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cannot receive json in js ajax request from php file

onclick onto a button the js starts an ajax request to the php file. The php file then gets all entries of one table of a database with php-loop. adding them to an array then the php file parse it to a JSON and echos it back to the ajax request. on success the ajax request should output an alert but neither do i get an error nor a alert.
After adding some changes according to the comments. it is now showing the error message 2 error messages randomly:
Fehler: {"readyState":0,"responseText":"","status":0,"statusText":"error"}
Fehler: {"readyState":4,"responseText":"<br />\n<b>Fatal error</b>: Uncaught Error: Cannot use object of type mysqli_result as array in C:\\xampp\\htdocs\\integration\\get.php:32\nStack trace:\n#0 C:\\xampp\\htdocs\\integration\\get.php(13): getLevel1()\n#1 {main}\n thrown in <b>C:\\xampp\\htdocs\\integration\\get.php</b> on line <b>32</b><br />\n","status":200,"statusText":"OK"}
php request (mysqli attributes are left out on purpose):
$func = $_POST['func'];
if ($func == "getLevel1"){
getLevel1();
}
$result = array();
function getLevel1(){
// Create connection
$conn = new mysqli(servername, username, password, dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, name FROM capability_level1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$result[] = '<button onclick="capability('. $row["id"] .')">' . $row["name"]. '</button></br>';
}
echo json_encode($result);
} else {
echo json_encode("0 results");
}
$conn->close();
}
js ajax call:
async function getLevel1() {
return $.ajax({
type: "POST",
dataType: "json",
url: "get.php",
data: {
func: "getLevel1"
},
success: function(data) {
alert(JSON.stringify(data));
console.log(data);
},
error: function(data) {
alert("Fehler: "+ JSON.stringify(data));
}
});
}

You need to put the json encoding when you have a complete array to be encoded: after the while:
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$result[] = '<button onclick="capability('. $row["id"] .')">' . $row["name"]. '</button></br>';
}
echo json_encode($result);
} else {
Also note that you probably have to change your data type to Json (and send a json to php) to be able to return it. Actually your Ajax is waiting for text to be returned (based on the data type)
For your further error: it is related to the point that you are fetching the rows from mysql using the wrong function. See this question for more details on how to fix it.

AJAX function is not getting the correct response from PHP?

I have written a script i JQuery and PHP,
After the success return from PHP, AJAX function should catch a success response but I am not getting that.
Below is the code:
$.ajax({
url :"script_admin-add-category.php",
method :"POST",
data :{lExpensesId:lcl_ExpensesId},
success:function(data){
//if(data=="ok"){
if(data=="YES"){
alert("EMAIL");
}else{
alert(data);
}
//}
//if(data=="ok"){
//alert("Expenses Id already exists!");
//}else{
//alert(data);
//}
}
});
and here is the php code
//Check connection
if(!$conn){
die("Connection Failed: " .mysqli_connect_error());
}else{
//echo "helloooo";
if(isset($_POST["lExpensesId"])){
$lExpensesId = $_POST["lExpensesId"];
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
if($query_result = mysqli_query($conn, $Lquery)){
if(mysqli_num_rows($query_result )){
echo 'YES';
}else{
//echo "Proceed";
}
}else{
echo "Not Okay";
}
}else{
}
}
I can see the echo value on browser and alert value also. But if condition is not working for success function???

Try set correct data type for returned data.
$.ajax({
url: 'script_admin-add-category.php',
method: 'POST',
data: {lExpensesId: lcl_ExpensesId},
dataType: 'text',
success: function (data) {
if (data === 'YES') {
alert('EMAIL')
} else {
alert(data)
}
}
})

