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 insert data not returning response - PHP & MySQL

I am working on a scanner reader, so I used ajax when the code is read by the scanner, it should insert data to the database. The problem is the data is not inserting.
Inside the script / Ajax - query is the variable I used to get the data (name)
var query = $('#scanned-QR').val();
fetch_customer_data(query);
$(document).on('keyup', '#scanned-QR', function(){
var query = $(this).val();
fetch_customer_data(query);
});
function fetch_customer_data(query = '')
{
$.ajax({
url:"validScan.php",
method: 'GET',
data:{query:query},
dataType: 'json',
success:function(data) {
console.log(data);
if (data.status == '1') {
decoder.stop();
alert('Sucess!');
}
else if(data.status=='0'){
decoder.stop();
alert('Fail!');
}
},
error:function(err){
console.log(err);
}
});
}
My Input/Textarea
<textarea id="scanned-QR" name="scanQR" readonly></textarea>
MySQL
<?php
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
$link = mysqli_connect("localhost","root","");
mysqli_select_db($link, "schedule");
$query = $_GET['query'];
$res = mysqli_query($link,"INSERT INTO attendance (name) VALUES ('$query')");
if (mysqli_num_rows($res) > 0) {
$respose = array('status'=>'1');//1 for success
echo json_encode($respose );
} else {
$respose = array('status'=>'0');//0 for fail
echo json_encode($respose );
}
mysqli_close($link);
?>

For insert query, result will return as boolean, So mysqli_num_rows($res) won't accept boolean argument. mysqli_num_rows() expects parameter 1 to be mysqli_result
So you can simply check by below, whether it is inserted or not:
if ($res) {
$respose = array('status'=>'1');//1 for success
echo json_encode($respose);
exit;
} else {
$respose = array('status'=>'0');//0 for fail
echo json_encode($respose);
exit;
}
mysqli_close($link);

You should use exit try following code :
ini_set('display_errors', 1);
ini_set('display_startup_errors', 1);
error_reporting(E_ALL);
$link = mysqli_connect("localhost","root","");
mysqli_select_db($link, "schedule");
$query = $_GET['query'];
$res = mysqli_query($link,"INSERT INTO attendance (name) VALUES ('$query')");
if (mysqli_num_rows($res) > 0) {
$respose = array('status'=>'1');//1 for success
echo json_encode($respose );
exit;
} else {
$respose = array('status'=>'0');//0 for fail
echo json_encode($respose );
exit;
}
mysqli_close($link);
exit;

mysqli_num_rows() is for getting the number of rows returned from a SELECT query. You need to check the number of affected rows instead.
You should also be using a prepared statement, and I also recommend that you set up MySQLi to throw errors. I also prefer the object-oriented approach.
<?php
// Configure MySQLi to throw exceptions on failure
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
// Init connection
$link = new mysqli("localhost", "root", "", "schedule");
$response = [];
// Prepare the statement and execute it
$stmt = $link->prepare("INSERT INTO attendance (name) VALUES (?)");
$stmt->bind_param("s", $_GET['query']);
$stmt->execute();
// Check the number of inserted rows
if ($stmt->affected_rows) {
$response['status'] = 1;
} else {
$response['status'] = 0;
}
// Close the statement and connection
$stmt->close();
$link->close();
echo json_encode($response);

How to check successful data Submit using MySQL in a PHP file

I'm submitting a form using MySQL command inside a PHP file. I'm able to insert the data without any problem.
However, I also, at the same time, want to display the user a "Thank you message" on the same page so that he/she knows that the data has been successfully registered. On the other hand I could also display a sorry message in case of any error.
Therein lies my problem. I've written some lines in Javascript to display the message in the same page. However, I'm stuck on what (and how) should I check for success and failure.
I'm attaching my code below.
Can you please help me on this with your ideas?
Thanks
AB
HTML Form tag:
<form id="info-form" method="POST" action="form-submit.php">
form-submit.php:
<?php
require("database-connect.php");
$name = $_POST['name'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$sql = "INSERT INTO tbl_details ".
"(name,email_id,mobile_number) ".
"VALUES ".
"('$name','$email','$mobile')";
mysql_select_db('db_info');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
return false;
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
submit-logic.js:
$(function ()
{
$('form').submit(function (e)
{
e.preventDefault();
if(e.target === document.getElementById("info-form"))
{
$.ajax(
{
type:this.method,
url:this.action,
data: $('#info-form').serialize(),
dataType: 'json',
success: function(response)
{
console.log(response);
if(response.result == 'true')
{
document.getElementById("thankyou_info").style.display = "inline";
$('#please_wait_info').hide();
document.getElementById("info-form").reset();
}
else
{
document.getElementById("thankyou_info").style.display = "none";
document.getElementById("sorry_info").style.display = "inline";
$('#please_wait_info').hide();
}
}
}
)};
});
}

