load JSON data attributes from external php page - javascript

Hi I try to load a JSON data from a php link in a js file whith this
$.getJSON('link to rempl.php', function(data) {
cn =data.nom;
});
this is the code of rempl.php
<?php
header('Content-Type: text/plain; charset=utf-8');
require_once ('connectDB.php');
if ($_POST['nom']== 0){
header("Location: http://localhost/geopp/editer.php");
}else{
$id = $_POST['nom'];
}
$sql = "SELECT id , id_archi , nom , archithect , adresse , date_construction , dateconst , resume FROM patri where id=".$id.";";
$req = mysqli_query ($link,$sql);
$feature = array();
//echo "lon\tlat\ttitle\tdescription\ticon";
while ($row = mysqli_fetch_assoc($req)) {
$res['id_archi'] = $row['id_archi'];
$res['id'] = $row['id'];
$res['date'] = $row['dateconst'];
$res['nom'] = $row['nom'];
$res['archithect'] = $row['archithect'];
$res['adresse'] = $row['adresse'];
$res['date_construction'] = $row['date_construction'];
$res['resume'] = $row['resume'];
$feature[] = json_encode($res);
}
echo implode(', ',$feature);
header("Location: http://localhost/geopp/empl.php");
?>
but when I alert the variable cn it show me undefined and thanks for helping me.

Related

Success function in ajax not returning anything in console

Here is my code, and while running it's not giving anything in the console.
This is how I am trying to check the data. If the data correctly I want the mentioned console in success code. But if it is not then I want else code to run. But the if-else conditions are not working properly. I am including PHP code and ajax code which I have tried. Am I doing it right?
<?php
$host = "dpydaldermt01.******.com";
$username = "test";
$password = "Test";
$database_name = "test";
$conn = mysqli_connect($host, $username, $password, $database_name) or die("Connection failed: " . mysqli_error());
$sql = "select ID, user_email from ci_iwp_wp_users limit 10";
$result = mysqli_query($conn, $sql);
$users = array();
?>
<script>
(function($) {
<?php
while($row = mysqli_fetch_assoc($result)) {
$email = $row['user_email'];
?>
var mail = "<?php echo $email ?>";
$.ajax({
type:'POST',
url:'http://bluepages.ibm.com/BpHttpApisv3/wsapi?byInternetAddr='+mail,
dataType:'someData',
success: function(data) {
if(data === '# rc=0, count=0, message=Success') {
console.log(data);
}
}
});
<?php
$users[]=$row;
}
?>
});
</script>
<?php
echo json_encode($users);
?>
Just Remove dataType:'someData', from your code because it always request and response in json so you dont have to declare separately.

How to handle PHP session variables on AJAX call?

When I try to get a Session variable into a php script called with AJAX to make a MySQL query it gives me ""
I've tried passing the id using a hidden_input filled with php at the loading of my web page and the function worked! but when I try to catch the session variable directly from the script called by the AJAX it stops working again :(
This is the code where I set my php variables, I call this script directly from a html form.
<?php
if(isset($_POST['empresa']))
{
if(isset($_POST['usuario']))
{
if(isset($_POST['password']))
{
$empresa = $_POST["empresa"];
$usuario = $_POST["usuario"];
$password = $_POST["password"];
$host = "localhost";
$bd = "nominet_Directorio_Web_Beta";
$us = "nominet_Marvin2";
$pas = "NominetBD2019!";
error_reporting(0);
$con = new mysqli($host, $us, $pas, $bd);
if($con->connect_errno)
{
echo "Error de conexión al servidor de base de datos...";
exit();
}
mysqli_set_charset('utf8');
$query = "SELECT `Tbl_Usuarios`.`Id`, `Tbl_Usuarios`.`Fk_Empresa`, `Tbl_Usuarios`.`Tipo_Usuario` FROM `Tbl_Usuarios` INNER JOIN `Tbl_Empresas` ON `Tbl_Usuarios`.`Fk_Empresa` = `Tbl_Empresas`.`Id` WHERE `Tbl_Usuarios`.`Usuario` = '" . $usuario . "' AND `Tbl_Usuarios`.`Password` = '" . $password . "' AND `Tbl_Empresas`.`Razon_Social` = '" . $empresa . "'";
//$query = "SELECT `Tbl_Administradores`.`Id` FROM `Tbl_Administradores` WHERE `Tbl_Administradores`.`Usuario` = '" . $usuario . "' AND `Tbl_Administradores`.`Password` = '" . $password . "'";
$resultado = mysqli_query($con, $query);
$res= mysqli_fetch_array($resultado);
if($res["Id"] > 0)
{
session_start();
$_SESSION["Id"] = $res["Id"];
$_SESSION["Empresa"] = $res["Fk_Empresa"];
$_SESSION["Usuario"] = $usuario;
$_SESSION["Tipo_Usuario"] = $res["Tipo_Usuario"];
header("Location: ../SISTEMA/");
}
else
{
header("Location: ../?error=0");
}
}
else
{
header("Location: ../?resp=error1");
}
}
else
{
header("Location: ../?error=2");
}
}
else
{
header("Location: ../?error=3");
}
?>
The code below it's my JavaSCript (JQuery) function where I call my php script.
function contactosGeneral(){
//var empresa = $('#hidden1').val();
var funcion = "contactosGeneral";
$.ajax({
url: "/PHP/PRUEBA.PHP",
type: "POST",
data: {funcion: funcion},
error: function(xhr){
window.location.href = "../REPORTES/?resp=0";
},
success: function(respuesta) {
var arreglo = JSON.parse(respuesta);
$('#p_contactos').html(arreglo[0]["resp"]);
}
});
}
this is my php script, here I call the DataBase.
<?php
$funcion = $_POST['funcion'];
switch($funcion){
case 'contactosGeneral':
break;
}
function contactosGeneral(){
require("../../ABRIR_CON.php");
//$empresa = $_POST['empresa'];
session_start();
$empresa = $_SESSION["Empresa"];
$sql = 'SELECT COUNT(`Tbl_Contactos`.`Id`) AS "resp" FROM `Tbl_Contactos` WHERE `Tbl_Contactos`.`Fk_Empresa` = ' . $empresa;
$query = mysqli_query($con, $sql);
$json = array();
while($row = mysqli_fetch_array($query))
{
$json[]=array('resp'=>$row['resp']);
}
$resources_JSON_array = json_encode($json);
echo $resources_JSON_array;
require("../../CERRAR_CON.php");
}
?>
I know there's a lot I can improve in my code, but i'm not here for that reason, just help my with my question. thanks :)
As #Barman said, I never called my php function contactosGeneral() into my switch(){} code :B

