Develop Reference

ajax php post with sessions wont give results

I am trying to get pre set session variables through ajax and display in my modal. its like a cart. But when i try to get them there is no result at all when i read the ajax result from chrome inspect network or the modal. below is my code. what am i doing wrong here?
<script>
$("#cart-button").click(function(){
$.ajax({
url: "includes/cart-read.php",
success: function(data){
console.log(data);
alert(data);
$('#modal-body').empty().append(''+data+'');
}
});
});
</script>
and in cart-read.php
if(isset($_SESSION['dices'])){
foreach ($_SESSION['dices'] as $dice){
$msg = $dice;
echo json_encode($msg);
}
}
session dices is an array with simple numbers. such as 4, 5, 6.

Make sure you have the following things in your code
Check if you have added session_start(); below after
Try putting an else case to your session check,
Try using an array to store and display the results.
<?php
session_start();
$result = array();
$msg = array();
if(isset($_SESSION['dices'])){
foreach ($_SESSION['dices'] as $dice){
array_push($msg,$dice);
}
if(sizeof($msg) > 0)
{
$result['status'] = true;
$result['message'] = $msg;
}
else
{
$result['status'] = false;
$result['message'] = 'No values';
}
}
else
{
$result['status'] = false;
$result['message'] = 'Session not set';
}
echo json_encode($result);
?>

JQuery form submission generates a new form

I have a JQuery script that submits user input to a PHP script in the same file, and then displays the result of what the PHP script does with the input. That part works fine. The issue that I’m having is that, upon submission, the JQuery script (at least, I think it's the script) also generates a new submission box below the original.
I’m not sure why. I thought at first that it was an issue with the input type, with the asynchronous part, or even with where I had the form in the overall code, but none of those seem to be playing any role. I'm still a beginner and I'm just not seeing the issue.
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"></script>
<form id = "my_form">
verb <input type = "text" id ="word1"/>
<input type = "submit"/></form>
<div id="name"></div>
<script>
$(document).ready(function(){
$("#my_form").on('submit', function(e)
{
e.preventDefault();
var verb = $ ("#word1").val();
var tag = "#Latin ";
var url = "http://en.wiktionary.org/wiki/"+verb+tag;
$.ajax({
url: "Parser.php",
data: {"verb": verb},
type: "POST",
async: true,
success: function(result){
$("#name").html(result);
$("#name").append(url);
}
});
});
});</script>
RESULT:
PHP
<?php
$bank = array();
function endsWith($haystack, $needle) {
return $needle === "" || (($temp = strlen($haystack) - strlen($needle)) >= 0 && strpos($haystack, $needle, $temp) !== false);
}
function check_end_array($str, $ends)
{
foreach ($ends as $try) {
if (substr($str, -1*strlen($try))===$try) return $try;
}
return false;
}
function db_connect() {
static $connection;
if(!isset($connection)) {
$connection = mysqli_connect('127.0.0.1','username','password','Verb_Bank');
}
if($connection === false) {
return mysqli_connect_error();
}
return $connection;
}
function db_query($query) {
$connection = db_connect();
$result = mysqli_query($connection,$query);
return $result;
}
function db_quote($value) {
$connection = db_connect();
return "'" . mysqli_real_escape_string($connection,$value) . "'";
}
$y = false;
if (isset($_POST['verb'])){
$y=db_quote($_POST['verb']);
echo $y;
echo "\n";
$m = db_query("SELECT `conjugation` FROM normal_verbs WHERE (" . $y . ") LIKE CONCAT('%',root,'%')");
if($m !== false) {
$rows = array();
while ($row = mysqli_fetch_assoc($m)) {
$rows[] = $row;
}
}
foreach ($rows as $key => $value){
if (in_array("first",$value)==true){
echo "first conjugation verb\n";}
$y = $_POST["verb"];
$x = $y;
foreach ($bank as $key => $value)
(series of IF-statements)
}}?>

As Roamer-1888 says's the problem lies in server side, you are returning a html which has a input too. You need to change your code to return only the result string which you append to the div. Else if this is not possible doing at server side as it might require you to change lot of code, then you can strip off the input element from the result and then append it to the div. Like below.
success: function(result){
var div = document.createElement('div');
div.innerHTML = result;
$(div).find('input').remove();
$("#name").html(div.innerHTML);
$("#name").append(url);
}

