Develop Reference

Returning an itterable object to Ajax from php

How can I iterate through the object sent back from the php script below in my jquery/ajax call?
I tried result[0] whiche gave me back c. That means that I'm being returned a string.What code should I write to be returned company1 etc. ?
DROP DATABASE IF EXISTS test;
CREATE DATABASE test;
USE test;
CREATE TABLE company(
name VARCHAR(255),
id INT PRIMARY KEY AUTO_INCREMENT
);
INSERT INTO company(name) VALUES( 'company1');
INSERT INTO company(name) VALUES( 'company2');
INSERT INTO company(name) VALUES( 'company2');
$.ajax({
type: "GET",
url: "manager-get-company.php",
success: function (response) {
//iterate through response here
//console.log(response[0]; -> log Company1
//console.log(response[1]; -> log Company2
}
});
<?php
//manager-get-company.php
$hostname = 'localhost';
$username = 'root';
$password = '';
$database_name = 'test';
$con = mysqli_connect($hostname,$username,$password,$database_name);
//Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
exit();
}
$sql= mysqli_query($con, 'SELECT name FROM company');
while($row = mysqli_fetch_array($sql)){
echo $row['name'];
}

In ajax add dataType : "json",
$.ajax({
type: "GET",
dataType : "json",
url: "manager-get-company.php",
success: function (response) {
//iterate through response here
if(!response.error){
console.log(response[0]);
}
}
});
In php
$result = [];
//Check connection
if (mysqli_connect_errno()) {
$result['error'] = "Failed to connect to MySQL: " . mysqli_connect_error();
}else{
$sql= mysqli_query($con, 'SELECT name FROM company');
while($row = mysqli_fetch_array($sql)){
$result[] = $row['name'];
}
}
echo json_encode($result);
Edited : use $result['error'] not result['error']

Query result in the error response in the JQuery function

I have a problem, clicking in a < tr > of a table I call a javascript function which in turn calls a function in php to get data in a database. The click on the table row works, sql works, and from the console.log command I know there is an answer in reponseText. but it does not work and I get the error back, I'll post the code. I hope you can help me.
file config.php
//database credentials
define('DBHOST','localhost');
define('DBUSER','root');
define('DBPASS','');
define('DBNAME','toor');
try{
//create PDO connection
$db = new PDO("mysql:host=".DBHOST.";charset=utf8mb4;dbname=".DBNAME, DBUSER, DBPASS);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
} catch(PDOException $e) {
//show error
echo '<p class="bg-danger">'.$e->getMessage().'</p>';
exit;
}
//include the user card, pass in the database connection
include($_SERVER['DOCUMENT_ROOT'].'cards.php');
$card = new card($db);
file cards.php
<?php
class card
{
private $_db;
function __construct($db){
$this->_db = $db;
}
public function view_card_id($id)
{
$rows = array();
$statement = $this->_db->prepare('SELECT * FROM card_details WHERE card_id = :card_id');
$statement->execute(array(':card_id' => $id));
$numrows = $statement->fetch(PDO::FETCH_ASSOC);
if($numrows < 1) {
$this->error = "Error";
return false;
} else {
$statement->bindColumn("card_id", $cid);
$statement->bindColumn("a", $a);
$statement->bindColumn("b", $b);
$rows[] = array('card_id' => $numrows['card_id'], 'a' => $numrows['a'], 'b' => $numrows['b']);
while ($row = $statement->fetch(PDO::FETCH_ASSOC)) {
$row = array('card_id' => $cid, 'a' => $a, 'b' => $b);
$rows[] = $row;
}
return $rows;
}
}
file index.php
<?php
//include config
require_once($_SERVER['DOCUMENT_ROOT'].'config.php');
?>
html code ....
<script src="/js/Cards.js" type="text/javascript"></script>
file Cards.js
$('#table-cards tr').click(function() {
var id = $(this).find("a").text();
$.ajax({
type: 'POST',
url: '/classes/cardsFunc.php',
dataType:'text',
data: {functionname: "view_card", id: id },
success: function(response){
//Use response
alert("Server echo: "+response);
console.log(response);
},
error: function(msg){
console.log(msg);
alert("Error: "+msg);
}
});
});
In the Cards.js file, once the $ .ajax function is called, it does not return to success but to error, but in the console.log I see the array of the executed query under the responseText entry.
that is, in the error response, I see the result of the query, which in theory should be in the response of success.
I also tried to use
$.post('/classes/cardsFunc.php', { functionname: 'view_card', id: id }, function(data){
});
but nothing
file cardsFunc.php
<?php
//include config
require_once($_SERVER['DOCUMENT_ROOT'].'config.php');
if(isset($_POST['functionname']) && $_POST['functionname'] == "view_card"){
$card_view = $card->view_card_id($_POST['id']);
print json_encode($card_view);
}
?>
thank you for the time you have dedicated to me
I noticed that if I recreate the connection to the db in the file cardsFunc.php, everything works, but I do not understand why, since everything is in the config.php file.
Like this:
file cardsFunc.php
<?php
//database credentials
define('DBHOST','localhost');
define('DBUSER','root');
define('DBPASS','');
define('DBNAME','toor');
try{
//create PDO connection
$db = new PDO("mysql:host=".DBHOST.";charset=utf8mb4;dbname=".DBNAME, DBUSER, DBPASS);
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
} catch(PDOException $e) {
//show error
echo '<p class="bg-danger">'.$e->getMessage().'</p>';
exit;
}
//include the user card, pass in the database connection
include($_SERVER['DOCUMENT_ROOT'].'cards.php');
$card = new card($db);
if(isset($_POST['functionname']) && $_POST['functionname'] == "view_card"){
$card_view = $card->view_card_id($_POST['id']);
print json_encode($card_view);
}
?>