How to display values in JS from a PHP array that was posted from jQuery

I have a button called rename that when pushed, executes in jQuery a rename.php file. Inside that php file the program selects data from a mysql, creates an array with that data, and processes an array in to json_encode($array);. How can I then get that json encoded array and echo it out into javascript?
I'm trying to echo the array out so that javascript displays my images src's.
This is my second line of ajax so I just wrote the javascript out as if it were php because I'm not sure of the commands or structure in js.
$.ajax
(
{
url:"test4.php",
type: "GET",
data: $('form').serialize(),
success:function(result)
{
/*alert(result);*/
document.getElementById("images_to_rename").innerHTML = foreach(jArray as array_values)
{
"<img src=\""array_values['original_path']"/"array_values['media']"/>";
}
}
}
);
and my jQuery php file:
<?php
include 'db/mysqli_connect.php';
$username = "slick";
if(empty($_GET['image_name']))
{
echo '<div class="refto" id="refto">image_name is empty</div>';
}
else
{
echo '<div class="refto" id="refto">image_name is not empty</div>';
foreach($_GET['image_name'] as $rowid_rename)
{
//echo '<br><p class="colourful">rowid_refto: '.$rowid_refto.'</p><br>';
$active = 1;
$command = "Rename";
$stmt = $mysqli->prepare("UPDATE ".$username." SET active=?, command=? WHERE rowid=?");
$stmt->bind_param("isi", $active, $command, $rowid_rename);
$stmt->execute();
$stmt->close();
}
//go to database, get parameters of sort
$command = "Rename";
$active = 1;
$stmt = $mysqli->prepare("SELECT original_path, media FROM " . $username . " WHERE active=? and command=?");
$stmt->bind_param("is", $active, $command);
$stmt->execute();
$result = $stmt->get_result();
while($row = $result->fetch_assoc())
{
$arri[] = $row;
}
foreach($arri as $rows) //put them into workable variables
{
$rowt = $rows['original_path'];
$rowy = $rows['media'];
//echo 'rows[\'original_path\'] = '.$rows['original_path'].''.$rows['media'].'';
}
$stmt->close();
echo json_encode($arri);
?>
<script type="text/javascript">
var jArray= <?php echo json_encode($arri); ?>;
</script>
<?php
}
echo "something2";
?>
My PHP file is a jQuery url:"test4.php", type: "GET", and is not the main file. The main file is called main.php and the test4.php is something that's called in jQuery when the user clicks on rename.
Somebody suggested console log so here's what chrome says:
<div class="refto" id="refto">image_name is not empty</div>[{"original_path":"Downloads","media":"shorter.jpg"},{"original_path":"Album 2","media":"balls.jpg"}] <script type="text/javascript">
var jArray= [{"original_path":"Downloads","media":"shorter.jpg"},{"original_path":"Album 2","media":"balls.jpg"}];
</script>
something2
Your ajax php file isn't render in browser, then the variable jArray is undefined. With your case, let return php file to json and you can get it as variable in result.
<?php
include 'db/mysqli_connect.php';
$username = "slick";
if (empty($_GET['image_name'])) {
//echo '<div class="refto" id="refto">image_name is empty</div>';
} else {
//echo '<div class="refto" id="refto">image_name is not empty</div>';
foreach ($_GET['image_name'] as $rowid_rename) {
//echo '<br><p class="colourful">rowid_refto: '.$rowid_refto.'</p><br>';
$active = 1;
$command = "Rename";
$stmt = $mysqli->prepare("UPDATE " . $username . " SET active=?, command=? WHERE rowid=?");
$stmt->bind_param("isi", $active, $command, $rowid_rename);
$stmt->execute();
$stmt->close();
}
//go to database, get parameters of sort
$command = "Rename";
$active = 1;
$stmt = $mysqli->prepare("SELECT original_path, media FROM " . $username . " WHERE active=? and command=?");
$stmt->bind_param("is", $active, $command);
$stmt->execute();
$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
$arri[] = $row;
}
foreach ($arri as $rows) { //put them into workable variables
$rowt = $rows['original_path'];
$rowy = $rows['media'];
//echo 'rows[\'original_path\'] = '.$rows['original_path'].''.$rows['media'].'';
}
$stmt->close();
echo json_encode($arri);
//stop render here
die();
}
?>
php file need return only json string. Then we'll get result as json variable in javascript.
And Js:
$.ajax
(
{
url:"test4.php",
type: "GET",
data: $('form').serialize(),
success:function(result)
{
var jArray = JSON.parse(result);
/*alert(result);*/
var txt = "";
jArray.forEach(function(array_values){
txt += `<img src=\""array_values.original_path."/"array_values.media"/>`;
})
document.getElementById("images_to_rename").innerHTML = txt;
}
}
);

