Using the following I have tried a hundred different ways and all I get is the failure alert
Please could anyone spot my silly beginners mistake
myy.php
$con = mysqli_connect('adress','dbase','pass','table');
$result = mysqli_query("SELECT mycoat FROM $table");
$response = array();
while($array = mysqli_fetch_row($result)) {
$response[] = $array;
}
echo json_encode($response);
JavaScript:
function myfunction()
{
$.ajax({
url: 'myy.php',
dataType: 'json',
success: function() {
alert('success');
},
error: function() {
alert('failure');
}
});
}
Try this, You have missed db connection link in mysqli_query
$result = mysqli_query($con, "SELECT mycoat FROM $table");
instead of,
$result = mysqli_query("SELECT mycoat FROM $table");
Related
Currently, I made script, which after onclick event,sending question to the database and showing data in console.log( from array ). This all works correctly, but.. I want to show data from array in the different position in my code. When I try to use DataType 'json' and then show some data, then it display in my console.log nothing. So, my question is: How to fix problem with displaying data? Is it a good idea as you see?
Below you see my current code:
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
//console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "text",
data: {id:id},
success: function(data){
console.log(data);
}
});
});
});
:
public function GetPlayer($id){
$id = $_GET['id'];
$query = "SELECT name,surname FROM zawodnik WHERE id='".$id."'";
$result = $this->db->query($query);
if ($result->num_rows>0) {
while($row = $result->fetch_assoc()){
$this->PlayerInfo[] = $row;
}
return $this->PlayerInfo;
}else {
return false;
}
}
:
$info = array();
$id = $_GET['id'];
$vv = new AddService();
foreach($vv->GetPlayer($id) as $data){
$info[0] = $data['name'];
$info[1] = $data['surname'];
}
echo json_encode($info);
I think it would be better to change the line fetch_all in mysqli to rm -rf. That information in the DB is all obsolete, or completely not true.
Try this:
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<button class="profile" data-id="1">Click</button>
<script
src="https://code.jquery.com/jquery-3.3.1.min.js"
integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8="
crossorigin="anonymous"></script>
<script>
$(document).ready(function(){
$(".profile").click(function(){
var id = $(this).data('id');
console.log(id);
$.ajax({
method: "GET",
url: "../functions/getDataFromDB.php",
dataType: "json",
data: {id:id},
success: function(data){
console.log(data);
$.each(data, function(idx, item) {
console.log(item.surname);
});
}
});
});
});
</script>
</body>
</html>
PHP side:
<?php
class AddService {
public function GetPlayer($id) {
if (filter_var($id, FILTER_VALIDATE_INT) === false) {
return false;
}
$query = "SELECT name, surname FROM zawodnik WHERE id={$id}";
$result = $this->db->query($query);
if ($result->num_rows <= 0) {
return false;
}
// assumming you are using mysqli
// return json_encode($result->fetch_all(MYSQLI_ASSOC));
// or
WHILE ($row = $result->fetch_assoc()) {
$data[] = $row;
}
return json_encode($data);
}
}
if (isset($_GET['id'])) {
$id = $_GET['id'];
$vv = new AddService();
// you don't need foreach loop to call the method
// otherwise, you are duplicating your results
echo $vv->GetPlayer($id);
}
I know that this question is already answered a lot, but even the previous responses from php are working, this response cannot work and i cannot find the reason for this issue.
Although php was send the response succesfully, ajax cannot display without refreshing the page first.
Here is my jquery.ajax code in the file helpers.js:
function likeButton(commentId, userId) {
$.ajax({
url: "requests.php",
type: "POST",
data: {
like: "likeUp",
commentId: commentId,
userId: userId
},
success: function(response) {
$("#comment_body").append(response);
}
});
}
Here is my php code in requests.php:
if(isset($_POST['like'])) {
if($_POST['like'] == "likeUp") {
$commentId = $_POST['commentId'];
$userId = $_POST['userId'];
$sql = "SELECT gaming_comment_like FROM gaming_comments WHERE gaming_comment_id='$commentId'";
$result = mysqli_query($conn, $sql);
if($row = mysqli_fetch_assoc($result)) {
$gaming_comment_like = $row['gaming_comment_like'];
}
$gaming_comment_like = $gaming_comment_like + 1;
$sql_update = "UPDATE gaming_comments SET gaming_comment_like='$gaming_comment_like' WHERE gaming_comment_id='$commentId'";
$result_update = mysqli_query($conn, $sql_update);
exit();
}
}
here is the eventhandler that calling the likeButton function, which is in a php file:
<p><img src='like.png' class='like_button' onclick='likeButton(".$gaming_comment_id.", ".$user_id.");'>$gaming_comment_like</p>";
I'm trying to convert a PHP variable to a JS variable so I can use it in a game I'm making. When I check the map code it is just undefined. Thanks in advance. FYI the PHP works.
