I am currently migrating an already built web application to MVC, and I'm figuring out that I'm too newbie to do some kind of changes. There are some ajax calls that are freaking me out. I'll try to be as clear as possible, but due to my inexperience I'm not sure if I won't let some important information by the way.
The point is in the old application, things go this way:
In the php code:
if ($action_user == 'show_alerts') {
$list = array();
$query = "SELECT alert_type FROM alert_contact WHERE NOT
deleted AND user_email=" . typeFormat($email);
$result = mysqli_query($db, $query) or die('Error in query "'.$query . '": ' . mysqli_error($db));
while ($db_field = mysqli_fetch_assoc($result)) {
$list[] = $db_field['alert_type'];
}
echo json_encode($list);
In the jquery code:
$.ajax({
type: 'POST',
url: 'userpost.php',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json'
Here comes my problem, and since I don't have an userpost.php file anymore, I have to send it to the index.php and call my users component by a get petition, which I don't like, but I coudn't find another way to do it. And, what is even worse, I don't know at all how ajax is getting the variables that it needs. It must be a pretty basic mistake, but I recognize my skills at this point are't so good. That's what I'm doing in my version:
In the php code:
if ($action_user == 'show_alerts') {
$list = ModelUser::getAlertContact($act_email);
echo json_encode($list);//I predict that ajax don't reach this line, but not sure
}
In the jquery code:
$.ajax({
type: 'POST',
url: 'index.php?option=users',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json',
success: function(data) {
alert ('gotcha');
$.each(alertsarray, function(index, value) {
if ($.inArray(value, data) === -1) {
$("#sub" + value).prop("checked", false);
$('#alert' + value).removeClass("list_alert_sub");
}
else {
$("#sub" + value).prop("checked", true);
$('#alert' + value).addClass("list_alert_sub");
}
});
},
error: function(data) {
alert("¡Error (ajax)!");
}
});
Any help would be appreciated, and if there's some more information I've missed, please let me know. Thanks in advance.
UPDATE:
I've been making some progress but don't seem to find a real solution. Now I know that the url has to be the controller, so I'm using 'components/userpost/controller.php' as it, and it reaches the ajax call, cause the success alert is showing up. The problem is the MVC way, because I send ajax to the controller, but since I don't have a reload in the page, all the includes are failing so they are obviously not being loaded, and I'm getting errors like this:
PHP Warning: include(components/userpost/model.php): failed to open
stream: No such file or directory in
/var/www/html/viewer_mvc/components/userpost/controller.php on line 3,
referer: http://localhost/viewer_mvc/index.php
Really hope you guys can show me where am I failing, and if there's a special way to do these thing in MVC.
For the JQuery call it makes a POST request to index.php?option=users with JSON data. The form with the ID userForm is serialized using the Jquery serialize method.
The .serialize() method creates a text string in standard URL-encoded notation. It can act on a jQuery object that has selected individual form controls
$.ajax({
type: 'POST',
url: 'index.php?option=users',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json'
Now for your PHP sample
if ($action_user == 'show_alerts') {
$list = ModelUser::getAlertContact($act_email);
echo json_encode($list);//I predict that ajax don't reach this line, but not sure
}
This code will be looking for variables that probably don't exist anymore if it is a different file i.e. is there an $action_user variable?
To start reimplementing it you will need to add the logic so that it checks the POST variable if your not using the framework code. So if you have a form element with the name 'name' then that will be available in your PHP script POST variable
$_POST['name']
[How to call a PHP function in MVC using AJAX]
$.ajax({
type: 'POST',
url: 'save-user.php',
data: { fname: "manish", email: "manishkp#com", role:"admin"},
success: function(data) {
console.log(data);
if(data == 'error')
{
$('#Register_error').text('Must Be filled...');
$('#Register_error').show();
}
else {
$('#Register_error').hide();
$('#Register_success').text('Successfully submit');
$('#Register_success').show();
}
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<?php
$fname = $_POST['fname'];
$email = $_POST['email'];
$role = $_POST['role'];
if(!empty($fname) && !empty($email) && !empty($role))
{
#MYSQL CONNECTION QUERY #
echo"success";
}
else{
echo "error";
}
?>
Related
I am trying to validate list of dynamic text fields.
