(data == "OK")
{
$('div.result').html('The new Question has been created');
}
else
{
$('div.result').html('Sorry, the new Question has not been created');
}
},
error : function(xhr, textStatus, errorThrown ) {
if (textStatus == 'timeout') {
this.tryCount++;
if (this.tryCount <= this.retryLimit) {
//try again
$.ajax(this);
return;
}
return;
}
if (xhr.status == 500) {
alert("Error 500: "+xhr.status+": "+xhr.statusText);
} else {
alert("Error: "+xhr.status+": "+xhr.statusText);
}
},
complete : function(xhr, textStatus) {
}
});
EDIT: as per comment, in step 2 (the method that is called with AJAX) replace
if($s)
{
return Redirect::route('questions.index') ->with('flash', 'The new Question has been created');
}
with
return ($s) ? Response::make("OK") : Response::make("FAIL");
EDIT 2:
To pass validation errors to the ajax-returned-results, you cannot use
return Response::make("FAIL")
->withInput()
->withErrors($s->errors());
as in your GIST. Instead you have to modify the suggested solution to work on JSON response instead of a plain text OK/FAIL. That way you can include the errors in the response and still benefit from the AJAX call (not having to refresh the page to retrieve the $errors from session). Check this post on the Laravel Forum for a working solution - you will get the idea and be able to fix your code.
Hi I’m quite new to jquery -ajax and I’d like some help please to join it with CI.
I have followed this tutorial on Submitting a Form with AJAX and I’d like to add this functionality to my CodeIgniter site. What I’d like to do is when the user submits the form, if there are any validation errors to show the individually on each input field (as in native ci process), or if this is not possible via validation_errors() function. If no errors occured to display a success message above the form.
Here's my code so far:
my view
// If validation succeeds then show a message like this, else show errors individually or in validation_errors() in a list
<div class="alert alert-success">Success!</div>
<?php echo validation_errors(); //show all errors that ajax returns here if not individualy ?>
<?php echo form_open('admin/product/add, array('class' => 'ajax-form')); ?>
<p>
<label for="product_name">Product *</label>
<input type="text" name="product_name" value="<?php echo set_value('product_name', $prod->product_name); ?>" />
<?php echo form_error('product_name'); ?>
</p>
<p>
<label for="brand">Brand</label>
<input type="text" name="brand" value="<?php echo set_value('brand', $prod->brand); ?>" />
<?php echo form_error('brand'); ?>
</p>
...
my controller
public function add($id){
// set validation rules in CI native
$rules = $this->product_model->rules;
$this->form_validation->set_rules($rules);
if ($this->form_validation->run() === true) {
// get post data and store them in db
$data = $this->input_posts(array('product_name', 'brand', 'category_id', 'description'));
$this->product_model->save($data, $id);
// no errors - data stored - inform the user with display success-div
} else {
// validation failed - inform the user by showing the errors
}
//load the view
$this->load->view('admin/products/add', $data);
}
and here’s the js script
$(document).ready(function () {
$('form.ajax-form').on('submit', function() {
var obj = $(this), // (*) references the current object/form each time
url = obj.attr('action'),
method = obj.attr('method'),
data = {};
obj.find('[name]').each(function(index, value) {
// console.log(value);
var obj = $(this),
name = obj.attr('name'),
value = obj.val();
data[name] = value;
});
$.ajax({
// see the (*)
url: url,
type: method,
data: data,
success: function(response) {
console.log(response); // how to output success or the errors instead??
}
});
return false; //disable refresh
});
});
How should I pass my validation results (either success or the post errors) throught the ajax request and display them on my view??
From some little research I did I've found that you can use a single controller, that holds both the native proccess and the ajax request (instead of using 2 controllers), but my main difficulty is, I don't understand how the results of the validation will pass through the js script and display them on my view?? Please note that I don't want to display anything on an alert box, instead show the results on a div or the errors individualy(if possible).