Ajax returns success but doesn't change the database

I'm developing a small script of js to edit a profile in the way facebook used to be (click a button, edit and save without reloading the page). The problem is that when I run it, the ajax function returns sucess but akes no changes on the database. The function os js is this:
$('.savebtn').click(function(){
var editdata = $(".editbox").val();
var parameter = $(this).closest("td").find("#parameter").text();
var datastring = "data="+editdata+"&parameter="+parameter;
var $t = $(this);
console.log(datastring);
$.ajax({
type: "POST",
url: BASE_URL + "/API/update_profile.php",
data: datastring,
cache: false,
success: function()
{
$t.closest('td').find('.curr_value').html(editdata);
$t.closest('td').find('.curr_value').hide;
console.log(editdata);
$(this).prev(".edit").hide();
$(this).prev(".curr_value").show();
$(this).prev('.edit_link').show();
$(this).hide();
}
});
});
(Ignore the $t thing, somehow this works like this, but not if I use $(this))
Ajax executes the code for sucess but doesn't update anything on the database.
The PHP code for the database is:
<?php
include_once("../../config/connect_db.php");
include_once("../../database/cliente.php");
$parameter = $_POST['parameter'];
$data = $_POST['data'];
$id = $_SESSION['id'];
var_dump($_POST);
try {
updateProfile($parameter, $data, $id);
}
catch (PDOException $e)
{
echo 'Caught exception: ', $e->getMessage(), "\n";
}
?>
function updateProfile($parameter, $data, $id)
{
global $conn;
$stmt = $conn->prepare("UPDATE biofood.users
SET ? = ?
WHERE id = ?");
$stmt->execute(array($parameter, $data. $id));
}
EDIT: As pointed out, this could be a problem with trying to pass a column name as a parameter. Changed the code to the following, but with no sucess:
function updateProfile($parameter, $data, $id)
{
global $conn;
$query = "UPDATE biofood.users
SET $parameter = $data
WHERE id = $id";
$stmt = $conn->prepare($query);
$stmt->execute();
}

This line:
$stmt->execute(array($parameter, $data. $id));
I think should be
$stmt->execute(array($parameter, $data, $id));
(notice the comma after $data)

This might not solve your problem, but it might give you a better indication on where your problem is.
First, you are not checking whether it works or not as your updateProfile function returns nothing.
Modify your updateProfile function, so that it returns the number of rows affected. (BTW this is a safer way to write your function. If you can check or limit the value of $parameter prior to calling this function, it will be less prone to SQL injection.)
function updateProfile($parameter, $data, $id)
{
global $conn;
$stmt = $conn->prepare("UPDATE biofood.users SET $parameter = ? WHERE id = ?");
$stmt->execute(array($data, $id));
return $stmt->rowCount(); // # of rows affected
}
In the script that calls this function, get the value and send it back as a response. We'll send back a JSON.
$response = array();
try {
$response['success'] = updateProfile($parameter, $data, $id);
} catch (PDOException $e) {
echo 'Caught exception: ', $e->getMessage(), "\n";
}
header('Content-Type: application/json');
echo json_encode($response);
In your JavaScript file, make the following change:
$.ajax({
type: "POST",
url: BASE_URL + "/API/update_profile.php",
data: datastring,
cache: false,
success: function (data) {
if (data.success) {
$t.closest('td').find('.curr_value').html(editdata);
$t.closest('td').find('.curr_value').hide;
console.log(editdata);
$(this).prev(".edit").hide();
$(this).prev(".curr_value").show();
$(this).prev('.edit_link').show();
$(this).hide();
}
},
dataType: 'json'
});

Get data from database using PHP, JQuery and AJAX in JSON format

I'm having trouble getting data from my database. My goal is to get all groups from my database and return them in JSON (in an alert box or whatever).
Now it won't convert to JSON and I am getting weird response text from the ajax call. If you need anything else to solve this problem, please do not hesitate to ask.
Here is what I did.
PHP
$servername = "redacted";
$username = "redacted";
$password = "redacted";
$dbname = "redacted";
if(isset($_POST['action']) && !empty($_POST['action'])) {
$action = $_POST['action'];
switch($action) {
case 'getGroups' : getAllGroups();break;
}
}
function getAllGroups() {
$mysqli = new mysqli($servername, $username, $password, $dbname);
$query = $mysqli->query("SELECT * FROM groups");
while($row = $query->fetch_object()) {
$result[] = $row;
}
echo "{\"results\":";
echo json_encode($result);
echo "}";
$mysqli->close();
}
JS
function getPosts() {
$.ajax({
url: 'functions.php',
data: {action: 'getGroups'},
type: 'post',
success: function(output) {
var result = JSON.parse(output);
result = result.resultaten;
alert(result);
}
});
}
getPosts();
Thanks in advance,
Mistergrave.

