So I am working on this form that contains both form and files data. I want to submit it through Ajax. So when it can't pass form Validation, use won't loose the whole entry.
create.php
<script type="text/javascript">
$("#add-product-form").submit(function(e){
e.preventDefault();
var formData = new FormData(this);
console.log(formData);
var url="products/ajax_add_single_product";
$.ajax({
type:"post",
url:"<?php echo base_url() ?>"+url,
data:formData,
dataType:'json',
cache: false,
contentType: false,
processData: false,
error: function (jqXHR, textStatus, errorThrown) {
console.log(jqXHR,textStatus,errorThrown);
error = jqXHR.responseJSON.error;
$(".submit-message").html(error);
console.log(error);
$("html, body").animate({ scrollTop: 0 }, 200);
},
success: function (data, textStatus, jqXHR) {
message = jqXHR.responseJSON.success;
$(".submit-message").html(message);
location.href = "/products";
}
})
})
</script>
controller
public function create(){
$data\['title'\] = "Add Product";
$data\['categories'\] = $this->category_model->get_categories();
$data\['children'\] = $this->category_model->get_child_cats(0);
$data\['vendors'\] = $this->vendor_model->get_vendors();
$data\['attributes'\] = $this->product_model->get_attributes();
$this->load->view('templates/header', $data);
$this->load->view('products/create');
$this->load->view('templates/footer');
}
public function ajax_add_single_product(){
$this->form_validation->set_error_delimiters('<div class="error">', '</div>');
$this->form_validation->set_rules('productname', 'Product Name', 'required');
$this->form_validation->set_rules('partnumber', 'Part Number', 'required|is_unique\[items.itemSKU\]');
$this->form_validation->set_rules('catID', 'Category', 'required', array('required'=>"You need to pick a %s"));
header('Content-Type: application/json');
if ($this->form_validation->run() === FALSE)
{
$this->output->set_status_header(400);
$errors = validation_errors();
echo json_encode(['error'=>$errors]);
}
else
{
$this->output->set_status_header(200);
$imageData = $this->images_upload();
$this->product_model->create_product($imageData);
echo json_encode(['success'=>'Record added successfully.']);
}
}
With above code, when entry can't pass form validation, It will give me form validation errors. When the entry is success, it will insert data to database as I expected. but it will still give me a error. seems like i am not getting JSON data, and getting html.
So turns out, there was one line in the create_product() function in the model. After insert data to tables, I put redirect('/products')after, that's why I am getting the html source code of page products.
Related
Hello All, I have written a one demo code to check weather the given input user is exist in the list or not using ajax and validate.js, when I'm running the code all functions are executing fine but insted of getting response message in succes function it is jumping to the error function and giving undefined response as sent form php.
Here is my code:
$.validator.addMethod("checkUserExists", function(value, element){
alert("input checking");
var inputElem = $('#hl_form :input[name="username"]'),
data = { "username" : inputElem.val(),"check": "userCheck"},
eReport = ''; //error report
$.ajax(
{
type: "POST",
url: "services.php",
async: true,
dataType: 'json',
data: data,
success: function(result)
{
alert(result);
console.log(result);
if (result.status !== 'true')
{
return '<p>This User is already registered.</p>';
}else{
return true;
}
},
error: function(xhr, textStatus, errorThrown)
{
//alert('ajax loading error... ... '+url + query);
return false;
}
});
}, 'User Alread exist in the DB');
My Validate.js Rules and and Message are
Validate Method Rule
username: {
required: true,
checkUserExists:true
}
Validate.js Method Message
username: {
required: "Please enter your Username",
checkUserExists: "User... already exist"
},
My Php Code (service.php)
<?php
header('Content-Type: application/json');
class form_services{
public $sql;
public $returnResult = array();
function checkUser($requestedUser) {
$registeredUser = array('xyz', 'abc', 'def', 'ghi', 'jkl');
if( in_array($requestedUser, $registeredUser) ){
$returnResult["status"] = 'false';
}else{
$returnResult["status"] = 'true';
}
return $returnResult;
}
} //Class Ends Here
$checkRequest = $_POST['check'];
$frmServices = new form_services();
$data = '';
switch ( $checkRequest) {
case 'userCheck': $requestedUser = $_REQUEST['username'];
$data = $frmServices->checkUser( $requestedUser);
echo json_encode($data);
break;
default: echo json_encode($data);
break;
}
?>
Please help me in resolving my issue, i'm getting undefined result in ajax call from php cod.
I'm trying to create a forum and jquery throws an 'illegal invocation' error.
Here is my jquery code:
$('#formSumbit').on('submit', function(e) {
e.preventDefault();
$.ajax({
url: 'data-get.php',
type: 'POST',
data: new FormData(this),
contentType: false,
dataType: 'json',
success: function(value) {
var serialize = $.parseJSON(value);
if (serialize.success == 'false') {
$('.alert').fadeIn().delay(3000).fadeOut();
$('.alert-msgText').html(serialize.datamsg);
}
}
});
});
And here is my PHP code:
<?php
$user = $_POST['user'];
$msg = $_POST['message'];
if(empty($user)&&empty($message)) {
$data = array(
'success' => 'false',
'datamsg' => 'Please fill the textboxes'
);
echo json_encode($data);
} else {
mysqli_query($con,"INSERT INTO forums(name,message) VALUES ('$user','$msg')");
$data = array(
'success' => 'true',
'datamsg' => 'Done!'
