issue with ajax doesnt continue - php

I have one issue with ajax when I get echo success on my php. I don't get execute the first if
if(text[0]==="success")
Code inside php
if(count($error)==0)
{
$user = ORM::for_table('usuario')->create();
$user->username = $username;
$user->contrasenia = password_hash($password, PASSWORD_DEFAULT);
$user->email = $email;
$user->admin = $is_admin;
$user->save();
echo "success";
}
else
{
$error = json_encode($error);
echo $error;
}
My code in ajax
[![$("#create-button").on('click', function(event){
//cancels the form submission
event.preventDefault();
submitForm();
});
function submitForm()
{
var dataString = $("#userForm").serialize();;
$.ajax({
dataType: "json",
type: "POST",
url: "/altausers",
data: dataString,
success: function(text)
{
console.log("hola");
console.log(text);
if(text\[0\]==="success")
{
alert("hola");
//$("#error").addClass('hidden');
}
else if(text.length > 0)
{
$("#error").removeClass('hidden');
texterror = "<ol type='disc'>";
$.each(text,function(index,value)
{
texterror+="<li>"+value+"</li>";
});
texterror+="</ol>";
document.getElementById("error").innerHTML = texterror;
}
}
});
}
Image console
What is the problem?
Could say me which it is problem?
I have that convert the message success to json too

Replace the success echo with
echo json_encode(["success"]);
or leave echo the same and replace if in ajax
if(text == "success")

The problem could be that it is an array, so you can try this:
$.ajax({
url: "items.php",
async: false,
type: "POST",
dataType: "JSON",
data: { "command" : "getItems" }
}).success(function( response ) {
alert( response.fruits.apple );
alert(Object.keys(response).length);
});
And if you want to see the results:
Object.keys( response ).forEach(function( key ) {
console.log('key name: ', key);
console.log('value: ', response[key]);
});

Related

jquery onSubmit() problems after returning data from php

I am trying to validate my forms by using jQuery and php .. What I am trying to achieve is pass my form inputs to process.php in the background, check if inputs pass my validation code, then return true or false back to my jQuery checkForm() function .. so far the below code is now working ..
function checkForm() {
jQuery.ajax({
url: "process.php",
data: {
reguser: $("#reguser").val(),
regpass: $("#regpass").val(),
regpass2: $("#regpass2").val(),
regemail: $("#regemail").val()
},
type: "POST",
success: function(data) {}
});
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<form action="register.php" method="post" onSubmit="return checkForm()">
send all input's in php to validate all the data.
function checkForm() {
$.ajax({
url : "process.php",
type: "POST",
data: $('#yourForm').serialize(),
dataType: "JSON",
success: function(data)
{
if(data.status)
{
alert("Success");
}
else
{
for (var i = 0; i < data.inputerror.length; i++)
{
alert(data.inputerror[i] + "|" + data.error_string[i]);
}
}
}
});
};
validating the data you've sent through Ajax.
public function process()
{
$this->_validate_all_data();
echo json_encode(array("status" => true));
}
private function _validate_all_data()
{
$data = array();
$data['error_string'] = array();
$data['inputerror'] = array();
$data['status'] = true;
if(empty($POST['reguser']))
{
$data['inputerror'][] = 'reguser'; // input name
$data['error_string'][] = 'Reguser is required'; // message for validation
$data['status'] = false;
}
if($data['status'] === false)
{
echo json_encode($data);
exit();
}
}

Can't Get AJAX Success Message To Work

Racking my brains for hours with this. I have the following PHP AJAX script:
<script type="text/javascript">
$(document).ready(function(){
$("#submitValue").click( function(){
var uemail=$("#uemail").val();
var uage=$("#uage").val();
var city=$("#city").val();
var urname=$("#urname").val();
$.ajax({
type: "POST",
url:"acctUpdate.php",
data: "uemail=" + uemail +"&uage="+ uage +"&city="+ city +"&urname="+urname +"&uname="+"<?php echo $memName; ?>" +"&uID="+"<?php echo $memID; ?>" +"&acctDB="+"profile" ,
dataType: "dataString",
success: function(data){
$('#results').html('Success').delay(1000).fadeOut();
}
});
});
});
</script>
I am trying to get the message 'Success' to populate this span element;
<span id="results"></span>
But just can't seem to get it to work.
The PHP is as follows (the table is updated just fine);
if($_POST['acctDB'] == 'profile') {
$uemail = $DB->real_escape_string($_POST['uemail']);
$uage = $DB->real_escape_string($_POST['uage']);
$city = $DB->real_escape_string($_POST['city']);
$urname = $DB->real_escape_string($_POST['urname']);
$uname = $DB->real_escape_string($_POST['uname']);
$uID = $DB->real_escape_string($_POST['uID']);
mysqli_query($DB, 'UPDATE profile SET memEmail="'.$uemail.'", memAge="'.$uage.'", memCity="'.$city.'", memRealName="'.$urname.'" WHERE memID="'.$uID.'" AND memUname="'.$uname.'" ') or die(mysqli_error($DB));
}
Anyone be of assistance please?
dataType: "dataString"
Please comment this part and it will work.
if($_POST['acctDB'] == 'profile') {
$uemail = $DB->real_escape_string($_POST['uemail']);
$uage = $DB->real_escape_string($_POST['uage']);
$city = $DB->real_escape_string($_POST['city']);
$urname = $DB->real_escape_string($_POST['urname']);
$uname = $DB->real_escape_string($_POST['uname']);
$uID = $DB->real_escape_string($_POST['uID']);
mysqli_query($DB, 'UPDATE profile SET memEmail="'.$uemail.'", memAge="'.$uage.'", memCity="'.$city.'", memRealName="'.$urname.'" WHERE memID="'.$uID.'" AND memUname="'.$uname.'" ') or die(mysqli_error($DB));
echo 'yes';
}
// add echo 'yes'; at php submit page.
change the script as follows
$.ajax({
type: "POST",
url:"acctUpdate.php",
data: "uemail=" + uemail +"&uage="+ uage +"&city="+ city +"&urname="+urname +"&uname="+"<?php echo $memName; ?>" +"&uID="+"<?php echo $memID; ?>" +"&acctDB="+"profile" ,
// dataType: "dataString",
dataType : "text",
success: function(data){
$('#results').html(data).delay(1000).fadeOut();
return false;
}
});
return false;
In php file change this
$qry = mysqli_query($DB, 'UPDATE profile SET memEmail="'.$uemail.'", memAge="'.$uage.'", memCity="'.$city.'", memRealName="'.$urname.'" WHERE memID="'.$uID.'" AND memUname="'.$uname.'" ') or die(mysqli_error($DB));
if($qry)
echo "Success";
}
When you make an ajax call and you pass the values to a php file, you will also need to return the response.
So at the end of your query if everything is completed successful you will do something like this:
return Response::json(array(
'success' => true,
'message' => trans('admin.update_success'),
), 200);
And your ajax cal looks something like this:
$("#submitValue").click(function(e){
var uemail=$("#uemail").val();
var uage=$("#uage").val();
var city=$("#city").val();
var urname=$("#urname").val();
$.ajax({
url: 'acctUpdate.php',
type: 'POST',
dataType: 'json',
data: "uemail=" + uemail +"&uage="+ uage +"&city="+ city +"&urname="+urname +"&uname="+"<?php echo $memName; ?>" +"&uID="+"<?php echo $memID; ?>" +"&acctDB="+"profile" ,
dataType: "dataString",
})
.done(function(response) {
alert(response.message)
})
.fail(function(response) {
if (response.status == 400) {
var output = '<ul>';
var errors = $.parseJSON(response.responseText).errors;
$.each(errors, function(id, message) {
output += '<li>' + message[0] + '</li>'
});
output += '</ul>'
alert(output)
} else {
alert('UnknownError');
}
})
e.preventDefault();
})
So to recap:
You make the ajax call
The php file will process the data
You pass the response back to the 'done function)
And hier you can make anything you want with your response.
I have just as example inserted an alert message
Hope this helps.
Sorry but the code I have provider for you is for a Laravel framework and I suppose you are not using it. So you don't have the 'Respone::' class.
In the php file:
//if the query has success
$return = array();
$return['responseCode'] = 1;
$return['responseHTML'] = 'Success'
echo json_encode( $return );
And the ajax call:
$("#submitValue").click(function(e){
var uemail=$("#uemail").val();
var uage=$("#uage").val();
var city=$("#city").val();
var urname=$("#urname").val();
$.ajax({
url: 'acctUpdate.php',
type: 'POST',
dataType: 'json',
data: "uemail=" + uemail +"&uage="+ uage +"&city="+ city +"&urname="+urname +"&uname="+"<?php echo $memName; ?>" +"&uID="+"<?php echo $memID; ?>" +"&acctDB="+"profile" ,
})
.done(function(response) {
if( response.responseCode == 0 ) {
//alert(respomse.responseHTML)
$('#results').html(response.responseHTML);
} else if( response.responseCode == 1 ) {
//alert(res.responseHTML)
$('#results').html(response.responseHTML);
}
})
e.preventDefault();
})
So I don't use the response anymore but I will just return an array with the response.