Validation needs an AJAX call to interact with server.
At the backend I have written just one php file that reads the input request data and performs operation. Below is the example.
abc.js
row_count = 6
for (i = 1; i <=row_count; i++) {
id = "#val"+i.toString() ;
$(id).change(function(){
input_val="random";
$.ajax({
url:"url.php",
type:post,
async:true,
dataType: 'json',
data : {temp:input_val},
success:function(result){},
error: function (request, status, error) {}
});
});
}
url.php
<?php
$random_val = $_POST['temp'];
$cmd = 'systemcommand '.$random_val;
$flag = exec($cmd);
if ($flag == 0){
echo json_encode(array("status"=>'Fail'));
}
else{
echo json_encode(array("status"=>'Success'));
}
?>
It works fine when the row_count = 1 (Just one text field) but fails when the input is more than 1.
When the count is more than 1, the php script is not able to read the request data(The key in JSON data "temp"). it is blank in that case.
Any lead or help should be appreciated.
Thanks
Your javascript bit needs some adjusting, because you do not need to define an ajax for every single element. Use events based on a class. Also, since input behave differently than select, you should setup two different event class handlers.
function validateAjax ( element ) {
var input_val = element.val();// get the value of the element firing this off
$.ajax({
url: "url.php",
type: 'post',
async: true,
dataType: 'json',
data : { temp: input_val },
success: function(result) {
// check your result.status here
},
error: function (request, status, error) { }
});
}
$(".validate_change").on("change",function() { // for selects
validateAjax( $(this) );
});
$(".validate_input").on("input",function() { // for text inputs
validateAjax( $(this) );
});
And for your select or input you add that appropriate class.
<select class="validate_change" name="whatever"><options/></select>
<input class="validate_input" name="blah">
PS
I really worry about this code you have:
$cmd = 'systemcommand '.$random_val;
$flag = exec($cmd);
So, you are just executing anything that is coming in from a webpage POST var??? Please say this website will be under trusted high security access, and only people using it are trusted authenticated users :-)
I'm currently having some trouble posting a variable on jQuery for MySQLi SELECT on a PHP page.
Code of jQuery:
$("#carta1").click(function()
{
cartaId = document.getElementById("carta1").value;
console.log(cartaId);
ajaxGetResults = $.ajax({
context: this,
type: "POST",
url: "darResposta.php",
data: {'cartaId' : cartaId},
cache: false,
dataType: "json"
})
.done(function(data){
$('#3').html(data);
console.log("Avançou para a terceira parte");
$("#2").hide();
$("#3").show();
})
.fail(function(){
console.log('Erro ao buscar dados');
$("#2").hide();
$("#3").show();
$('#3').html("Deu erro");
});
});
Code of PHP:
if(!$conn)
{
echo "Falhou a ligação à base de dados";
}
else
{
if(isset($_POST['cartaId']))
{
$cartaId = $_POST['cartaId'];
$res = mysqli_query($conn,"
SELECT cartaNome, cartaDescricao
FROM tarot_cartas
WHERE cartaId = ".$cartaId
);
$data = array();
while($row = mysqli_fetch_assoc($res))
{
$data=$row;
}
echo json_encode($data);
}
}
Tried several approaches to this problem such as putting the $cartaId outside the if statement with a direct $_POST, and nothing happened.
Would appreciate if you could shed some light on this problem.
Thanks for taking the time to read and suggest a solution.
use below code
data: { 'cartaId' : cartaId },
instead of
data: {"data":JSON.stringify({'cartaId' : JSON.stringify(this)})},
Solution:
1.Remove this
data: {"data":JSON.stringify({'cartaId' : JSON.stringify(this)})},
2.Replace This one
data: { cartaId: cartaId }
Hope it works....