Canvas will not save server side PHP

I have a canvas which I need to save to a directory and store the URL in a database.
When I save the file without storing the URL in the database it works fine, and vice versa.
However, when I put the two together and specify the PHP file through AJAX, for some reason it doesn't recognise the session variable?
When I try to call the "success" on AJAX, nothing shows up. I get no response.
This could possibly be an easy fix! I think I've been staring at this code for too long.
JavaScript:
function doodleSave() {
var canvas = document.getElementById("doodle-canvas");
var canvasData = canvas.toDataURL("image/png");
$.ajax({
url:'doodleupload.php',
type:'POST',
data:{ data:canvasData },
success: function(response){
alert(response);
//echo what the server sent back...
}
});
}
PHP:
<?php
session_start();
/* AUTOMATED VARIABLES */
$url = md5(uniqid(rand(), true));
$unique_user_id = $_SESSION['unique_user_id'];
$unique_post_id = md5(uniqid(rand(), true));
$timestamp = time();
$nature = "doodle";
$imageUrl = $upload_dir.$url.'.png';
$upload_dir = "images/external/doodles/";
$img = $_POST['data'];
$img = substr($img,strpos($img,",")+1);
$data = base64_decode($img);
$file = $upload_dir . $url . ".png";
$success = file_put_contents($file, $data);
echo $success ? $file : 'Unable to save the file.';
require_once 'php/connect.php';
try
{
$stmt = $pdo->prepare("INSERT INTO posts (unique_user_id, unique_post_id, nature, image_url, timestamp) VALUE (:unique_user_id, :unique_post_id, :nature, :image_url, :timestamp)");
$stmt->bindParam(":unique_user_id",$unique_user_id);
$stmt->bindParam(":unique_post_id",$unique_post_id);
$stmt->bindParam(":nature",$nature);
$stmt->bindParam(":image_url",$imageUrl);
$stmt->bindParam(":timestamp",$timestamp);
if($stmt->execute())
{
echo "File in database";
}
else
{
echo "Not in database";
}
}
catch(PDOException $e){
echo $e->getMessage();
}
?>
Move $upload_dir at the top, as you are calling it before you initialize it.
$upload_dir = "images/external/doodles/";
$url = md5(uniqid(rand(), true));
$unique_user_id = $_SESSION['unique_user_id'];
$unique_post_id = md5(uniqid(rand(), true));
$timestamp = time();
$nature = "doodle";
$imageUrl = $upload_dir.$url.'.png';

Export HTML5 form data to CSV?

Any ideas on Exporting HTML5 Form data to CSV please?
I can only use HTML5 or Javascript or both.
Thanks
You could just display the form data in CSV format in the browser (maybe in a popup) and have the user do a File -> Save Page (assuming it doesn't have to be fully automated).
OK, this is what I'm going to have to do: 1. If there IS internet connection ... then save the data to MySQL Table. 2. Using PHP run a script on the MySQL DB to export the data to CSV.
I'll use something like this for example:
<?php
$host = 'localhost';
$user = 'mysqlUser';
$pass = 'myUserPass';
$db = 'myDatabase';
$table = 'products_info';
$file = 'export';
$link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error());
mysql_select_db($db) or die("Can not connect.");
$result = mysql_query("SHOW COLUMNS FROM ".$table."");
$i = 0;
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_assoc($result)) {
$csv_output .= $row['Field']."; ";
$i++;
}
}
$csv_output .= "\n";
$values = mysql_query("SELECT * FROM ".$table."");
while ($rowr = mysql_fetch_row($values)) {
for ($j=0;$j<$i;$j++) {
$csv_output .= $rowr[$j]."; ";
}
$csv_output .= "\n";
}
$filename = $file."_".date("Y-m-d_H-i",time());
header("Content-type: application/vnd.ms-excel");
header("Content-disposition: csv" . date("Y-m-d") . ".csv");
header( "Content-disposition: filename=".$filename.".csv");
print $csv_output;
exit;
?>

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