<script>
var mapCode;
var used;
var active;
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: $mapCode,
used: $used,
active: $active,
},
dataType: "text",
});
}
</script>
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
mysqli_select_db($conn, $dbname);
// Check connection
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
// echo "Connected successfully";
$query = "SELECT mapCode FROM mapCodes";
$result = mysqli_query($conn, $query);
$mapCode = mysqli_fetch_row($result);
$query1 = "SELECT used FROM mapCodes";
$result1 = mysqli_query($conn, $query1);
$used = mysqli_fetch_row($result1);
$query2 = "SELECT active FROM mapCodes";
$result2 = mysqli_query($conn, $query2);
$active = mysqli_fetch_row($result2);
mysqli_close($conn);
?>
I understand that the PHP Code is hideous but it works and I'm going to 'pretty it up' later when the whole thing is working
If the file extension is .php and not .js then this should work
<script>
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: "<?php echo $mapCode; ?>",
used: "<?php echo $used; ?>",
active: "<?php echo $active; ?>",
},
dataType: "text",
});
}
</script>
If you have .js file then declare javascript variable before including your js in .php file
<script>
var mapCode = "<?php echo $mapCode; ?>";
var used = "<?php echo $used; ?>";
var active = "<?php echo $active; ?>";
</script>
then in .js file you will get easily
<script>
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
mapCode: mapCode,
used: used,
active: active,
},
dataType: "text",
});
}
</script>
You only need to use <?php echo $mapCode;?> instead $mapCode. .... php variables can't be reed whithout open Php tag
My current project is actually dealing with lots of ajax calls,
here is the simplified version of what I use to communicate with server:
// php
// needed functions
function JSONE(array $array)
{
$json_str = json_encode( $array, JSON_NUMERIC_CHECK );
if (json_last_error() == JSON_ERROR_NONE)
{
return $json_str;
}
throw new Exception(__FUNCTION__.': bad $array.');
}
function output_array_as_json(array $array)
{
if (headers_sent()) throw new Exception(__FUNCTION__.': headers already sent.');
header('Content-Type: application/json');
print JSONE($array);
exit();
}
// pack all data
$json_output = array(
'mapCode' => $mapCode,
'used' => $used,
'active' => $active
);
// output/exit
output_array_as_json( $json_output );
// javascript
function _fetch()
{
return $.ajax({
url: 'getMapCode.php', // url copied from yours
type: 'POST',
dataType: 'json',
success: function(data, textStatus, req){
console.log('server respond:', data);
window.mydata = data;
},
error: function(req , textStatus, errorThrown){
console.log("jqXHR["+textStatus+"]: "+errorThrown);
console.log('jqXHR.data', req.responseText);
}
});
}
window.mydata = null;
_fetch();
I have not tested this, but let me know I'll fix it for you.