The jq you have will post a variable to the url.
To debug you should first check in a console (i use firebug) for mozilla) if the request is sent. Using firebug you can see the names of the POST variables you send.
Following this you should check what values get received on the server side by doing
var_dump($_POST);
Finally get the correct variable into your query. You can also debug the query by viewing the log file or, depending on whether you are using a framework, something like CI:
db->last_query();
I've been trying to figure out what I have done wrong but when I use my JavaScript Console it shows me this error : Cannot read property 'success' of null.
JavaScript
<script>
$(document).ready(function() {
$("#submitBtn").click(function() {
loginToWebsite();
})
});
</script>
<script type="text/javascript">
function loginToWebsite(){
var username = $("username").serialize();
var password = $("password").serialize();
$.ajax({
type: 'POST', url: 'secure/check_login.php', dataType: "json", data: { username: username, password: password, },
datatype:"json",
success: function(result) {
if (result.success != true){
alert("ERROR");
}
else
{
alert("SUCCESS");
}
}
});
}
</script>
PHP
$session_id = rand();
loginCheck($username,$password);
function loginCheck($username,$password)
{
$password = encryptPassword($password);
if (getUser($username,$password) == 1)
{
refreshUID($session_id);
$data = array("success" => true);
echo json_encode($data);
}
else
{
$data = array("success" => false);
echo json_encode($data);
}
}
function refreshUID($session_id)
{
#Update User Session To Database
session_start($session_id);
}
function encryptPassword($password)
{
$password = $encyPass = md5($password);
return $password;
}
function getUser($username,$password)
{
$sql="SELECT * FROM webManager WHERE username='".$username."' and password='".$password."'";
$result= mysql_query($sql) or die(mysql_error());
$count=mysql_num_rows($result) or die(mysql_error());
if ($count = 1)
{
return 1;
}
else
{
return 0;;
}
}
?>
I'm attempting to create a login form which will provide the user with information telling him if his username and password are correct or not.
There are several critical syntax problems in your code causing invalid data to be sent to server. This means your php may not be responding with JSON if the empty fields cause problems in your php functions.
No data returned would mean result.success doesn't exist...which is likely the error you see.
First the selectors: $("username") & $("password") are invalid so your data params will be undefined. Assuming these are element ID's you are missing # prefix. EDIT: turns out these are not the ID's but selectors are invalid regardless
You don't want to use serialize() if you are creating a data object to have jQuery parse into formData. Use one or the other.
to make it simple try using var username = $("#inputUsername").val(). You can fix ID for password field accordingly
dataType is in your options object twice, one with a typo. Remove datatype:"json", which is not camelCase
Learn how to inspect an AJAX request in your browser console. You would have realized that the data params had no values in very short time. At that point a little debugging in console would have lead you to some immediate points to troubleshoot.
Also inspecting request you would likely see no json was returned
EDIT: Also seems you will need to do some validation in your php as input data is obviously causing a failure to return any response data
Try to add this in back-end process:
header("Cache-Control: no-cache, must-revalidate");
header('Content-type: application/json');
header('Content-type: text/json');
hope this help !
i testet on your page. You have other problems. Your postvaribales in your ajax call are missing, because your selectors are wrong!
You are trying to select the input's name attribute via ID selector. The ID of your input['name'] is "inputUsername"
So you have to select it this way
$('#inputUsername').val();
// or
$('input[name="username"]').val();
I tried it again. You PHP script is responsing nothing. Just a 200.