#J Salaria as i understood your question you are having problem with jquery AJAX and PHP code as you are not getting you desired result. There are different ways to send the data through jquery ajax which i will be explain in detail.
$_POST["lExpensesId"] are you getting this ID from a HTML <form> ?.Because here i'll be showing you 3 different practiced ways to send data through ajax..
NOTE: YOUR CODE IS VULNERABLE TO SQL INJECION. I'LL BE ALSO SHOWING YOU THE METHODS TO OVERCOME.IF YOU WANT TO LEARN MORE ABOUT SQL INJECTION CLICK ON THIS LINK SQL INJECTION LINK
HTML FORM CODE :
<form action="" id="send_lExpensesId_form" method="post">
<input type="text" name="lExpensesId" id="lExpensesId" >
<input type="submit" name="submit" >
</form>
FIRST WAY FOR SENDING DATA THIS IS THOUGH HTML <FORM>
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var form_serialize = $(this).serialize();
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:form_serialize,
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
SECOND WAY FOR SENDING DATA THIS IS THOUGH HTML <FORM>
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var form_serialize = new FormData($(this)[0]);
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:form_serialize,
contentType: false,
processData: false,
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
THIRD WAY FOR SENDING DATA THIS IS USED WHEN A LINK CLICKED OR TO DELETED THROUGH ID OR CLASS
<script>
$(document).ready(function(){
$("#send_lExpensesId_form").submit(function(e){
e.preventDefault();
var lcl_ExpensesId = $("#lExpensesId").val();
$.ajax({
type:'POST',
url:'script_admin-add-category.php',
data:{lExpensesId:lcl_ExpensesId},
success:function(data){
if(data == "YES"){
alert("EMAIL");
}else{
alert(data);
}
}
});
});
});
</script>
HERE IT THE PHP CODE WITH mysqli_real_escape_string(); AGAINST SQL INJECTION
<?php
$servername = "localhost";
$username = "root";
$password = "admin";
$dbname = "demo";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST["lExpensesId"])){
$lExpensesId = mysqli_real_escape_string($conn, $_POST["lExpensesId"]);
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
if($query_result = mysqli_query($conn, $Lquery)){
if(mysqli_num_rows($query_result )){
echo 'YES';
}else{
echo "Proceed";
}
}else{
echo "Error".mysqli_connect_error();
}
}
?>
HERE IT THE OTHER PHP CODE WITH MYSQLI->PREPARED WHICH IS BETTER AGAINST SQL INJECTION
<?php
// WITH MYSQLI PREPARED STATEMENT AGAINST SQL INJECTION
$sql = $conn->stmt_init();
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId =?";
if($sql->prepare($Lquery)){
$sql->bind_param('i',$lExpensesId);
$sql->execute();
$sql->store_result();
if($sql->num_rows > 0){
echo 'YES';
}else{
echo "Proceed";
}
}
else
{
echo "Error".mysqli_connect_error();
}
?>
I HOPE YOU GOT ANSWERE FOR YOU QUESTION IF YOU HAVE OTHER DOUBTS FEEL FREE AND COMMENT BELOW

All methods are known and many thanks for assistance. My question is that Why I am not able to get proper return from PHP. Below is my code:
var lcl_ExpensesId = $("#IExpensesId").val();
//alert(lcl_ExpensesId);
$.ajax({
url :"script_admin-add-category.php",
method :"POST",
data :{lExpensesId:lcl_ExpensesId},
success:function(data){
if(data=="ok"){
alert("Inserted");
}else{
alert(data);
}
}
});
ob_start();
/------------------FUNCTION TO READ ACCOUNTS DROPDOWN EXPENSES LIST -----------------------/
require_once 'db_config.php';
$newlist = fxn_CONFIGURATION();
$HOST = $newlist[0];
$DBNAME = $newlist[1];
$UNAME = $newlist[2];
$PSWD = $newlist[3];
$conn = mysqli_connect($HOST, $UNAME, $PSWD, $DBNAME);
//Check connection
if(!$conn){
die("Connection Failed: " .mysqli_connect_error());
}else{
if(isset($_POST["lExpensesId"])){
$lExpensesId = $_POST["lExpensesId"];
$Lquery = "SELECT ExpensesId FROM tblexpensestype WHERE ExpensesId = '$lExpensesId'";
$query_result = mysqli_query($conn, $Lquery);
if(mysqli_num_rows($query_result) > 0){
echo "ok";
}else{
echo "Proceed";
}
}
}
mysqli_close($conn);
ob_flush();
As, i am using this AJAX in my one of input keyup method so whatever I will type, each and everytime, it will execute PHP script. I am having a item as FOOD in databse. When i type "F", I got Proceed, "O" - Proceed, "O" - Proceed, "D" - ok....
When I type D, i should get "Inserted" instead of Ok....
This is my doubt that why i m getting this????

