This is my ajax call.
$(document).on('click','#Quote_create_value',function(){
$.ajax({
type : 'GET',
url : '../../../protected/config/ajax.php',
success : function(response){
$("#Quote_template_value").html(response);
}
});
});
I have many methods in ajax.php. Each and every method throws some response.
<?php
function respose()
{
$query = "select * from quote where template IS NOT NULL";
$result = mysql_query($query, $con);
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="'.$row['template'].'">' . $row['template'] . '</option>';
}
$query1 = "select * from template";
$data = mysql_query($query1,$con);
while ($row = mysql_fetch_assoc($data)) {
echo json_encode($row);
}
}
function result()
{
}
?>
But i want to get response from one method [ie. from response()].
How can this be done?
You could include a selector in the ajax request data. Like this for example:
$(document).on('click','#Quote_create_value',function(){
$.ajax({
type : 'GET',
url : '../../../protected/config/ajax.php',
data: "function=result",
success : function(response){
$("#Quote_template_value").html(response);
}
});
});
Then in your PHP code, a simple if-statement will check which one to output.
if(isset($_GET['function'])) {
if($_GET['result'] == 'result') {
// do result stuff
} elseif($_GET['function'] == 'response') {
// do response stuff
}
}
Related
I am using ajax call for sending value to php file and get response back into first page,(work fine), but when i am getting the result in loop form this response nothing will be return on first page.
My first page:
$.ajax({
url: "action.php",
method: "post",
data: {
'trader': selected_trader
},
dataType: 'JSON',
success: function (response) {
$('#date').html(response['date']);
$('#symbol').html(response['symbol']);
$('#status').html(response['status']);
$('#alert').html(response['alert']);
$('#company_name').html(response['company_name']);
}
})
My 2nd page:
if (isset($_POST['trader'])) {
$trader = $_POST['trader'];
$sql = "Select * from current_trader where status='$trader' ";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
$data["date"] = date('Y:m:d', strtotime($row['date']));
$data["symbol"] = $row['symbol'];
$data["company_name"] = $row['company_name'];
$data["alert"] = $row['alert'];
$data["status"] = $row['status'];
echo json_encode($data);
}
}
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 have read through dozens of similar questions on this website, and am having a lot of trouble trying to understand what is wrong with my code. I am trying to dynamically update select boxes based on the value of another one, but for some reason cannot seem to get any type of response data back once I post to PHP with Ajax.
JAVASCRIPT:
function toggleHiddenDIV()
{
var dc = document.getElementById("datacenter");
var dcVal = dc.options[dc.selectedIndex].value;
// Check if Datacenter selection has no Value selected
if(dcVal != '')
{
document.getElementById("hidden-options").style.display="block";
$.ajax({
type: "POST",
url: "handler.php",
data: { 'action_type': 'update_inventory_fields', id: dcVal },
success: function(response)
{
alert(response);
}
});
}
else
{
document.getElementById("hidden-options").style.display="none";
}
};
</script>
PHP:
if ($_POST['action_type'] == "update_inventory_fields")
{
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["id"])) { return; }
}
$result = mysql_query("SELECT id, ip, block FROM ipv4 WHERE datacenter = " . $_POST["id"]);
$data = array();
while($row = mysql_fetch_array($result, true))
{
$data[] = $row;
};
return json_encode($data);
}
Don't call return (since you're not returning a function); just echo then content onto the page:
echo json_encode($data);
Change to this...no need to return, just echo, since youre outside of a function call
if ($_POST['action_type'] == "update_inventory_fields")
{
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["id"])) { return; }
}
$result = mysql_query("SELECT id, ip, block FROM ipv4 WHERE datacenter = " . $_POST["id"]);
$data = array();
while($row = mysql_fetch_array($result, true))
{
$data[] = $row;
};
echo json_encode($data);
}
If the php code you posted is inside a function than you need to use echo functionname();
If the php code is not in a function, then just use echo json_encode($data);
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']";
Trying to pass data to the server but it keeps returning a "Parameter Missing"
So either the data is not being passed to the PHP script or I am doing something wrong.
Here is the jQuery:
function quickJob(obj) {
var quickJobNumber = $(obj).text();
//alert(quickJobNumber)
$.ajax({
type: "GET",
url: "quickJobCB.php",
data: quickJobNumber,
success: function(server_response)
{
$("#message").removeClass().html(server_response);
}
});
}
Ok....when tracing the issue I created an alert as seen below. The alert is producing the expected results.
Here is the PHP script:
<?php
require_once("models/config.php");
// Make the connection:
$dbc = #mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if (!$dbc) {
trigger_error('Could not connect to MySQL: ' . mysqli_connect_error());
}
if (isset($_GET['quickJobNumber'])) {
$quickJobNumber = trim($_GET['quickJobNumber']);
$quickJobNumber = mysqli_real_escape_string($dbc, $quickJobNumber);
$query = "SELECT * FROM projects WHERE projectNumber = '" . $quickJobNumber . "'";
$result = mysqli_query($dbc, $query);
if ($result) {
if (mysqli_affected_rows($dbc) != 0) {
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo $row['projectName'];
}
} else {
echo 'No Results for :"' . $_GET['quickJobNumber'] . '"';
}
}
} else {
echo 'Parameter Missing';
}
?>
<?php include("models/clean_up.php"); ?>
data: quickJobNumber,
should be
data: { 'quickJobNumber': quickJobNumber },
You'll need to pass the data either as a query string like so
data: "quickJobNumber="+quickJobNumber,
or a map like so
data: data { quickJobNumber: quickJobNumber },
If you want to use the GET request, use $.get
$.get("/get_request.php", { quickJobNumber: "myAjaxTestMessage"},
function(data){
console.log("WOW! Server was answer: " + data);
});
In php
<?php
if(isset($_GET['quickJobNumber'])){
header('Content-Type: application/json; charset=utf-8');
echo json_encode(array('answer'=>'Hello user!'));
}
?>
If you want to use the POST request, use $.post
$.post("/post_request.php", { quickJobNumber: "myAjaxTestMessage"},
function(data){
console.log("WOW! Server was answer: " + data);
});
In php
<?php
if(isset($_POST['quickJobNumber'])){
header('Content-Type: application/json; charset=utf-8');
echo json_encode(array('answer'=>'Hello user!'));
}
?>
P.S. or you can use $_REQUEST in php.