Develop Reference

Convert a Ajax Json String to a JS Object for my Ajax Get Script

I ideally want the Ajax result to be converted from Jsonstring to OBJ Thank You in advance.
I know the AJAX GET script is working becuase when I alert the Ajax Post result I see the Contents in json string format as below.
alert(JSON.stringify(data));
[{"id":"1","username":"jiten","name":"Jitensingh\t","email":"jiten93mail”},{“id":"2","username":"kuldeep","name":"Kuldeep","email":"kuldeemail”}]
I want the AJAX GET result data converted to look like this in OBJ format like below.
{id:31,name:"Mary",username:"R8344",email:"wemail}];
PHP/SQL CODE with the Json encoded Array
<?php
include "../mytest/config.php";
$return_arr = array();
$sql = "SELECT * FROM users ORDER BY NAME";
$result = $conn->query($sql);
//Check database connection first
if ($conn->query($sql) === FALSE) {
echo 'database connection failed';
die();
} else {
while($row = $result->fetch_array()) {
$id = $row['id'];
$username = $row['username'];
$name = $row['name'];
$email = $row['email'];
$return_arr[] = array(
"id" => $id,
"username" => $username,
"name" => $name,
"email" => $email);
}
// Encoding array in JSON format
echo json_encode($return_arr);
}
?>
php echo _encode array above returns below Json string format
[{"id":"1","username":"jiten","name":"Jitensingh\t","email":"jiten93mail”},{“id":"2","username":"kuldeep","name":"Kuldeep","email":"kuldeemail”}]
I am looking for something like below.( top half of the script)
<script>
$(document).ready(function(){
$.ajax({
url: 'ajaxfile.php',
type: 'get',
dataType: 'JSON',
success: function(result){
var data =(JSONstring convert to OBJ(result);
//-----The top half of script -------------
$.each(data, function( i, person ) {
if(i == 0) {
$('.card').find('.person_id').text(person.id);
$('.card').find('.person_name').text(person.name);
$('.card').find('.person_username').text(person.username);
$('.card').find('.person_email').text(person.email);
} else {
var personDetailCloned = $('.card').first().clone();
personDetailCloned.find('.person_id').text(person.id);
personDetailCloned.find('.person_name').text(person.name);
personDetailCloned.find('.person_username').text(person.username);
personDetailCloned.find('.person_email').text(person.email);
$('.card-container').append(personDetailCloned);
}
});
});
</script>
I will need help with the closing tags as above is just an example

The solution is:
success: function(result){
data =(result);
There was no need o convert the data to OBJ or anything ( blush). Then the code on the 2nd half of the Ajax script will receive the data and populate. Thanks to all contributors.

Ajax request : get text from database and display in div

My goal is to perform an AJAX request when clicking on a button to retrieve "name" and "story" stored in my database. Each button will get info of another hero.
I'm working on multiple files.
With my current code (which is the closer to what seems to be correct in my mind) the switchHeroInfo will always change the text to "TestName" and "StoryName" instead of "Gertrude" "An old lady"(stored in database).
Can you enlight me on what may be the cause of my struggles?
the php file for connecting to database : connect_database.php
<?php
try
{
$bdd = new PDO('mysql:host=localhost;dbname=biomass;charset=utf8', 'root', '', array(PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION));
}
catch(Exception $e)
{
die('Error : '.$e->getMessage());
}
?>
The Javascript part :
$(document).ready(function()
{
$(".hero_portrait").click(function()
{
var index = $(this).data("id");
$.ajax(
{
type: "POST",
url: "../php/get_data.php",
data: {newIndex:index},
success: function(data)
{
// Display {"nick":"Gertrude","0":"Gertrude","story":"Vieille folle senile","1":"Vieille folle senile"}
alert(data);
//Display : undefined
alert(data.story);
$("#hero_name").html(data.nick);
$("#hero_story").html(data.story);
},
error: function()
{
alert("Request failure");
}
});
});
});
The php file : get_data.php
<?php
$tempValue = $_POST['newIndex'];
$sql = $bdd->prepare('SELECT * FROM heroes WHERE ID = :indexValue');
$sql->bindParam(":indexValue", $tempValue, PDO::PARAM_STR);
$sql->execute();
while($data = $sql->fetch())
{
?>
<script>
$heroNameTemp = <?php echo json_encode($data["name"]); ?>;
$heroStoryTemp = <?php echo json_encode($data["story"]); ?>;
</script>
<?php
}
$sql->closeCursor();
?>
Finally the HTML relative to the current problem:
<div id="squad_portraits">
<div class="hero_portrait" id="1"></div>
<div class="hero_portrait" id="2"></div>
<div class="hero_portrait" id="3"></div>
<div class="hero_portrait" id="4"></div>
</div>
<div id="hero_info">
<h2 id="hero_name">Hero_Name</h2>
<p id="hero_story"> Hero_Description</p>
</div>
If i switch my sql request :
$tempValue = $_POST['newIndex'];
$sql = $bdd->prepare('SELECT * FROM heroes WHERE ID = :indexValue');
to this
$tempValue = 4;
$sql = $bdd->prepare('SELECT * FROM heroes WHERE ID = 4');
AND add the following to my HTML file
<?php include("../php/get_data.php"); ?>
everything works but my index will always be "4".

There are several issues and missunderstandings in your code.
first, in ajax change data to this:
data: {newIndex:index}, // remove the (), simple syntax misstake
This should solve the sql problem.
Now the get_data.php:
<?php
// including db connection is missing here for $bdd
// You should add a test here, wether you've received any and correct data in 'newIndex'
if(empty($_POST['newIndex']) {
// throw an error, send that back to ajax (with a fitting http response code) and exit script
http_response_code(400);
$error = new stdClass();
$error->msg = "Parameter newIndex was missing";
header('Content-Type: application/json'); // tell the browser that there is some json coming!
echo json_encode($error);
exit;
}
$tempValue = $_POST['newIndex'];
// only select the values you need (name, story)
$sql = $bdd->prepare('SELECT name, story FROM heroes WHERE ID = :indexValue');
$sql->bindParam(":indexValue", $tempValue, PDO::PARAM_STR);
$sql->execute();
$data = $sql->fetch(); // if you only expect/need one row, no while is needed
// echo ONE json string as plain string:
header('Content-Type: application/json'); // tell the browser that there is some json coming!
echo json_encode($data);
Now you receive content of $data as a json as param (data) in the success-callback of your ajax, which you can use like this:
success: function(data){
$("#hero_name").html(data.name);
$("#hero_story").html(data.story);
}
Finally let's change storing the item's id from the id attribute to the data-attribute:
In html:
<div class="hero_portrait" data-id="1"></div>
<div class="hero_portrait" data-id="2"></div>
and in javascript change
var index = $(this).attr("id");
to
var index = $(this).data("id"); // $(this).attr("data-id"); will also work

How to store PHP array in MySQL?

I tried to store a JavaScript array into MySQL table using PHP, see the following script below.
I first converted the string using JSON.stringify and passed it into a PHP file via AJAX request.
I then converted it to a PHP array, and after after I inserted those arrays using serialize();.
Finally, it properly stored using localhost but it does not work on my live server.
sample.Js
$.ajax(
{
type: "POST",
url: "save_plan_ajax.php",
data: {plan: plan,totalInvesment: totalInvesment, location: JSON.stringify(locationsArr), cost: JSON.stringify(cost), personalised: JSON.stringify(personalised)},
success: function(data){
//alert(data);
}
}
)
In this above script I passed the JavaScript array into save_plan_ajax.php using JSON.stringify.
save_plan_ajax.php
<?php
session_start();
include 'config.php';
if(isset($_POST)){
$planname = $_POST['plan'];
$cost = json_decode($_POST['cost'], true);
$personalised = json_decode($_POST['personalised'], true);
$locations = json_decode($_POST['location'], true);
$userid = $_SESSION['BIID'];
$constriant = $_SESSION['CONSTRAINT'];
$created_date = date("Y-m-d H:i:s");
$total_invs = $_POST['totalInvesment'];
$query = "INSERT INTO `plans` (
`refid` ,
`userid` ,
`plan_name` ,
`cost` ,
`locations`,
`personalised` ,
`total_invs`,
`constriant` ,
`created_date` ,
`stat`
)
VALUES (
NULL , '$userid', '$planname', '".serialize($cost) ."', '".serialize($locations)."','".serialize($personalised)."', '$total_invs', '$constriant', '$created_date', 'A'
)";
$res = $GLOBALS['Db']->Insert($query);
if($res){
echo $res;
}
else{
echo "Error";
}
}
?>
In the above script, the record stores correctly at local server, but in this same script insert N; in server.. how do I fix this error, is the above way correct?
In the MySQL database table I have set the cost, location and personalized fields datatype as LONGTEXT.

Using AJAX to send form information to another page using a button

Hello I have two files that are supposed to be connected to one another. I want to send an AJAX request to another page that uses a sql query to send form information.
The application that I'm trying to create is a questionnaire with eight questions, each questions has four answers paired together with the same id (qid) and each answer has a value from the database. After you answer eight questions you will see a button that sends an AJAX request to the page test.php, (named submitAJAX).
The problem is that although my connection with AJAX is working, the values from the form are not being sent to my database. Previously I thought that the problem may lie with the form page, but now I I think the problem lies in this file:
test.php (file with json)
<?php
$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) or die ("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');
if(count($_GET) > 0){
$answerPoint = intval($_GET['radiobtn']);
$qid = intval($_GET['qid']);
$tid = intval($_GET['tid']);
$sql2 = "INSERT INTO result (qid, points, tid) VALUES ($qid, $answerPoint, $tid)";
$connect->query($sql2);
$lastid = $connect->insert_id;
if($lastid>0) {
echo json_encode(array('status'=>1));
}
else{
echo json_encode(array('status'=>0));
}
}
?>
I think that the problem may lie in the row where: if($lastid>0) {
$lastid should always be more than 0, but whenever I check test.php I get this message: {"status":0} What's intended is that I get this message: {"status":1}
<html>
<head>
<meta charset="utf-8">
</head>
<body>
<?php
$localhost = "localhost";
$username = "root";
$password = "";
$connect = mysqli_connect($localhost, $username, $password) or die ("Kunde inte koppla");
mysqli_select_db($connect, 'wildfire');
$qid = 1;
if(count($_POST) > 0){
$qid = intval($_POST['qid'])+1;
}
?>
<form method="post" action="">
<input type="hidden" name="qid" id="qid" value="<?=$qid?>">
<?php
$sql1 = mysqli_query($connect,"SELECT * FROM question where answer != '' && qid =".intval($qid));
while($row1=mysqli_fetch_assoc($sql1)){
?>
<input type='radio' name='answer1' class="radiobtn" value="<?php echo $row1['Point'];?>">
<input type='hidden' name='tid' class="tid" value="<?php echo $row1['tid'];?>">
<?php echo $row1['answer'];?><br>
<?php
}
?>
<?php if ($qid <= 8) { ?>
<button type="button" onclick="history.back();">Tillbaka</button>
<button type="submit">Nästa</button>
<?php } else { ?>
<button id="submitAjax" type="submit">Avsluta provet</button>
<?php } ?>
</form>
<script src="https://code.jquery.com/jquery-1.11.3.js"></script>
<script type="text/javascript">
function goBack() {
window.history.go(-1);
}
$(document).ready(function(){
$("#submitAjax").click(function(){
if($('.radiobtn').is(':checked')) {
var radiobtn = $('.radiobtn:checked').val();
var qid = $('#qid').val();
var answer = $('input[name=answer1]:radio').val();
$.ajax(
{
type: "GET",
url: 'test.php',
dataType: "json",
data: "radiobtn="+radiobtn+"&qid="+qid,
success: function (response) {
if(response.status == true){
alert('points added');
}
else{
alert('points not added');
}
}
});
return false;
}
});
});
</script>
</body>
The values that I want to send to my database from test.php are:
qid(int), tid(int), Point(int)
There is a database connection, and my test.php file's sql query should work, but its not sending form information. Is there something that I need to rewrite or fix to make it work?

First, your data parameter in the AJAX call is not using the correct syntax. You're missing brackets. It should look like:
data: JSON.stringify({ radiobtn: radiobtn, qid: qid }),
Second, I'd suggest using POST instead of GET:
type: "POST",
which means that you need to look for your data in $_POST['radiobtn'] and $_POST['qid'] on test.php. NOTE: you should check for the key you expect using isset() before assigning the value to a variable, like so:
$myBtn = isset($_POST['radiobtn']) ? $_POST['radiobtn'] : null;
Third, for testing, use a console.log() inside your condition that checks for the checkbox being checked in order to verify that condition is working as expected.
if($('.radiobtn').is(':checked')) {
console.log('here');
UPDATE:
Fourth: You should specify the content type in your AJAX call, like so:
contentType: "application/json; charset=utf-8",

After you execute your query that inserts the result you can use a sql statement to select the last insert id. Try something like
$sql2 = "INSERT INTO result (qid, points, tid) VALUES ($qid, $answerPoint, $tid)";
$connect->query($sql2);
$result = $connect->query("SELECT LAST_INSERT_ID()");
$row = $result->fetch_row();
$lastid = $row[0];
That should return the correct last insert id, if that was where your error was occurring.

mysqli_insert_id() returns the ID generated by a query on a table with a column having the AUTO_INCREMENT attribute.
In your SQL, you are providing the ID yourself, there is no auto-increment. So you should get 0 from $connect->insert_id, because the function returns zero if there was no previous query on the connection or if the query did not update an AUTO_INCREMENT value.
For your purpose, you can use the return value of mysqli_query() instead, which returns TRUE on success and FALSE on failure.
if($connect->query($sql2)) {
echo json_encode(array('status'=>1));
}
else{
echo json_encode(array('status'=>0));
}

ajax call to php to get database rows

The following is in my file.js
function mainget(){
$.ajax({
type: 'GET',
url: 'example.php',
data:json,
success:function(data){
}
});
}
example.php
<?php
$con = mysqli_connect('address','DATBASE','pass','futureday');
$result = mysql_query("SELECT * FROM $futureday");
$array = mysql_fetch_row($result);
echo json_encode($array);
?>
I have been struck with this for the past 2 days. I have tried inserting alert as first line of function mainget , which is successful, but after that I get nothing.

You are using data property in AJAX call to indicate the json data type. It is an invalid one. Use dataType to provide the data type. data property is used to pass the datas. And also put quotes to the values like:
dataType:'json'
Also change your example.php file. There you are using mysqli_connect to connect the database, then mysql_* to execute and fetch operations. It is not correct. Use either mysqli_* or mysql_*. Edit as:
<?php
$con = mysqli_connect('address','DATBASE','pass','futureday');
$result = mysqli_query("SELECT * FROM $futureday");
$response = array();
while($array = mysqli_fetch_row($result)){
$response[]=$array;
}
echo json_encode($response);
?>

Use this
$mysqli = new mysqli('address','DATBASE','pass','futureday');
$query = "SELECT * FROM $futureday";
$results=$mysqli->query($query) ;
$res=$mysqli->fetch_array(MYSQLI_ASSOC);
echo json_encode($res);