The above problem is resolved by using exit() statement in PHP as I am getting five ↵↵↵↵↵ after my values and it means I am having 5 lines of html without closing ?>. So the best way to resolve the issue is to use exit() in PHP as per need

Pass a variabile from php to js via JSON

At the lessons i learn how to pass result from a sql request to js via JSON/AJAX. I need the value of the row from this request in my js but it doesnt work. Via console i have an error: Uncaught SyntaxError: Unexpected token <
part of PHP:
<?php
//get all the course from db and reply using json structure
//connection to db
$mysqli = new mysqli("localhost", "root", "", "my_hyp");
$id = $_POST['id'];
if (mysqli_connect_errno()) { //verify connection
exit(); //do nothing else
}
else {
# extract results mysqli_result::fetch_array
$query = " SELECT * FROM course WHERE course_category='$id'";
//query execution
$result = $mysqli->query($query);
//if there are data available
if($result->num_rows >0)
{
$myArray = array();//create an array
while($row = $result->fetch_array(MYSQL_ASSOC)) {
$myArray[] = array_map('utf8_encode', $row);
}
$response = array();
$response['rows'] = $row;
$response['query'] = $myArray;
echo json_encode($response);
}
//free result
$a=num_rows;
$result,$a->close();
//close connection
$mysqli->close();
}
?>
first part of Script:
$.ajax({
method: "POST",
//dataType: "json", //type of data
crossDomain: true, //localhost purposes
url: "./query/cate_has_courses.php", //Relative or absolute path to file.php file
data: {id: i},
success: function(response) {
console.log(JSON.parse(response));
var course=JSON.parse(response.query);
var row=JSON.parse(response.rows);

Seem you use JSON.parse in wrong way.
The JSON.parse must be done one time only. The result of JSON.parse is store in course the the access to the data is due by response.query or respons.row .. and so on and not by JSON.parse(respose.query)

d3.js How to pass parameter to php for query condition

I attempt to combine d3, mysql php Tutorial.
I want to use mysql to store data, and use d3 table to display the result.
Following the tutorial I successfully connected the sql, and display it.
However, in my example, the where condition of sql in queryData.php is hard encoded.
As show below: WHERE pathwayID='1643685' && symbol='VIF'
I need to pass the parameter '1643685' and 'VIF' from d3 file to php file, how should I do?
And how should I modify queryData.php, thanks.
d3 file
d3.json("queryData.php", function(error, jsonData) {
....
});
queryData.php
<?php
error_reporting(E_ALL ^ E_DEPRECATED);
// load in mysql server configuration (connection string, user/pw, etc)
include 'mysqlConfig.php';
// connect to the database
#mysql_select_db($database) or die( "Unable to select database");
//Query
$myquery = "
SELECT `pathwayID`, `proteinID`, `uniprotID`, `symbol`, `displaySymbol`, `reactomeID`, `cellularLocation` FROM `protein` WHERE pathwayID='1643685' && symbol='VIF'
";
$result = mysql_query($myquery);
if ( ! $result ) {
echo mysql_error();
die;
}
$data = array();
for ($x = 0; $x < mysql_num_rows($result); $x++) {
$data[] = mysql_fetch_assoc($result);
}
echo json_encode($data);
mysql_close();
?>
mysqlConfig.php
<?php
error_reporting(E_ALL ^ E_DEPRECATED);
$username="root"; //replace with your mySql username
$password=""; //replace with your mySql password
$database="pathway"; //replace with your mySql database name
$host="localhost"; //replace with the name of the machine your mySql runs on
$connection=mysql_connect($host,$username,$password);
?>

