i have a javascript code that requests a php page to provide it with list of names that are currently online and update a Table, but i have a problem sending it back in form of an array, someone told me that this is usually done using XML, but i dont know how to start.
javascript Post method:-
$.post( "updateTable.php", POSTdata,
function( data ) {
$("#mytable").last().append('<tr><td>'+data+'</td></tr>');
}
);
the php file:-
include("connect.php");
$query1 = "SELECT * FROM formtable";
$result_id = mysql_query($query1, $global_dbh)
or die ("display_db_query:" . mysql_error());
while ($table_array = mysql_fetch_object ($result_id))
{
$rows[] = $table_array;
}
foreach ($rows as $temp ) {
if ($temp->isOnline==1)
$newRow[] = $temp->name;
}
echo "$newRow";
mysql_close($global_dbh);
Please excuse any syntax or semantics in my code, i am a beginner.
How can i populate my table using ajax callback function, and in what form the data will arrive there, and how can i use xml to help me.
Many thanks in advance.
A quick example of json:
var table = $("#mytable").last();
$.ajax({
type: 'post',
url: "updateTable.php",
dataType: 'json',
data: POSTdata,
success: function(data){
jQuery.each(data, function(i, row){
//console.log(row);
table.append('<tr><td>'+row.name+'</td></tr>');
});
}
});
and in php file, instead of :
echo "$newRow";
replace with:
echo json_encode($newRow);
That's it!
Related
I'm currently having some trouble posting a variable on jQuery for MySQLi SELECT on a PHP page.
Code of jQuery:
$("#carta1").click(function()
{
cartaId = document.getElementById("carta1").value;
console.log(cartaId);
ajaxGetResults = $.ajax({
context: this,
type: "POST",
url: "darResposta.php",
data: {'cartaId' : cartaId},
cache: false,
dataType: "json"
})
.done(function(data){
$('#3').html(data);
console.log("Avançou para a terceira parte");
$("#2").hide();
$("#3").show();
})
.fail(function(){
console.log('Erro ao buscar dados');
$("#2").hide();
$("#3").show();
$('#3').html("Deu erro");
});
});
Code of PHP:
if(!$conn)
{
echo "Falhou a ligação à base de dados";
}
else
{
if(isset($_POST['cartaId']))
{
$cartaId = $_POST['cartaId'];
$res = mysqli_query($conn,"
SELECT cartaNome, cartaDescricao
FROM tarot_cartas
WHERE cartaId = ".$cartaId
);
$data = array();
while($row = mysqli_fetch_assoc($res))
{
$data=$row;
}
echo json_encode($data);
}
}
Tried several approaches to this problem such as putting the $cartaId outside the if statement with a direct $_POST, and nothing happened.
Would appreciate if you could shed some light on this problem.
Thanks for taking the time to read and suggest a solution.
use below code
data: { 'cartaId' : cartaId },
instead of
data: {"data":JSON.stringify({'cartaId' : JSON.stringify(this)})},
Solution:
1.Remove this
data: {"data":JSON.stringify({'cartaId' : JSON.stringify(this)})},
2.Replace This one
data: { cartaId: cartaId }
Hope it works....
The jq you have will post a variable to the url.
To debug you should first check in a console (i use firebug) for mozilla) if the request is sent. Using firebug you can see the names of the POST variables you send.
Following this you should check what values get received on the server side by doing
var_dump($_POST);
Finally get the correct variable into your query. You can also debug the query by viewing the log file or, depending on whether you are using a framework, something like CI:
db->last_query();
I am currently migrating an already built web application to MVC, and I'm figuring out that I'm too newbie to do some kind of changes. There are some ajax calls that are freaking me out. I'll try to be as clear as possible, but due to my inexperience I'm not sure if I won't let some important information by the way.
The point is in the old application, things go this way:
In the php code:
if ($action_user == 'show_alerts') {
$list = array();
$query = "SELECT alert_type FROM alert_contact WHERE NOT
deleted AND user_email=" . typeFormat($email);
$result = mysqli_query($db, $query) or die('Error in query "'.$query . '": ' . mysqli_error($db));
while ($db_field = mysqli_fetch_assoc($result)) {
$list[] = $db_field['alert_type'];
}
echo json_encode($list);
In the jquery code:
$.ajax({
type: 'POST',
url: 'userpost.php',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json'
Here comes my problem, and since I don't have an userpost.php file anymore, I have to send it to the index.php and call my users component by a get petition, which I don't like, but I coudn't find another way to do it. And, what is even worse, I don't know at all how ajax is getting the variables that it needs. It must be a pretty basic mistake, but I recognize my skills at this point are't so good. That's what I'm doing in my version:
In the php code:
if ($action_user == 'show_alerts') {
$list = ModelUser::getAlertContact($act_email);
echo json_encode($list);//I predict that ajax don't reach this line, but not sure
}
In the jquery code:
$.ajax({
type: 'POST',
url: 'index.php?option=users',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json',
success: function(data) {
alert ('gotcha');
$.each(alertsarray, function(index, value) {
if ($.inArray(value, data) === -1) {
$("#sub" + value).prop("checked", false);
$('#alert' + value).removeClass("list_alert_sub");
}
else {
$("#sub" + value).prop("checked", true);
$('#alert' + value).addClass("list_alert_sub");
}
});
},
error: function(data) {
alert("¡Error (ajax)!");
}
});
Any help would be appreciated, and if there's some more information I've missed, please let me know. Thanks in advance.