How to send an array of objects to PHP with Ajax and update the MySQL server

I have this array of objects:
var desks = [];
function Desk(id, name, x, y) {
var desk = {};
desk.id = id;
desk.name = name;
desk.x = x;
desk.y = y;
desks.push(desk);
}
And I have a MySQL table with desk_id, desk_name, desk_x and desk_y columns.
When I click a button I want to update the values in the table with the Desk objects.
Here's basically what I'm trying achieve:
JS
$("#btn").click(function() {
$.ajax({
method: "POST",
url: "phpfile.php",
data: desks
})
});
PHP
<?php
if ($_SERVER['REQUEST_METHOD'] === 'POST') {
require "../db.php";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$desks = $_POST("desks");
for($i = 0; $i < sizeof($desks); $i++) {
$sql = "UPDATE desks SET desk_x=$desks[i]->x WHERE desk_id=$desks[i]->id";
$conn->query($sql);
$sql = "UPDATE desks SET desk_y=$desks[i]->y WHERE desk_id=$desks[i]->id";
$conn->query($sql);
}
$conn->close();
}
?>
I don't know whats wrong with it. Any help is appreciated.

Do like this
JS
$("#btn").click(function() {
$.ajax({
method: "POST",
url: "phpfile.php",
data: {desks:desks}
})
});
PHP
$desks = $_POST("desks");
$sql = "UPDATE desks SET desk_x=$desks->x,desk_y=$desks->y WHERE desk_id=$desks->id";
$conn->query($sql);

You need the stringify the array when you send it.
JS
$("#btn").click(function() {
$.ajax({
method: "POST",
url: "phpfile.php",
data: { "desks" : JSON.stringify(desks) }
})
});
PHP
$desks = json_decode($_POST["desks"]);
foreach($desks as $key => $value) {
$sql = "UPDATE desks SET desk_x=".$value->x.",desk_y=".$value->y." WHERE desk_id=".$value->id;
$conn->query($sql);
}

Ajax triggers error even when success

This is the first time I worked with Ajax in combination with PHP and I'm having a hard time getting it to work.
I have this piece of Ajax code:
$.ajax({
url: '/dev/php/login.php',
type: 'POST',
dataType: 'json',
error: function(date) {
console.log(date);
alert("Het emailadres of wachtwoord is onjuist");
},
data: {email:email,password:password},
success: function(response) {
console.log(response);
alert(response);
}
})
With this PHP code:
<?php
$servername = "blur";
$username = "blur";
$dbpassword = "blur";
$dbname = "blur";
$emailadres = $_POST['email'];
$password = $_POST['password'];
// Create connection
$conn = new mysqli($servername, $username, $dbpassword, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$encodedpass = md5($password);
$sql = "SELECT first_name, last_name FROM account WHERE `email`='$emailadres' AND `password`='$encodedpass'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "Name: " . $row["first_name"]. " " . $row["last_name"] . " WEEEE";
}
}
else {
if (!function_exists('http_response_code'))
{
function http_response_code($newcode = NULL)
{
static $code = 200;
if($newcode !== NULL)
{
header('X-PHP-Response-Code: '.$newcode, true, $newcode);
if(!headers_sent())
$code = $newcode;
}
return $code;
}
}
}
$conn->close();
?>
For some reason the error function in Ajax always gets called, even when the response text echoes me the name from the database. It looks like this:

Why ajax code to connect with php is not working?

I wrote the ajax in the JavaScript function. That code is
function getValidate(checkID)
{
alert(checkID);
$.ajax({
type: 'post',
url: 'checkval.php',
datatype: 'json',
data: {checkID : checkID},
success: function (response) {
if (response === "OK"){
alert("Validation Successed.");
}else if(response === "NG"){
alert("Check Already Exists.");
}
},
error : function(err, req) {
alert("Error Occurred");
}
});
}
this code is outputs only "Error Occurred".
the connected php script is
<?php
echo("welcome");
$check = $_POST['checkID'];
$host = 'localhost';
$database = 'database';
$username = 'root';
$password = 'root';
$dbc = mysqli_connect($host,$username,$password,$database);
$checkno = $check;
$sql = "select claimno from check_details where checkno = $checkno";
$result = mysqli_query($dbc,$sql);
$rows = mysqli_num_rows($result);
if($rows != 0)
{
echo "NG";
}
else
{
echo "OK";
}
?>
at a time of calling the JavaScript function php file not executed......
please give me the idea to success it...........

Try this :
if($rows != 0)
{
$return = "NG";
}
else
{
$return = "OK";
}
echo json_encode($return);

Also you set datatype to json so response data must be json type
$.ajax({ dataType:"json"});
In php, store result in one variable and return json_encode
<?php
echo("welcome");
$check = $_POST['checkID'];
$host = 'localhost';
$database = 'database';
$username = 'root';
$password = 'root';
$dbc = mysqli_connect($host,$username,$password,$database);
$checkno = $check;
$sql = "select claimno from check_details where checkno = '$checkno'"; //use single quote
$result = mysqli_query($dbc,$sql);
$rows = mysqli_num_rows($result);
if($rows != 0)
{
$res = "NG";
}
else
{
$res = "OK";
}
echo json_encode($res);
?>

You are getting error occurred, because of below code. Try logging something meaningful to triag this.
error : function(err, req) {
alert("Error Occurred");
}
Please try below code to get a clue of the error
error: function(xhr, status, error) {
var err = eval("(" + xhr.responseText + ")");
alert(err.Message);
}
Reference: Take a look at this query

It seems that the php script is not working well.
Debug it with try and catch and see what output is comming.