EDIT I did some changes to my application, here's the code so far:
the controller
public function manage($id = NULL){
$this->load->library('form_validation');
$data['categ'] = $this->category_model->with_parents();
//fetch a single product or create(initialize inputs empty) a new one
if (isset($id) === true) {
$data['prod'] = $this->product_model->get($id);
$data['attr'] = $this->attribute_model->get_by('product_id', $id, null, true);
} else {
$data['prod'] = $this->product_model->make_new();
$data['attr'] = $this->attribute_model_model->make_new();
}
if (isset($_POST['general_settings'])) {
if ($this->form_validation->run('product_rules') === true) {
// get post inputs and store them in database
$data = $this->product_model->input_posts(array('product_name', 'brand', 'category_id', 'general_description'));
$this->product_model->save($data, $id);
$status = true;
} else {
// validation failed
$status = validation_errors();
}
if ( $this->input->is_ajax_request() ) {
echo json_encode($status);
exit;
}
redirect('admin/product');
}
//if (isset($_POST['attributes_settings'])) { the same thing here }
// load the view
$this->load->view('admin/products/manage', $data);
}
and the js
success: function(response) {
//console.log(response);
if (data.status === true) {
$('#ajaxResults').addClass('alert alert-success').html(response);
} else {
$('#ajaxResults').addClass('alert alert-error').html(response);
};
}
But I'm having some issues
Although I get the error messages from validation_errors() as an alert-error when there are no errors I get the true in an alert-error too, insted of alert-success.
2.how should I return the success message too? eg. a message saying "Saves were done!".
Althought in a non-ajax-request the data are stored in the database, in case fo ajax the don't store. Any ideas What may be wrong???
HTML:
<div id="ajaxResults"></div>
Javascript ajax:
success: function(response) {
$('#ajaxResults').text(response);
}
this script you've wrote is only if the validation succeeds, right?
Wrong. The code in "success" gets executed any time you get a response back from the server (assuming the HTTP header is 200). Does your javascript knows if the server has any error for you? No.
You need your JavaScript to recognize if the validation failed or succeeded. You have many ways to do that. One of these could be sending the message to display followed by a 0 or 1.
So your PHP will looks like:
return "0 " . $errorMessage;
and
return "1 " . $successMessage;
and your javascript should then recognize, with if statement and substring, if the message starts with 0 or with 1.
Use this way i hope this will work for you
<script type='text/javascript'>
var base_url = '<?=base_url()?>';
function ajax_call()
{
var ids = $("#all_users").val();
$.ajax({
type:"POST",
url: base_url+"expense/home/get_expense",
data: "userid=" + ids,
success: function(result){
$("#your_div_id").html(result);
}
});
}
</script>
Basically whats happening is I have a php form to send an email. There are several different validation steps all along the way, which all work great. What i'd like to do is have a javascript alert pop up if there is a validation error when the form is submitted. I have this working with the following php:
// Validate email
if(!filter_var($EmailFrom, FILTER_VALIDATE_EMAIL))
{
echo "<script language=javascript>alert('Please Use a Valid Email Address')</script>";
exit;
}
The alert pops up but the page redirects to domain.com/sendemail.php which leaves the user with a blank page. I'd really like to have the alerts pop up without reloading the page. How would I do this?
You can use ajax to accomplish this. But if you don't want to use ajax, instead doing an exit on error, you could redirect back to the form page along with a query string parameter.
header("Location: http://www.example.com/form.php?error=1");
And on your form page you could put the script withing php if. Something like
<?php if(isset($_GET['error']) && $_GET['error']==1): ?>
<script>
....
</script>
<?php endif; ?>
That would achieve what you are looking for. In fact you can perform multiple checks and set error based on your checks. But I would still suggest Ajax will give a better user experience.
Edit: Super easy solution, use jQuery form plugin : http://jquery.malsup.com/form/
I do something similar in some of my web apps, you might find it useful.
I do my validation server side and if I encounter an error I do this :
json_die(array(
'status' => 'error',
'message'=> 'Your error message'
));
and for success :
json_die(array(
'status' => 'success',
'message'=> 'Your success message'
));
The json_die is function is :
function json_die($array) {
header("content-type: application/json");
die(json_encode($array, true));
}
Then on the front end I do something like this:
$.post('/your_url', {
'your': vars
}, function (r) {
if(r.status == 'success') {
alert(r.message);
} else if (r.status == 'error') {
alert(r.message);
//handle error
} else {
alert('server exploded / no connection');
}
},'json');
This is a script I used in some forms I created to validate them fast. It's very simple and effective, I hope it helps you.
<script type="text/javascript">
function validate(){
emailfilter = /^([a-zA-Z0-9_\.\-])+\#(([a-zA-Z0-9\-])+\.)+([a-zA-Z0-9]{2,4})+$/;
condition = 1;
mensaje = "Complete: ";
//Validate 1st input (Check if empty)
if (document.formname.forminput1.value.length==0){
condition = 0;
msg = msg + "-Input 1 is empty "
}
//Validate 1nd input (Check email)
if (!emailfilter.test(document.formname.forminput1.value)) {
condition = 0;
msg = msg + "-Input 1 has a invalid email adresss "
}
if (condition == 0){
alert(msg)
document.formname.forminput1.focus()
return 0;
}
//send
alert("Form sended.");
document.formname.submit();
}
</script>