UPDATE:
I've been making some progress but don't seem to find a real solution. Now I know that the url has to be the controller, so I'm using 'components/userpost/controller.php' as it, and it reaches the ajax call, cause the success alert is showing up. The problem is the MVC way, because I send ajax to the controller, but since I don't have a reload in the page, all the includes are failing so they are obviously not being loaded, and I'm getting errors like this:
PHP Warning: include(components/userpost/model.php): failed to open
stream: No such file or directory in
/var/www/html/viewer_mvc/components/userpost/controller.php on line 3,
referer: http://localhost/viewer_mvc/index.php
Really hope you guys can show me where am I failing, and if there's a special way to do these thing in MVC.
For the JQuery call it makes a POST request to index.php?option=users with JSON data. The form with the ID userForm is serialized using the Jquery serialize method.
The .serialize() method creates a text string in standard URL-encoded notation. It can act on a jQuery object that has selected individual form controls
$.ajax({
type: 'POST',
url: 'index.php?option=users',
data: $('#userForm').serialize(),
cache: false,
dataType: 'json'
Now for your PHP sample
if ($action_user == 'show_alerts') {
$list = ModelUser::getAlertContact($act_email);
echo json_encode($list);//I predict that ajax don't reach this line, but not sure
}
This code will be looking for variables that probably don't exist anymore if it is a different file i.e. is there an $action_user variable?
To start reimplementing it you will need to add the logic so that it checks the POST variable if your not using the framework code. So if you have a form element with the name 'name' then that will be available in your PHP script POST variable
$_POST['name']
[How to call a PHP function in MVC using AJAX]
$.ajax({
type: 'POST',
url: 'save-user.php',
data: { fname: "manish", email: "manishkp#com", role:"admin"},
success: function(data) {
console.log(data);
if(data == 'error')
{
$('#Register_error').text('Must Be filled...');
$('#Register_error').show();
}
else {
$('#Register_error').hide();
$('#Register_success').text('Successfully submit');
$('#Register_success').show();
}
}
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<?php
$fname = $_POST['fname'];
$email = $_POST['email'];
$role = $_POST['role'];
if(!empty($fname) && !empty($email) && !empty($role))
{
#MYSQL CONNECTION QUERY #
echo"success";
}
else{
echo "error";
}
?>
I really have never done this before and I am getting frustrated because I'm not sure how it fits together. I have a function that I want to call my php (one php file selects info from a database and the second inserts into the database)... I need to use ajax in the way my site is setup but I don't know how to pass data from and to the php files.
In first .js file:
q1LoadVar();
This is my ajax function in second .js file that I have so far (not working):
//ajax code has been edited here since original post:
function q1LoadVar() {
alert("called"); //works!
$.get( "q1LoadVar1.php", function( data ) {
console.log(data); //nothing happens!
// alert(data); //nothing happens!
}, "json" );
}
And here is the code I have in q1LoadVar1.php that I want to select data back from and be able to populate a text area in my html:
/*works when I type this file path directly into the url;
but the file is not communicating back to the ajax function on the
.js file that is calling it*/
<?php
$config = parse_ini_file('../config.ini');
$link = mysqli_connect('localhost',$config['username'],$config['password'],$config['dbname']);
if(mysqli_connect_errno()){
echo mysqli_connect_error();
}
echo '<script type="text/javascript">alert("working from php!");</script>';
$query = "SELECT * FROM Game1_RollarCoaster";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result)) {
$newRow[] = $row;
}
$json = json_encode($newRow);
echo $json; //works on php file directly!
/*while ($row = mysqli_fetch_array($result)) {
echo $row[Q1_AnswerChoosen];
}*/
mysqli_free_result($result);
mysqli_close($link);
?>
Can someone help me understand how to make this all work together? Thank you, Kristen
You can retrieve post data from ajax in php with
$_POST['action']
//in your case will return: test
To return data to ajax you need to use echo
If the success: callback function doesnt get called try to remove datatype: 'json'
I also think that you need to echo $newrow instead of $row.
If this still doesnt work you can catch the error with the error: callback function to see what is wrong.
Try to start with a simple request and work from there.
$(document).ready(function() {
$.ajax({
type: "POST",
url: "yourphp.php",
data: {simplestring: "hi"},
success: function(result){
alert(result);
}
});
});
and yourphp.php
<?php
$simplestring = $_POST['simplestring'];
echo $simplestring;
I am using Wordpress and I have a page where I have a drop down that when the links are clicked it will make an Ajax call and pass the data variable to PHP, at least that is what I'm attempting to do lol.