Per documentation: http://api.jquery.com/jquery.ajax/
dataType (default: Intelligent Guess (xml, json, script, or html))
Type: String
The type of data that you're expecting back from the server.
You are explicitly setting this to json but then returning a string. You should be returning json like you are telling the ajax script to expect.
<?php
require("database-connect.php");
$name = mysql_real_escape_string($_POST['name']);
$email = mysql_real_escape_string($_POST['email']);
$mobile = mysql_real_escape_string($_POST['mobile']);
$sql = "INSERT INTO tbl_details ".
"(name,email_id,mobile_number) ".
"VALUES ".
"('$name','$email','$mobile')";
mysql_select_db('db_info');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die(json_encode(array('result' => false, 'message' => 'Could not enter data: ' . mysql_error()));
}
echo json_encode(array('result' => true, 'message' => 'Entered data successfully'));
mysql_close($conn);
?>
I also added code to sanitize your strings, although mysql_* is deprecated and it would be better to upgrade to mysqli or PDO. Without sanitization, users can hack your database..
Nevertheless, returning json properly will ensure that your response in success: function(response) is an object, and response.result will be returned as expected, and you can use response.message to display the message where you want.

Error in $.ajax call

I have an $.ajax call in one of my pages that links to a simple php page.
I am getting my alert for the error: property. I am not getting anything back in the errorThrown variable or in the jqXHR variable. I have never done this kind of thing before and i am not seeing what is wrong with my page.
JQuery $.ajax call :
function jsonSync(json) {
$.ajax({
type: 'POST',
url: 'http://www.cubiclesandwashrooms.com/areaUpdate.php',
dataType: 'json',
data: json,
context: this,
success: function () {
},
error: function (jqXHR, textStatus, errorThrown) {
alert('Error has occured! \n ERR.INDEX: Sync failed, ' + jqXHR.responseText + ';' + textStatus + ';' + errorThrown.message);
return false;
}
});
And this is my PHP Page :
$JSON = file_get_contents('php://input');
$JSON_Data = json_decode($JSON);
//handle on specific item in JSON Object
$insc_area = $JSON_Data->{'insc_area'};
//mysqlite connection.open() equivilent
$insc_db = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if (mysqli_connect_errno($insc_db)) {
die('Could not connect: ' . mysql_error());
echo "Failed to connect to MySql: " . mysqli_connect_error();
//mysqli_close($insc_db);
}
//$insc_area.length equivilent
$insc_area_size = sizeof($insc_area);
//cycle through reult set
for ($i = 0; $i < $insc_area_size; $i++) {
//assign row to DataRow Equivilent
$rec = $insc_area[$i];
//get specific column values
$area = $rec->{'area'};
$id = $rec->{'srecid'};
//sqlcommand equivilent
$query = "SELECT * FROM insc_products WHERE id='$id' LIMIT 1";
$result = mysqli_query($insc_db, $query);
$num = mysqli_num_rows($result);
//dataReader.Read equivilent
while ($row = $result->fetch_array()) {
$query = "UPDATE insc_products SET area='$area' where id = '$id'";
$res = mysqli_query($insc_db, $query);
//checking if update was successful
if ($res) {
// good
error_log('user update done');
echo 'update was successful';
} else {
error_log('user update failed');
echo 'error in update';
}
}
}
echo 'testing php';

dataType: 'json'
means: give me json back. your PHP file isn't returning json formatted data
similar question: jQuery ajax call returns empty error if the content is empty
to buid a json response fill an array in the php file with the return information and use echo json_encode($array); at the end of the file. if you are using dataType:'json' because the code is copy/pasted, and you won't need the response to be in json format, simply remove this option...

Add following line in php file, $JSON_Data encode then it will work.
echo json_encode($JSON_Data);

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);
}