Develop Reference

JQuery is not working properly in my wordpress

I have button which pressed update row in database based on variable:
<input type="submit" id="wyslij" name="przycisk" value="<?php echo $checkboxstatus;?>">
and I try to make ajax connection in wordpress but its not working, i tried diffrent ways but still the same results. It's even not responding while clicking the button and not receving error.
Ajax:
jQuery(document).ready( function() {
jQuery("#wyslij").click( function(e) {
e.preventDefault();
checkboxstatus = <?php echo $checkboxstatus; ?>
jQuery.ajax({
type : "post",
url : czekboks.php,
data : {checkboxstatus},
})
})
})
</script>
czekboks.php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "obejrzaneodcinki";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
$query = " UPDATE uzytkownik
SET status = '$checkboxstatus'
WHERE id = '2115' ";
if(mysqli_query($conn,$query))
{
echo "good job";
echo "<br />";
}

How to fetch data from database and display it using AJAX in an app

I have not touched PHP or AJAX that much in about 8 years, so my memory on this is pretty low at the moment.
What I'm doing is fetching data from my database that works great.
Then I want to use AJAX to get the data from the PHP file.
My PHP file on my server is connecting to the database and fetching the table "Form".
This data is then going to be retrieved by another app trough AJAX.
I have a working PHP file, but how I should order this for the AJAX to fetch it nicely is a big question for me.
The things I have at the moment is:
PHP FILE:
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM form";
$result = $conn->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$name = $row['name'];
$country = $row['country'];
$email = $row['email'];
$need = $row['need'];
$available = $row['available'];
echo "name: " . $name . "<br> Country: " . $country . " <br> Email: " . $email . "<br> Need: " . $need . "<br> Available: " . $available;
$form_data = array();
array_push($form_data)
}
} else {
echo "Null results";
}
$conn->close();
?>
AJAX FILE:
$.ajax({
url: 'url',
data: "",
dataType: 'json',
success: function(data) {
}
});
The AJAX file is not complete since I'm still wondering about these PHP results.
Right now they are just stored in variables.
What is the best way to store these results to get an fetch for AJAX here?
Should I put the results in an array and then push that array into another array?
There can be many lines in forms, and I want 1 person with data: name, country etc to be in one array. Or is it stupid to have it in an array?
I hope someone can give a little clue and help me on my way here. Long time since I have been doing PHP and a little bit unclear about the best approach here.
Been searching for a while, but nothing made so much sense to me, so I'm coming here hoping someone can guide the way.
Happy Easter.

Create an array and save all your data, then json_encode to send the data to JavaScript.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "db_julekgwa";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM form";
$result = $conn->query($sql);
$rows = array();
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$name = $row['name'];
$country = $row['country'];
$email = $row['email'];
$need = $row['need'];
$available = $row['available'];
$rows[] = "{ name: " . $name . ", Country: " . $country . ", Email: " . $email . ", Need: " . $need . ", Available: " . $available . "}";
}
echo json_encode($rows);
} else {
echo json_encode([ "error" => "Null results"]);
}
$conn->close();
?>
In your ajax do the following
$.ajax({
url: 'file.php',
type: "GET",
dataType: 'json',
success: function(data) {
data.forEach(item => {
console.log(item);
});
},
error: function(data) {
console.log(data);
}
});

How to get php echo result in javascript

I have my php file on a server that retrieves data from my database.
<?php
$servername = "myHosting";
$username = "myUserName";
$password = "MyPassword";
$dbname = "myDbName";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name, description FROM tableName;";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$row_number = 0;
while($row = $result->fetch_assoc()) {
$row_number++;
echo $_GET[$row_number. ";". $row["id"]. ";". $row["name"]. ";". $row["description"]. "<br>"];
}
} else {
echo "0 results";
}
$conn->close();
?>
Unfortunately, I do not know how to receive data from a php file using javascript.
I would like the script in javascript to display the received data in the console in browser.
The script written in javascript is Userscript in my browser extension(tampermonkey) and php file is on my server.
I've tried to use ajax, unfortunately without positive results.
(the php script works as expected).
JS(not working):
$.ajax({
url: 'https://myserver.com/file.php',
type: 'POST',
success: function(response) {
console.log(response);
}
});