$.ajax({
type: 'POST',
url: 'secure/check_login.php',
dataType: "json",
data: 'username='+$("#inputUsername").val()+'&password='+$("#inputPassword").val(),
success: function(result) {
if (result.success != true){
alert("ERROR");
} else {
alert("HEHEHE");
}
}
});
Try to add following code on the top of your PHP script.
header("Content-type: appliation/json");
echo '{"success":true}';
exit;
You need to convert the string returned by the PHP script, (see this question) for this you need to use the $.parseJSON() (see more in the jQuery API).
i have a javascript code that requests a php page to provide it with list of names that are currently online and update a Table, but i have a problem sending it back in form of an array, someone told me that this is usually done using XML, but i dont know how to start.
javascript Post method:-
$.post( "updateTable.php", POSTdata,
function( data ) {
$("#mytable").last().append('<tr><td>'+data+'</td></tr>');
}
);
the php file:-
include("connect.php");
$query1 = "SELECT * FROM formtable";
$result_id = mysql_query($query1, $global_dbh)
or die ("display_db_query:" . mysql_error());
while ($table_array = mysql_fetch_object ($result_id))
{
$rows[] = $table_array;
}
foreach ($rows as $temp ) {
if ($temp->isOnline==1)
$newRow[] = $temp->name;
}
echo "$newRow";
mysql_close($global_dbh);
Please excuse any syntax or semantics in my code, i am a beginner.
How can i populate my table using ajax callback function, and in what form the data will arrive there, and how can i use xml to help me.
Many thanks in advance.
A quick example of json:
var table = $("#mytable").last();
$.ajax({
type: 'post',
url: "updateTable.php",
dataType: 'json',
data: POSTdata,
success: function(data){
jQuery.each(data, function(i, row){
//console.log(row);
table.append('<tr><td>'+row.name+'</td></tr>');
});
}
});
and in php file, instead of :
echo "$newRow";
replace with:
echo json_encode($newRow);
That's it!
I have a simple checkbox, on click it sends XHR to PHP page , php processes correctly and I use json_encode($response) to return. But instead of a simple true or false I get the source code for the page and it is causing a "parsererror" of course.
ajax call as follows
$.ajax({
type: "post",
url: "myprocessor.php",
dataType: 'json',
data: { "id" : idnumber, "action": "makeLive", "isLive" : "1" },
beforeSend: function(data) {
$("#ajaxInProgress").addClass('progress');
},
success: function(data) {
$("#status").removeClass().addClass((data.error === true) ? 'error' : 'success').text('Success! Appraiser is NO LONGER Live ').slideDown('slow');
},
error: function(data) {
$("#status").removeClass().addClass('error').text(' - Changing the Live status for this appraiser to "Not Live" has failed - APPRAISER IS STILL LIVE IN SYSTEM, please try again').slideDown('slow');
},
complete: function(data) {
$("#ajaxInProgress").removeClass('progress');
setTimeout(function() {
$("#status").slideUp('slow').removeClass();
},2000);
}
});
The php I post to is as follows:
if (isset($_POST['action'])) {
if($_POST['action']=='makeLive') {
$checkappraiser=mysql_query("SELECT * FROM table WHERE id='".mysql_real_escape_string($_POST['id'])."'");
if (mysql_numrows($checkappraiser)>0) {
$livesetting=mysql_result($checkappraiser,0,"live");
$livesetting=!$livesetting;
$runSql = mysql_query("UPDATE table SET live='$livesetting' WHERE id='".mysql_real_escape_string($_POST['id'])."'");
if(!$runSql) {
$return['error'] = true;
} else {
$return['error'] = false;
}
}
}
echo json_encode($return);
}
Any suggestions would be great.
I am getting the proper data passed
I am getting the correct data updated in DB
My response is coming back as a parser error because it is trying to parse the source code as a json array.
Just a quick check, do you put <?php at the beginning of your php file?
That, or you're doing something wrong in your webserver, not passing files through to php correctly. Does hitting the php file directly load the source or the result?
If you hit page.php, does it load the same thing as if you hit page.phP or pHP, etc? It matters to web server filters, depending on the web server...
If you use tomcat for java, for example... you can turn off case sensitivity for finding files, but it does not turn off case sensitivity for mapping files to filters or servlets, so .jsp would load the jsp servlet, but .jsP would not.