);
echo json_encode($data);
}
exit();
?>
When the textboxes are empty and i click the submit button, nothing seems to work and jquery throws an illegal invocation error. I don't understand what the problem is. Can you please help?
And thanks in advance!
1) You have a typo mismatch between your form and your JavaScript:
<form id="formSubmit" and $('#formSumbit') - it should be $('#formSubmit') to match the spellings.
2) Unless you are trying to upload files via this AJAX request, then you can simplify things by replacing data: new FormData(this), contentType: false, with just data: $(this).serialize(). This will get rid of the illegal invocation error.
3) Writing dataType: 'json' means that jQuery will automatically try to parse the data coming from the server as JSON, and convert it. Therefore, in your "success" function, value will already be parsed and converted to an object. In turn therefore, using $.parseJSON is not necessary. You can just access value.success directly, for instance.
Here's a fixed version:
$('#formSubmit').on('submit', function(e) {
e.preventDefault();
$.ajax({
url: 'data-get.php',
type: 'POST',
data: $(this).serialize(),
dataType: 'json',
success: function(value) {
if (value.success == 'false') {
$('.alert').fadeIn().delay(3000).fadeOut();
$('.alert-msgText').html(value.datamsg);
}
}
});
});
Working demo: https://jsfiddle.net/khp5rs9m/2/ (In the demo I changed your URL for a fake one, just so it would get a response, but you can see where I have altered it and left your settings in the commented-out part).
i have problem with ajax post submit
here is my script
function postad(actionurl) {
if (requestRunning) return false ;
if (! $("#editadform").valid()) {
validator.focusInvalid();
return false ;
}
$('#ajxsave').show() ;
requestRunning = true ;
var postData = $('#editadform').serializeArray();
$.ajax(
{
url : actionurl,
type: "POST",
data : postData,
success:function(data, textStatus, jqXHR)
{
$('#diverrors').html(data.errors) ;
$('#divalerts').html(data.alerts) ;
if (data.status=='success') { alert(data.status);
$('#siteid').val(data.siteid) ;
if ($('#adimager').val())
$('#divlmsg').html(data.alertimage) ;
$('#editadform').submit() ;
} else {
$('#ajxsave').hide() ;
}
},
error: function(jqXHR, textStatus, errorThrown)
{
$('#ajxsave').hide() ;
},
complete: function() {
requestRunning = false;
}
});
$('.btn').blur() // remove focus
return false ;
}
This works on if (textStatus=='success') {
but when the action fails, alert(data.status) shows Undefined.
Using FireBug, I can see that the data is correctly returned. Why is data.status "Undefined" then?
If you don't specify the dataType field of an $.ajax() call in jQuery, it formats the response as plain text. A workaround to your code would be to either include dataType: "JSON" into your $.ajax() parameters, or alternatively, in your success function, parse the plain text response as a JSON object, by using the folllowing:
data = JSON.parse(data); // You can now access the different fields of your JSON object
UPDATE:
yes i have not status field in action url, how to add data status field in php code?
When creating your PHP script that is intended to return the JSON data, you first need to build an array and then encode it as JSON.
So, suppose you have a PHP script that either succeeds and produces some data that you put into a $data variable, or fails, then the following style could be adopted:
<?php
// ^^ Do your PHP processing here
if($success) { // Your PHP script was successful?
$response = array("status" => "success!", "response" => $data);
echo json_encode($response);
}
else {
$reponse = array("status" => "fail", "message" => "something went wrong...");
echo json_encode($response);
}
?>
I have a view file which contains a button (link):
<a href id="savebutton" class="btn btn-warning">Save</a>
Somewhere else in this view I have also declared some hidden fields in a form that contain my userid and vacancyid.
echo form_input(dataHiddenArray('userid', $this->auth_user_id));
echo form_input(dataHiddenArray('vacancyid', $vacancydetails[0]->vacancy_id));
These hidden fields translate to:
<input type="hidden" value="2" class="userid">
<input type="hidden" value="1" class="vacancyid">
Now I want to be able to send these values to my controller (via AJAX) so that I can insert them in my database.
