getStudents.js
$("#submit").click(function(){
$.ajax({
url: "displayStudents.php?branchCode=1",
datatype:"JSON",
success: function(obj){
for(var i=0; i<obj.length; i++){
$("ul").append(obj[i].name)
}
}
})
});
displayStudents.php
<?php
include 'config.php';
$branchCode = $_GET['branchCode'];
$list = mysqli_query($con, "Select * from student where branchCode = '$branchCode' ORDER BY rollNo");
$result = array();
while($row = mysqli_fetch_array($list)){
$result[] = array("RollNo"=>$row['rollNo'], "Name"=>$row['name']);
}
header("Content-type: application/json");
echo json_encode($result);
?>
The php file 'displayStudents.php' is sending the json correctly but I'm not able to display the names of the students.
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);
}
My jQuery Ajax code:
$(".comment").click(function(){
var id = $(this).data("id");
var name = $("#username_"+id).val();
if(name==''){
alert("Please Fill All Fields");
}else{
$.ajax({
type: "POST",
url: "comments",
data:{username:name},
dataType: "JSON",
success: function(jsonStr){
$("#cmt_output").html(JSON.stringify(jsonStr));
}
});
}
});
My Php Code:
............
$con = mysqli_connect("localhost","root","","quotes") or die("Error".mysqli_error($con));
$name = $_POST['username'];
$user = get_userdetails();
$vw_id = $user->id;
$query = mysqli_query($con,"INSERT INTO tlb_comments(user_id,comments) values ('$vw_id','$name')");
$comments = mysqli_query($con,"SELECT * FROM tlb_comments");
$avatar = mysqli_query($con,"SELECT * FROM tlb_avatar WHERE `vw_id`='".$vw_id."'");
$avatar_select = mysqli_fetch_all($avatar);
$comment_select = mysqli_fetch_all($comments);
array_push($avatar_select,$comment_select);
echo json_encode($avatar_select);
i want the function(data) to get 2 attribute value from the database
This is the select option that display the display the address and contact value of the selected option in an input field
$('#recipient').change(function(){
var FULL_NAME = $(this).val();
$.ajax({
url:"load_data.php",
method:"POST",
data:{FULL_NAME:FULL_NAME},
success:function(data){
$('#address').val(data);
$('#contact').val(data);
}
});
});
and this is the load_data.php
<?php
$sql = "SELECT * FROM recipient";
$result = mysqli_query($connect, $sql);
while($row = mysqli_fetch_array($result))
{
$output1 = $row["ADDRESS"];
$output2 = $row["CONTACT"];
$arr = array($output1,$output2);
}
echo $output1,$output2;
?>
How do i pass the $output1 into $('#address').val(data) and $output2 into $('#contact').val(data)
$('#recipient').change(function(){
var FULL_NAME = $(this).val();
$.ajax({
url:"load_data.php",
method:"POST",
dataType: "json",
data:{FULL_NAME:FULL_NAME},
success:function(data){
$('#address').val(data['ADDRESS']);
$('#contact').val(data['CONTACT']);
}
});
});
and the load_data.php
<?php
$result = mysqli_query($connect, $sql);
while($row = mysqli_fetch_array($result))
{
$arr["ADDRESS"] = $row["ADDRESS"];
$arr["CONTACT"] = $row["CONTACT"];
}
echo json_encode($arr);
?>
I have table users.
id
login
password
I want to display the data in JSON format through php mysql
page: config.php
$rep = $db->query("SELECT * FROM users");
$array_user = array();
while($data = $rep->fetch()){
$array_user = $data;
}
echo json_encode($array_user);
?>
page listUsers.php
<div id="tab"></div>
<script>
$(document).on("ready",function(){
loadData();
});
var loadData = function(){
$.ajax({
type:"POST",
url:"config.php"
}).done(function(data){
console.log(data);
var users = JSON.parse(data);
for(var i in users){
$("#tab").append(users[i].login + "<br>");
}
});
}
</script>
but it appears to me "undefined".
Use this
<?php
$rep = $db->query("SELECT * FROM users");
$array_user = array();
while($data = $rep->fetch()){
$array_user[] = $data;
}
echo json_encode($array_user);
?>
<div id="tab"></div>
<script>
$(document).on("ready",function(){
loadData();
});
var loadData = function(){
$.ajax({
type:"POST",
url:"config.php"
}).done(function(data){
console.log(data);
var users = JSON.parse(data);
for(var i in users){
$("#tab").append(users[i].login + "<br>");
}
});
}
</script>
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']";