Finally, I solved this by using ajax to post parameter.
$.ajax({
url: "./php/querybyPathwayId.php",
type: "GET",
data: {
pathwaydbId: dbId
},
dataType: "json",
success: function (jsonData) {
operation(jsonData);
},
error: function () {
}
});
and modified the querycentance
$pathwayId = $_GET["pathwaydbId"];
$myquery = "
SELECT `pathwayID`, `proteinID`, `uniprotID`, `symbol`, `displaySymbol`,
`reactomeID`, `cellularLocation` FROM `protein` WHERE pathwayID='$pathwayId'
";

Using ajax to display new database inputs without refreshing the page

I am using ajax to post comments to a certain page, I have everything working, except for when the user posts a comment I would like it to show immediately without refreshing. The php code I have to display the comments is:
<?php
require('connect.php');
$query = "select * \n"
. " from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '$s_post_id' ORDER BY comments.id DESC";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)) {
$c_comment_by = $row['comment_by'];
$c_comment_content = $row['comment_content'];
?>
<div class="comment_box">
<p><?php echo $c_comment_by;?></p>
<p><?php echo $c_comment_content;?></p>
</div>
<?php } ?>
</div>
</div>
<?php
}
}
and the code I have to post comments is:
<?php
$post_comment = $_POST['p_post_comment'];
$post_id = $_POST['p_post_id'];
$post_comment_by = "Undefined";
if ($post_comment){
if(require('connect.php')){
mysql_query("INSERT INTO comments VALUES (
'',
'$post_id',
'$post_comment_by',
'$post_comment'
)");
echo " <script>$('#post_form')[0].reset();</script>";
echo "success!";
mysql_close();
}else echo "Could no connect to the database!";
}
else echo "You cannot post empty comments!"
?>
JS:
function post(){
var post_comment = $('#comment').val();
$.post('comment_parser.php', {p_post_comment:post_comment,p_post_id:<?php echo $post_id;?>},
function(data)
{
$('#result').html(data);
});
}
This is what I have for the refresh so far:
$(document).ready(function() {
$.ajaxSetup({ cache: false });
setInterval(function() {
$('.comment_box').load('blogpost.php');
}, 3000);.
});
Now what I want to do is to use ajax to refresh the comments every time a new one is added. Without refreshing the whole page, ofcourse. What am I doing wrong?

You'll need to restructure to an endpoint structure. You'll have a file called "get_comments.php" that returns the newest comments in JSON, then call some JS like this:
function load_comments(){
$.ajax({
url: "API/get_comments.php",
data: {post_id: post_id, page: 0, limit: 0}, // If you want to do pagination eventually.
dataType: 'json',
success: function(response){
$('#all_comments').html(''); // Clears all HTML
// Insert each comment
response.forEach(function(comment){
var new_comment = "<div class="comment_box"><p>"+comment.comment_by+"</p><p>"+comment.comment_content+"</p></div>";
$('#all_comments').append(new_comment);
}
})
};
}
Make sure post_id is declared globally somewhere i.e.
<head>
<script>
var post_id = "<?= $s_post_id ; ?>";
</script>
</head>
Your new PHP file would look like this:
require('connect.php');
$query = "select * from comments inner join blogposts on comments.comment_post_id = blogposts.id WHERE blogposts.id = '".$_REQUEST['post_id']."' ORDER BY comments.id DESC";
$result = mysql_query($query);
$all_comments = array() ;
while ($row = mysql_fetch_array($result))
$all_comments[] = array("comment_by" => $result[comment_by], "comment_content" => $result[comment_content]);
echo json_encode($all_comments);
Of course you'd want to follow good practices everywhere, probably using a template for both server & client side HTML creation, never write MySQL queries like you've written (or that I wrote for you). Use MySQLi, or PDO! Think about what would happen if $s_post_id was somehow equal to 5' OR '1'='1 This would just return every comment.. but what if this was done in a DELETE_COMMENT function, and someone wiped your comment table out completely?