My JS file looks like this:
$(function() {
var postData = {
"userid" : $("input.userid").val(),
"vacancyid" : $("input.vacancyid").val()
};
btnSave = $('#savebutton'),
ajaxOptions = {
cache: false,
type: 'POST',
url: "<?php echo base_url();?>dashboard/vacancy/saveVacancy",
contentType: 'application/json',
dataType: 'text'
};
btnSave.click(function (ev) {
var options = $.extend({}, ajaxOptions, {
//data : $(this).closest('form').serialize()
data: postData
});
ev.preventDefault();
// ajax done & fail
$.ajax(options).done(function(data) {
alert(data); // plausible [Object object]
//alert(data[0]); // plausible data
console.log(data); // debug as an object
}).fail(function (xhr, status, error) {
console.warn(xhr);
console.warn(status);
console.warn(error);
});
});
And my controller looks like this (it is not doing much because it doesn't return anything):
public function saveVacancy() {
//$this->load->model('user/usersavedvacancies_model');
/*$data = array(
'userid' => $this->input->post('userid'),
'vacancyid'=>$this->input->post('vacancyid')
);*/
echo $this->input->post('userid');
}
Minor changes to javascript
$(function () {
var postData = {
"userid": $("input.userid").val(),
"vacancyid": $("input.vacancyid").val()
};
btnSave = $('#savebutton'),
ajaxOptions = {
type: 'POST',
url: "<?php echo base_url('dashboard/vacancy/saveVacancy);?>",
dataType: 'json'
};
btnSave.click(function (ev) {
var options = $.extend({}, ajaxOptions, {
//data : $(this).closest('form').serialize()
data: postData
});
ev.preventDefault();
// ajax done & fail
$.ajax(options).done(function (data) {
console.log(data); // debug as an object
if (data.result === 'success') {
alert("Yeah, it saved userid " + data.userid + " to vacancy id " + data.vacancyid);
}
}).fail(function (xhr, status, error) {
console.warn(xhr);
console.warn(status);
console.warn(error);
});
});
});
In the controller
public function saveVacancy()
{
//assigning a more useable object name to the model during load
$this->load->model('user/usersavedvacancies_model', 'save_vacancy');
$data = array(
'userid' => $this->input->post('userid'),
'vacancyid' => $this->input->post('vacancyid')
);
//send data to model and model returns true or false for success or failure
$saved = $this->save_vacancy->doSaveId($data); //yes, I made up the method, change it
$result = $saved ? "success" : "failed";
echo json_encode(array('result' => $result, 'userid' => $data['userid'], 'vacancyid' => $data['vacancyid']));
}
You need to understand that $.ajax takes two methods i.e GET and POST and from the documentation you can see that default method is GET so Since you have not defined method as GET/POST probably the method is taken GET so first change define ajax method to POST as well as you need to be clear about dataType of ajax it may be one of JSON/html and default is json.
$.ajax({
method: "POST",
url: url,
data: data,
dataType:'html'
});
I guess this helped you can learn detail from
Learn more.
Ajax jquery always running error function, althought success function run and i can get session value,i can't run window.location="profile.php";
$(document).ready(function(){
$("#login").click(function(){
var username=$("#usern").val();
var password=$("#user").val();
$.ajax({
type: "POST",
url: "model/user.php",
data: {
user_log : username,
password : password
},
dataType: 'json',
error: function (xhr,textStatus,errorThrown) {
$("#error").html("<span style='color:#cc0000'>Error:</span> Invalid username and password. ");
},
success: function(json){
window.location="profile.php";
},
beforeSend:function()
{
$("#error").html("<img src='http://www.chinesecio.com/templates/base/images/loading.gif' /> Loading...")
}
});
return false;
});
});
user.php
<?php
ob_start();
session_start();
error_reporting(E_ALL & ~E_NOTICE & ~E_DEPRECATED);
require_once(dirname(__FILE__).'/../model/connect.php');
?>
<?php
global $pdo;
if(isset($_POST['user_log'])) {
// username and password sent from Form
$username=$_POST['user_log'];
$password=$_POST['password'];
$qr= "SELECT * FROM user where username='$username' AND password='$password'" ;
$stmt= $pdo->query($qr);
$row= $stmt->fetch(PDO::FETCH_ASSOC);
if($stmt->rowCount() > 0)
{
$_SESSION['id']=$row['id'];
$_SESSION['name_mem']=$row['username'];
$_SESSION['level_mem']=$row['level'];
}
header("location:../../../../index.php");
}
?>
Remove this line :
header("location:../../../../index.php");
If above doesn't work, omit this from ajax properties :
dataType: 'json',
you can use ajax like this,
<script>
$("#login").click(function(){
var username=$("#usern").val();
var password=$("#user").val();
$.ajax({
xhr: function() {
var xhr = new window.XMLHttpRequest();
//progress
xhr.upload.addEventListener("progress", function(e) {
//progress value : you can load progress bar in here
}, false);
return xhr;
},
type: "POST",
url: "model/user.php",
data: {'username' : username, 'password' : password},
dataType:json,
success: function(msg) {
//when success //200 ok
if(msg.status=="done"){
window.location="profile.php";
}else{
$("#error").html("<span style='color:#cc0000'>Error:</span> "+msg.massage);
}
},
error: function(jqXHR, textStatus, errorThrown) {
//when error: this statement will execute when fail ajax
}
});
});
</script>
server side code like this(inside user.php),
$username=$_POST['username'];
$password=$_POST['password'];
...........
//$status="fail" or "done"
//success must be always success
//$massage= "password or username not match"
$respond=array("success"=>"success","status"=>$status,"massage"=>$massage);
echo json_encode($respond);
exit;
I hope you useful this.