Simple ajax function in cakephp 2.x not working

I am new to cakephp and trying to implement AJAX . I have a view add.ctp in which I have written the following lines :
$('#office_type').change(function(){
var office_id = $('#office_type').val();
if(office_id > 0) {
var data = office_id;
var url_to_call = "http://localhost/testpage/officenames/get_office_names_by_catagory/";
$.ajax({
type: "GET",
url: url_to_call,
data = data,
//dataType: "json",
success: function(msg){
alert(msg);
}
});
}
});
And the function get_office_names_by_catagory() within OfficenamesController.php is:
public function get_office_name_by_catagory($type = '') {
Configure::write("debug",0);
if(isset($_GET['type']) && trim($_GET['type']) != ''){
$type = $_GET['type'];
$conditions = array("Officename.office_type"=> $type);
$recursive = -1;
$office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
}
$this->layout = 'ajax';
//return json_encode($office_names);
return 'Hello !';
}
But unfortunately, its not alerting anything ! Whats wrong ?
Could be caused by two issues:
1) In your js snippet, you are querying
http://localhost/testpage/officenames/get_office_names_by_catagory/.
Note the plural 'names' in get_office_names_by_category. In the PHP snippet, you've defined an action get_office_name_by_catagory. Note the singular 'name'.
2) You may need to set your headers appropriately so the full page doesn't render on an AJAX request: Refer to this link.
I think, you have specified data in wrong format:
$.ajax({
type: "GET",
url: url_to_call,
data = data, // i guess, here is the problem
//dataType: "json",
success: function(msg){
alert(msg);
}
});
To
$.ajax({
type: "GET",
url: url_to_call,
data: { name: "John", location: "Boston" }, //example
success: function(msg){
alert(msg);
}
});
You should specify the data in key:value format.
$('#office_type').change(function(){
var office_id = $('#office_type').val();
if(office_id > 0) {
var data = office_id;
var url_to_call = "http://localhost/testpage/officenames/get_office_name_by_catagory/"+office_id;
$.ajax({
type: "GET",
url: url_to_call,
success: function(msg){
alert(msg);
}
});
}
});
In your action
public function get_office_name_by_catagory($type = '') {
$this->autoRender = false;
Configure::write("debug",0);
if(!empty($type)){
$conditions = array("Officename.office_type"=> $type);
$recursive = -1;
$office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
}
$this->layout = 'ajax';
//return json_encode($office_names);
echo 'Hello !';
exit;
}
See what I have done is I have changed your request to function get_office_name_by_catagory, as there is one paramenter $type is already defined in the function, so if I have the request by /get_office_name_by_catagory/2 then you will find value in $type in action.
So no need to use $_GET and rest everything is fine!
Try this,
remove type from ajax and try.
$('#office_type').change(function(){
var office_id = $('#office_type').val();
if(office_id > 0) {
var data = office_id;
var url_to_call = yourlink +office_id;
**$.ajax({
url: url_to_call,
success: function(msg){
alert(msg);
}
});**
}
});
In your action
public function get_office_name_by_catagory($type = '') {
$this->autoRender = false;
Configure::write("debug",0);
if(!empty($type)){
$conditions = array("Officename.office_type"=> $type);
$recursive = -1;
$office_names = $this->Officename->find("all",array("conditions"=>$conditions,"recursive"=>$recursive));
}
$this->layout = 'ajax';
//return json_encode($office_names);
echo 'Hello !';
}