No need for that extra echos. Try with -
echo json_encode(array('results' => $result));
Instead of -
echo "{\"results\":";
echo json_encode($result);
echo "}";
No need for - if(isset($_POST['action']) && !empty($_POST['action'])) {
if(!empty($_POST['action'])) { - do the all.
Define $result first.
$result = array();
while($row = $query->fetch_object()) {
$result[] = $row;
}

Okay guys, I managed to solve everything. Apparently the php function couldn't find my credentials to log in to the database server because I defined them on top of the php file (and since javascript only executed the function, these credentials were undefined).
Solution:
I just copy-pasted the credentials at the start of each function so these were defined. And tadaah! It worked :).
Now I realize why the responseText was full of tables, because it started to return error tables about the connection.
I hope my explanation will help other people who have this issue as well.
Cheers, and thanks for all the helpfull answers,
Mistergrave.

use
echo json_encode(array('result'=>$result));
As it takes array as parameter. Check here
Just a note :
If are sure you will return json data, use dataType:json , so you wont need JSON.parse(output).

best option to get php array variable in Javascript produced by php script that requested through an ajax call

Currently I am trying to create a live search bar that only produce 5 results max and more option if there is over 5 results. So what I have done so far is a jquery ajax script to call a php script that runs asynchronously on key up in textbox I have.
I want to get the php array then I will code it further using javascript.
This is my code now:
Javascript code
<script type="text/javascript">
function find(value)
{
$( "#test" ).empty();
$.ajax({
url: 'searchDb.php',
type: 'POST',
data: {"asyn": value},
success: function(data) {
return $lala;
var lala = $lala;
$( "#test" ).html($lala);
}
});
}
</script>
SearchDb PHP code:
<?php
function searchDb($abc, $limit = null){
if (isset($abc) && $abc) {
$sql = "SELECT testa FROM test WHERE testa LIKE '%$abc%'";
if($limit !== null){
$sql .= "LIMIT ". $limit;
}
$result = mysql_query($sql) or die('Error, insert query failed') ;
$lists = array();
while ( $row = mysql_fetch_assoc($result))
{
$var = "<div>".$row["testa"]."</div>";
array_push($lists, $var);
}
}
return $lists;
}
$abc = $_POST['asyn'];
$limit = 6;
$lala = searchDb($abc);
print_r($lala);
?>
How can I get $lala

Have you considered encoding the PHP array into JSON? So instead of just echoing the array $lala, do:
echo json_encode($lala);
Then, on the Javascript side, you'll use jQuery to parse the json.
var jsonResponse = $.parseJSON(data);
Then you'll be able to use this jsonResponse variable to access the data returned.

You need to read jQuery .ajax and also you must view this answer it's very important for you
$.ajax({
url: 'searchDb.php',
cache: false,
type: 'post'
})
.done(function(html) {
$("#yourClass").append(html);
});
In your searchDb.php use echo and try this code:
function searchDb($str, $limit = null){
$lists = array();
if (isset($str) && !empty($data)) {
$sql = "SELECT testa FROM test WHERE testa LIKE '%$data%'";
if(0 < $limit){
$sql .= "LIMIT ". $limit;
}
$result = mysql_query($sql) or die('Error, insert query failed') ;
while ( $row = mysql_fetch_assoc($result))
{
$lists[] = "<div>".$row["testa"]."</div>";
}
}
return implode('', $lists);
}
$limit = 6;
$data = searchDb($_POST['asyn'], $limit);
echo $data;
?>

If you dont have or your page searchDb.php dont throw any error, then you just need to echo $lala; and you will get result in success part of your ajax function
ALso in your ajax funciton you have
//you are using data here
success: function(data) {
return $lala;
var lala = $lala;
$( "#test" ).html($lala);
}
you must try some thing like this
success: function(data) {
var lala = data;
$( "#test" ).html($lala);
}