When clicking on the link I check my browser and in the Network tab for the page I receive a variable for the data object in the html and the ajax post's to the php page but for some reason I can't get a value.
My HTML
<div class="category-submenu">
<ul>
<li>Corporate</li>
<li>Office1</li>
<li>Office2</li>
<li>Office3</li>
</ul>
</div>
My jQuery
$('.category-submenu a').click(function(){
$.ajax({
type: "POST",
url: "/load-team.php",
dataType: 'json',
data: {office: $(this).data('office')},
success: function(data) {
$.each( data, function(i, item) {
alert(data[i].start);
});
}
});
});
My PHP
<?php
$office = $_GET['office'];
$link = mysql_pconnect("localhost", "root", "root") or die("Could not connect");
mysql_select_db("somedb") or die("Could not select database");
$arr = array();
$query = mysql_query("SELECT first_name, last_name FROM ic_team_members WHERE office ='" . $office . "'");
while($obj = mysql_fetch_object($query)) {
$arr[] = $obj;
}
echo '{"members":'.json_encode($arr).'}';
?>
I'm sure there is some code missing or my syntax might be incorrect in some parts but I can't seem to find where, if any place.
Again I want to grab the data object from HTML element, pass it through Ajax into PHP and return the result as json object, which I can do but for some reason I think the error is in my PHP.
Any help would be appreciated.
You are passing it by POST, and therefore you need to receive it with POST:
$office = $_POST['office'];
Otherwise, use GET to send the ajax request:
$.ajax({
type: "GET",
...
});
I am trying to figure out how to retrieve data from a MySQL database using an AJAX call to a PHP page. I have been following this tutorial
http://www.ryancoughlin.com/2008/11/04/use-jquery-to-submit-form/
But i cant figure out how to get it to send back json data so that i can read it.
Right now I have something like this:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "ajax.php",
data: "code="+ code,
datatype: "xml",
success: function() {
$(xml).find('site').each(function(){
//do something
});
});
});
My PHP i guess will be something like this
<?php
include ("../../inc/config.inc.php");
// CLIENT INFORMATION
$code = htmlspecialchars(trim($_POST['lname']));
$addClient = "select * from news where code=$code";
mysql_query($addClient) or die(mysql_error());
?>
This tutorial only shows how to insert data into a table but i need to read data. Can anyone point me in a good direction?
Thanks,
Craig
First of all I would highly recommend to use a JS object for the data variable in ajax requests. This will make your life a lot simpler when you will have a lot of data. For example:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "ajax.php",
data: { "code": code },
datatype: "xml",
success: function() {
$(xml).find('site').each(function(){
//do something
});
});
});
As for getting information from the server, first you will have to make a PHP script to pull out the data from the db. If you are suppose to get a lot of information from the server, then in addition you might want to serialize your data in either XML or JSON (I would recomment JSON).
In your example, I will assume your db table is very small and simple. The available columns are id, code, and description. If you want to pull all the news descriptions for a specific code your PHP might look like this. (I haven't done any PHP in a while so syntax might be wrong)
// create data-structure to handle the db info
// this will also make your code more maintainable
// since OOP is, well just a good practice
class NewsDB {
private $id = null;
var $code = null;
var $description = null;
function setID($id) {
$this->id = $id;
}
function setCode($code) {
$this->code = $code;
}
function setDescription($desc) {
$this->description = $desc;
}
}
// now you want to get all the info from the db
$data_array = array(); // will store the array of the results
$data = null; // temporary var to store info to
// make sure to make this line MUCH more secure since this can allow SQL attacks
$code = htmlspecialchars(trim($_POST['lname']));
// query
$sql = "select * from news where code=$code";
$query = mysql_query(mysql_real_escape_string($sql)) or reportSQLerror($sql);
// get the data
while ($result = mysql_fetch_assoc($query)) {
$data = new NewsDB();
$data.setID($result['id']);
$data.setCode($result['code']);
$data.setDescription($result['description']);
// append data to the array
array_push($data_array, $data);
}
// at this point you got all the data into an array
// so you can return this to the client (in ajax request)
header('Content-type: application/json');
echo json_encode($data_array);
The sample output:
[
{ "code": 5, "description": "desc of 5" },
{ "code": 6, "description": "desc of 6" },
...
]
So at this stage you will have a PHP script which returns data in JSON. Also lets assume the url to this PHP script is foo.php.
Then you can simply get a response from the server by:
$('h1').click(function() {
$.ajax({
type:"POST",
url: "foo.php",
datatype: "json",
success: function(data, textStatus, xhr) {
data = JSON.parse(xhr.responseText);
// do something with data
for (var i = 0, len = data.length; i < len; i++) {
var code = data[i].code;
var desc = data[i].description;
// do something
}
});
});
That's all.
It's nothing different. Just do your stuff for fetching data in ajax.php as usually we do. and send response in your container on page.
like explained here :
http://openenergymonitor.org/emon/node/107
http://www.electrictoolbox.com/json-data-jquery-php-mysql/