Pass a value from controller JSON to view Ajax in CodeIgniter

The problem is I am getting the dialog box of false but the query is running fine and values are successfully added into the database. It should print true, but it is giving me a false. I checked through Firebug also the value res = 1 is going, but I don't know what is wrong in it.
My View:
$.ajax({
url: "<?php echo site_url('itemsController/additems'); ?>",
type: 'POST',
data: form_data,
success: function(msg) {
if(msg.res == 1)
{
alert(true);
}
else
{
alert(false);
}
}
});
Controller:
$result = array();
$this->load->model('itemsModel');
$query = $this->itemsModel->addItemstoDB($data);
if ($query){ //&& any other condition
$result['res'] = 1;
}
else
{
$result['res'] = 0;
}
echo json_encode($result); //At the end of the function.
}
}
Try setting your dataType to json so the data sent back from the server gets parsed as JSON.
$.ajax({
url: "<?php echo site_url('itemsController/additems'); ?>",
type: 'POST',
data: form_data,
dataType: 'json',
success: function(msg) {
if(msg.res == 1) {
alert(true);
}
else
{
alert(false);
}
}
});
Notice return.
if ($query){
$result['res'] = 1;
}
else
{
$result['res'] = 0;
}
return json_encode($result);//at the end of the function.
}

How to return an error callback in php?

I was wondering if I can return an error callback back to my jquery from my php page that I created, which will then display an alert based upon the actions that happen in my php page. I tried creating a header with a 404 error but that didn't seem to work.
Sample JQuery Code:
$(document).ready(function()
{
var messageid= '12233122abdbc';
var url = 'https://mail.google.com/mail/u/0/#all/' + messageid;
var encodedurl = encodeURIComponent(url);
var emailSubject = 'Testing123';
var fromName = 'email#emailtest.com';
var dataValues = "subject=" + emailSubject + "&url=" + encodedurl + "&from=" + fromName + "&messageID=" + messageid;
$('#myForm').submit(function(){
$.ajax({
type: 'GET',
data: dataValues,
url: 'http://somepage.php',
success: function(){
alert('It Was Sent')
}
error: function() {
alert('ERROR - MessageID Duplicate')
}
});
return false;
});
});
Sample PHP Code aka somepage.php:
<?php
include_once('test1.php');
include_once('test2.php');
if(isset($_GET['subject']))
{
$subject=$_GET['subject'];
}
else
{
$subject="";
}
if(isset($_GET['url']))
{
$url=$_GET['url'];
}
else
{
$url="";
}
if(isset($_GET['from']))
{
$from=$_GET['from'];
}
else
{
$from="";
}
if(isset($_GET['messageID']))
{
$messageID = $_GET['messageID'];
}
else
{
$messageID="";
}
$stoqbq = new test2($from, $messageID);
$msgID = new stoqbq->getMessageID();
if($msgID = $messageID)
{
header("HTTP/1.0 404 Not Found");
exit;
}
else
{
$userID = $stoqbq->getUser();
$stoqb = new test1($subject,$url,$messageID,$userID);
$stoqb->sendtoquickbase();
}
?>
-EDIT-
If you get the invalid label message when using json this is what I did to fix this problem:
Server Side PHP Code Part-
if($msgID == $messageID)
{
$response["success"] = "Error: Message Already In Quickbase";
echo $_GET['callback'].'('.json_encode($response).')';
}
else
{
$userID = $stoqbq->getUser();
$stoqb = new SendToQuickbase($subject,$url,$messageID,$userID);
$stoqb->sendtoquickbase();
$response["success"] = "Success: Sent To Quickbase";
echo $_GET['callback'].'('.json_encode($response).')';
}
Client Side JQuery Part-
$('#myForm').submit(function(){
$.ajax({
type: 'GET',
data: dataValues,
cache: false,
contentType: "application/json",
dataType: "json",
url: "http://somepage.php?&callback=?",
success: function(response){
alert(response.success);
}
});
return false;
});
You can a return a JSON response from your PHP with a success boolean.
if($msgID = $messageID)
{
echo json_encode(array('success' => false));
}
else
{
$userID = $stoqbq->getUser();
$stoqb = new test1($subject,$url,$messageID,$userID);
$stoqb->sendtoquickbase();
echo json_encode(array('success' => true));
}
and in your Javascript:
$.ajax({
type: 'GET',
dataType: 'json',
data: dataValues,
url: 'http://somepage.php',
success: function(response){
if(response.success) {
alert('Success');
}
else {
alert('Failure');
}
}
});
There is an accepted q/a with the same thing: How to get the jQuery $.ajax error response text?
Basically you need to grab the response message from the error callback function:
$('#myForm').submit(function(){
$.ajax({
type: 'GET',
data: dataValues,
url: 'http://somepage.php',
success: function(){
alert('It Was Sent')
}
error: function(xhr, status, error) {
alert(xhr.responseText);
}
});
return false;
});

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