How did i get you, you need to get the result from ajax request, to do it, you should first setup your php outputs your results, so the ajax can get outputed results from php like this:
<?php
ini_set('display_errors', 1);
error_reporting(E_ALL);
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
// Create connection
$conn = mysqli_connect($servername, $username, $password);
mysqli_select_db($conn, $dbname);
// Check connection
if (!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}
// echo "Connected successfully";
$query = "SELECT mapCode FROM mapCodes";
$result = mysqli_query($conn, $query);
$mapCode = mysqli_fetch_row($result);
$query1 = "SELECT used FROM mapCodes";
$result1 = mysqli_query($conn, $query1);
$used = mysqli_fetch_row($result1);
$query2 = "SELECT active FROM mapCodes";
$result2 = mysqli_query($conn, $query2);
$active = mysqli_fetch_row($result2);
mysqli_close($conn);
// Outputing results:
echo json_encode(array('mapCode'=>$mapCode[0], 'used'=>$used[0], 'active'=>$active[0]));
?>
Then in ajax, use success for listening return message after ajax finished:
<script>
var mapCode;
var used;
var active;
function downloadCode() {
$.ajax({
type: 'GET',
url: 'getMapCode.php',
data: {
/** Your data to send to server **/
},
dataType: "text",
success: function(data) { /** Here is data returned by php echo **/
var temp = $.parseJSON(data);
mapCode = temp['mapCode'];
used = temp['used'];
active = temp['active'];
}
});
}
</script>
Using Chain SELECT works great from SELECT to SELECT, I'm trying to do SELECT to INPUT.
My mainpage.php
<label>Manufacturer</label>
<select>My Select statement is here</select>
<label>Model</label>
<select name="modelname">My Select statement is fed from the select above</select>
<label>Rating</label>
<input name="rating"></input>
This is the jQuery I have in the <head> section on the mainpage.php
<script>
$(document).ready(function(){
$("select#modelname").change(function(){
var id = $("select#modelname option:selected").attr('value');
$.post("assets/configs/getdata.php", {id:id}, function(data){
$("input[name='rating']").html(data);
console.log(data);
});
});
});
</script>
and finally the getdata.php
<?php
include "db.php";
$modelid = $_POST[id];
$sql = "SELECT EfficiencyRating FROM AllModels WHERE ModelID = '$modelid' ";
$res = odbc_exec($cnn, $sql);
while ($row = odbc_fetch_array($res)) {
$row_array[] = $row['EfficiencyRating'];
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
?>
Using the console log when this message is returned, how can I fix this?
HP Warning: array_push() expects parameter 1 to be array, null given in assets\configs\getdata.php on line 12
Try with this.
$res = odbc_exec($cnn, $sql);
$return_arr = array();
while ($row = odbc_fetch_array($res)) {
$return_arr[] = $row['EfficiencyRating'];
}
echo json_encode($return_arr);
JS Part
// Slightly modify the Request
$.post("assets/configs/getdata.php", {id:id}, function(data){
// JSON Object
console.log(data);
$("input[name='rating']").val(data);
}, 'json');
You need to declare $return_arr before the while statement. Also, I personally feel what you are doing is just not right. The proper way would be this...
$res = odbc_exec($cnn, $sql);
$return_arr = array(); //<----------- Here
while ($row = odbc_fetch_array($res)) {
array_push($return_arr,$row['EfficiencyRating']);
}
echo json_encode($return_arr);
I have jquery pop form . It takes one input from the user ,mapping_key , Once the user enters the mapping key ,i make an ajax call to check if there is a user in the database with such a key.
This is my call .
Javascript:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
success: function(response) {
alert(response)
}
});
PHP:
$sql = "select first_name,last_name,user_email,company_name from registered_users where mapping_key = '$mapping_key'";
$res = mysql_query($sql);
$num_rows = mysql_num_rows($res);
if($num_rows == 0)
{
echo $num_rows;
}
else{
while($result = mysql_fetch_assoc($res))
{
print_r($result);
}
}
Now i want to loop through the returned array and add those returned values for displaying in another popup form.
Would appreciate any advice or help.
In your php, echo a json_encoded array:
$result = array();
while($row = mysql_fetch_assoc($res)) {
$result[] = $row;
}
echo json_encode($result);
In your javascript, set the $.ajax dataType property to 'json', then you will be able to loop the returned array:
$.ajax({
url : base_url+'ns/config/functions.php',
type: 'POST',
data : {"mapping_key":mapping_key} ,
dataType : 'json',
success: function(response) {
var i;
for (i in response) {
alert(response[i].yourcolumn);
}
}
});
change
data : {"mapping_key":mapping_key} ,
to
data: "mapping_key=" + mapping_key,
You have to take the posted mapping_key:
$mapping_key = $_POST['mapping_key'];
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = '$mapping_key'";
or this:
$sql = "select first_name,last_name,user_email,company_name from registered_users
where mapping_key = $_POST['mapping_key']";