The code within the loop is a little screwy
$_GET[$row_number. ";". $row["id"]. ";". $row["name"]. ";". $row["description"]. "<br>"]
that suggests a very oddly named querystring parameter which is not, I think, what was intended.
Instead, perhaps try like this:
<?php
$servername = 'myHosting';
$username = 'myUserName';
$password = 'MyPassword';
$dbname = 'myDbName';
$conn = new mysqli($servername, $username, $password, $dbname);
if( $conn->connect_error ) {
die( 'Connection failed: ' . $conn->connect_error );
}
$sql = 'select `id`, `name`, `description` from `tablename`;';
$result = $conn->query($sql);
if( $result->num_rows > 0 ) {
$row_number = 0;
while( $row = $result->fetch_assoc() ) {
$row_number++;
/* print out row number and recordset details using a pre-defined format */
printf(
'%d;%d;%s;%s<br />',
$row_number,
$row['id'],
$row['name'],
$row['description']
);
}
} else {
echo '0 results';
}
$conn->close();
?>
A full example to illustrate how your ajax code can interact with the db. The php code at the top of the example is to emulate your remote script - the query is more or less the same as your own and the javascript is only slightly modified... if you were to change the sql query for your own it ought to work...
<?php
error_reporting( E_ALL );
ini_set( 'display_errors', 1 );
if( $_SERVER['REQUEST_METHOD']=='POST' ){
ob_clean();
/* emulate the remote script */
$dbport = 3306;
$dbhost = 'localhost';
$dbuser = 'root';
$dbpwd = 'xxx';
$dbname = 'xxx';
$db = new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );
$sql= 'select `id`, `address` as `name`, `suburb` as `description` from `wishlist`';
$res=$db->query( $sql );
$row_number=0;
while( $row=$res->fetch_assoc() ){
$row_number++;
/* print out row number and recordset details using a pre-defined format */
printf(
'%d;%d;%s;%s<br />',
$row_number,
$row['id'],
$row['name'],
$row['description']
);
}
exit();
}
?>
<!DOCTYPE html>
<html lang='en'>
<head>
<meta charset='utf-8' />
<script src='//code.jquery.com/jquery-latest.js'></script>
<title>Basic Ajax & db interaction</title>
<script>
$( document ).ready( function(){
$.ajax({
url: location.href,
type: 'POST',
success: function( response ) {
console.log( response );
document.getElementById('out').innerHTML=response;
}
});
} );
</script>
</head>
<body>
<div id='out'></div>
</body>
</html>

Hi you can do it this way:
your php script:
if (isset($_POST["action"])) {
$action = $_POST["action"];
switch ($action) {
case 'SLC':
if (isset($_POST["id"])) {
$id = $_POST["id"];
if (is_int($id)) {
$query = "select * from alumni_users where userId = '$id' ";
$update = mysqli_query($mysqli, $query);
$response = array();
while($row = mysqli_fetch_array($update)){
.......
fill your response here
}
echo json_encode($response);
}
}
break;
}
}
Where action is a command you want to do SLC, UPD, DEL etc and id is a parameter
then in your ajax:
function getdetails() {
var value = $('#userId').val();
return $.ajax({
type: "POST",
url: "getInfo.php",
data: {id: value}
})
}
call it like this:
getdetails().done(function(response){
var data=JSON.parse(response);
if (data != null) {
//fill your forms using your data
}
})
Hope it helps

How can i choose a particular sheet and display the data in it?

How can I choose a particular sheet and display the data in it. I have used some of the functions in the code which library functions should I add to execute the code?
It is displaying all the Excel file data, but I want specific page or sheet to be displayed.
<?php
require 'Classes/PHPExcel/IOFactory.php';
// Mysql database
$servername = "localhost";
$username = "root";
$password = "123456";
$dbname = "import_db";
$inputfilename = $_POST['fileToUpload'];
$exceldata = array();
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
// Read your Excel workbook
try
{
$inputfiletype = PHPExcel_IOFactory::identify($inputfilename);
$objReader = PHPExcel_IOFactory::createReader($inputfiletype);
$objPHPExcel = $objReader->load($inputfilename);
}
catch(Exception $e)
{
die('Error loading file "'.pathinfo($inputfilename,PATHINFO_BASENAME).'": '.$e->getMessage());
}
// Get worksheet dimensions
$sheet = $objPHPExcel->getSheet(0);
//$sheetNames = $sheet->getSheetNames();
$highestRow = $sheet->getHighestRow();
$highestColumn = $sheet->getHighestColumn();
$count=$objPHPExcel->getSheetCount();
$names = $objPHPExcel->getSheetNames();
echo $count;
echo $names;
//echo $sheetNames;
// Loop through each row of the worksheet in turn
for ($row = 1; $row <= $highestRow; $row++)
{
// Read a row of data into an array
$rowData = $sheet->rangeToArray('A' . $row . ':' . $highestColumn . $row, NULL, TRUE, FALSE);
$exceldata[] = $rowData[0];
}
// Print excel data
echo "<table>";
foreach ($exceldata as $index => $excelraw)
{
echo "<tr>";
foreach ($excelraw as $excelcolumn)
{
echo "<td>".$excelcolumn."</td>";
}
echo "</tr>";
}
echo "</table>";
mysqli_close($conn);
?>

Change below code in try and it will work for you.
try
{
$sheetname = 'Data Sheet #2'; // Sheet name which you want to load
$objReader = PHPExcel_IOFactory::createReader($inputfiletype); // Create a new Reader of the type defined in $inputfiletype
$objReader->setLoadSheetsOnly($sheetname); // Advise the Reader of which WorkSheets we want to load
$objPHPExcel = $objReader->load($inputfilename); //Load $inputfilenameto a PHPExcel Object
}

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: