I have this function that is suppose to change the html of a span. It works the first time but anyclick after that it doesn't change anything. I know the query is working because the database is updating. Here's the code.
$('#availabilityicon').click(function() {
$.ajax({
type: "GET",
url: "includes/changeavailability.php",
success: function(msg) {
if(msg === "available") {
var vailspan = $('span#avail').html('<a id="availabilityicon" href="#"><img align="center" width="16px" height="16px" src="images/available.png" /></a>');
}
else if(msg === "unavailable") {
var availspan = $('span#avail').html('<a id="availabilityicon" href="#"><img align="center" width="16px" height="16px" src="images/unavailable.png" /></a>');
}
}
});
});
here is the php code
<?php
session_start();
$user = $_SESSION['username'];
include("dbcon.php");
$result = mysql_query("SELECT availability FROM user WHERE username='$user' ORDER BY id DESC") or die(mysql_error());
$row = mysql_fetch_assoc($result);
$availability = $row['availability'];
if($availability == 'yes') {
$query = mysql_query("UPDATE user SET availability='no' WHERE username='$user'") or die(mysql_error());
echo "unavailable";
}
elseif($availability == 'no' or $availability == "") {
$query = mysql_query("UPDATE user SET availability='yes' WHERE username='$user'") or die(mysql_error());
echo "available";
}
mysql_close($con);
?>
There's a spelling mistake in your javascript, you've put vailspan where you probably meant to put availspan after the if(msg === "available") { line.
If it's not that, try changing the click event to a live one:
$('#availabilityicon').live('click', function() {
just in case your javascript is overwriting that the #availabilityicon icon element and failing to re-attach the event to the new one
The "availabiltyicon" you click is being replaced with another with the same name. You can try using .live() or you can read up on event delegation.
Related
I am validating a sign In form through ajax. After successful validation the form is not redirecting to the required page.
Ajax Codes
function login_submit(){
var stat="";
$("#submit").val("Loging in...");
$.ajax({
type: "POST",
url: "php/login.php",
data: {
uname: $("#uname").val(),
pass : $("#pass").val()
},
success: function(result) {
if(result=="parent"){
window.location = "http://localhost:90/auction/augeo/admin/parent_admin/index";
}
else if(result == "sucess_normal"){
window.location.assign("../normal_admin");
}
else if(result == "deactivated account") {
window.location.assign("reactivate_account/");
}
else if(result == "banned account") {
window.location.assign("banned_account/");
}
else{
$("#submit").val("Login");
$("#error_msg").css({color: 'red'});
document.getElementById("error_msg").innerHTML= result;
stat = false;
}
}
});
if(!stat)
return false;
}
The php code
if(isset($_POST['uname']) && isset($_POST['pass'])){
$username = encode($_POST['uname']);
$password = encrypt(encode($_POST['pass']));
// check if entered username and password is in the database
$result = mysqli_query($conn,"SELECT * From admin_account where admin_account.username = '$username' AND admin_account.password = '$password' ");
if($row = mysqli_num_rows($result) == 1){
$found = mysqli_fetch_array($result);
if($found['state'] == 1){
$account_id = $found['account_id'];
setcookie("admin_id", $account_id, time() + (86400 * 30), "/");
$_SESSION['admin_id'] = $account_id;
$result1 = mysqli_query($conn,"SELECT role_id From admin where admin_id = '$account_id'");
$found1 = mysqli_fetch_array($result1);
$_SESSION['account_type'] = $found1['role_id'];
if($found1['role_id'] == "1"){
echo "parent";
//header("Location: http://localhost:90/auction/augeo/admin/parent_admin/index");
}else{
echo "sucess_normal";
}
}
elseif($found['state'] == 2){
echo "banned account";
}
else{
$_SESSION['deactivated_id'] = $found['account_id'];
echo "deactivated account";
}
}
else{
echo "Incorrect Username or Password";
}
}
I have tried all I could do but to no avail. I want to check if result=="parent" and if result=="parent" it should redirect to window.location = "http://localhost:90/auction/augeo/admin/parent_admin/index"; but instead it is echoing out parent.
You say "it is echoing out parent". But this should never happen with the AJAX code you supplied.
So I'm suspecting that you have a form that's running its own default submit, and that is what you're seeing.
You may want to check out this answer:
$('#idOfYourForm').submit(function() {
var $theForm = $(this);
// This is a button or field, right? NOT the form.
$("#submit").val("Logging in...");
$.post(
'php/login.php',
{
uname: $("#uname").val(),
pass : $("#pass").val()
}
).done(function(result) {
// check the result
alert("Server said: " + result);
});
// prevent submitting again
return false;
});
You get the button with
$("#submit")
This is ok, but if the button is defined as:
<input type="submit" id="submit" value="..." />
You'll get a subsequent submit of the form the button is defined in.
To avoid this, a far easier solution to the other suggested, is to not use a submit button at all. Instead, use a simple action button. These are two examples, the second of which is probably better because it is easier to design with bootstrap/HTML5/CSS...
<input type="button" id="submit" value="..." />
or better:
<button type="button" id="submit">...</button>
In case of slow server/network, you'll probably want to aid AJAX usability by disabling the button:
$("#submit").val("Logging in...").prop("disable", "disable");
This helps avoiding multiple submits when the server is slow and the user impatient.
I'm creating a follow button, more or less like the twitter one.
You click the button, and you follow the user.
You click again, and you unfollow the user.
I have done this code
HTML
<div data-following="false" class='heart canal'><i class='fa fa-heart awesome'></i></div>
AJAX
$(document).ready(function() {
$(".heart.canal").click(function() {
if($(".heart").attr("data-following") == '0'){
$(".heart").attr('data-following', '1');
} else if($(".heart").attr("data-following") == '1'){
$(".heart").attr('data-following', '0');
}
var usuario = $(".left h4").attr("data-id");
var seguidor = $("#user_account_info .profile_ball").attr("data-id");
var seguir = $(".heart").attr("data-following");
$.ajax({
type: "POST",
url: "./include/php/follow.php",
data: { user: usuario, follower: seguidor, follow: seguir },
success: function(response) {
if(response == '0'){
$(".heart").addClass("like");
} else if(response == '1'){
$(".heart").removeClass("like");
}
}
});
return false;
});
});
PHP
<?php
$dsn = "mysql:host=localhost;dbname=tapehd;charset=utf8";
$usuario = "root";
$contraseña = "";
$conexion = new PDO($dsn, $usuario, $contraseña);
$resultado = null;
$sql = "";
$user = $_POST["user"];
$seguidor = $_POST["follower"];
$follow = $_POST["follow"];
if($follow == '0'){
$sql = "INSERT INTO seguidores(id_canal, id_seguidor) VALUES('$user', '$seguidor')";
} else if($follow == '1'){
$sql = "DELETE FROM seguidores WHERE id_canal = '$user' AND id_seguidor= '$seguidor'";
}
if($conexion){ $resultado = $conexion->query($sql); }
return $follow;
?>
The problem is, everytime I click the button, I only insert data in the database. I mean, I only create follows.
When I click twice, it doesnt remove the follow.
Is there anyway to insert data when data-following = true and remove it when data-following = false ?
UPDATED
I have changed the boolean false and true for 2 strings, 0 and 1. But it doesn't work anyway.
There are numerous problems here. For one, like #Mark said, you need to understand that when sending ajax requests to PHP, you are sending strings. Also, in your JS, you are binding a click function to the .heart.canal, but then the function changes all elements with that class rather than the actual clicked element. Lastly, once you send the right information to PHP you need to print your results in order to see it in ajax.
Try the following:
JS:
$(document).ready(function () {
$(".heart.canal").click(function () {
var $heart = $(this);
if ($heart.data("following")) {
$heart.data("following", false)
} else {
$heart.data("following", true);
}
var usuario = $(".left").find("h4").data("id");
var seguidor = $("#user_account_info").find(".profile_ball").data("id");
$.ajax({
type: "POST",
url: "follow.php",
data: {user: usuario, follower: seguidor, follow: $heart.data("following")},
success: function (result) {
if (result) {
console.log("true");
} else {
console.log("false");
}
}
});
return false;
});
});
PHP:
$user = (int)$_POST["user"];
$seguidor = (int)$_POST["follower"];
$follow = ($_POST["follow"] === 'true') ? true : false;
if ($follow) {
// insert
} else {
// delete
}
print $follow;
I am working on a voting system in jquery. I have it where a user can vote up or if they change their mind vote down and it deducts from the upvote and puts it in on the down vote. But my problem is I cant get both numbers to refresh when a vote is selected so it just uses the original number instead of the updated number.
Vote Page
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/
libs/jquery/1.3.0/jquery.min.js"></script>
<script type="text/javascript">
$(function() {
$(".vote").click(function() {
var id = $(this).attr("id");
var name = $(this).attr("name");
var dataString = 'id=' + id;
var parent = $(this);
if (name == 'up') {
$.ajax({
type: "POST",
url: "up_vote.php",
data: dataString,
cache: false,
success: function(html) {
parent.html(html);
}
});
} else {
$.ajax({
type: "POST",
url: "down_vote.php",
data: dataString,
cache: false,
success: function(html) {
parent.html(html);
}
});
}
return false;
});
});
</script>
<?php
$sql=mysql_query("SELECT * FROM uploads LIMIT 9");
while($row=mysql_fetch_array($sql))
{
$msg=$row['title'];
$mes_id=$row['id'];
$up=$row['up'];
$down=$row['down'];
?>
<a href="" class="vote" id="
<?php echo $mes_id; ?>" name="up">
<?php echo $up; ?> up
</a>
<div class='down'>
<a href="" class="vote" id="
<?php echo $mes_id; ?>" name="down">
<?php echo $down; ?>
</a>
</div>
<div class='box2' >
<?php echo $msg; ?>
</div>undefined</div>undefined
<?php } ?>
The up_vote.php page..
(down_vote.php is exactly the same as up_vote except it just changes up to down.)
<?php
include("config.php");
$ip = $_SERVER['REMOTE_ADDR'];
if ($_POST['id']) {
$id = $_POST['id'];
$id = mysql_escape_String($id);
//Verify IP address in Voting_IP table
$ip_sql = mysql_query("select ip from votes where img_id='$id' and ip='$ip'");
$count = mysql_num_rows($ip_sql);
if ($count == 0) {
// Update Vote.
$sql = "UPDATE uploads SET up=up+1 WHERE id='$id'";
mysql_query($sql);
// Insert IP address and Message Id in Voting_IP table.
$sql_in = "insert into votes (id,img_id,ip,type) values ('','$id','$ip','up')";
mysql_query($sql_in);
} else {
//if already voted change it..
$result = mysql_query("SELECT * FROM votes WHERE img_id='$id' AND ip='$ip'");
while ($row = mysql_fetch_array($result)) {
$vote_type = $row['type'];
}
if ($vote_type == 'down') {
$up = mysql_query("UPDATE uploads SET up=up+1 WHERE id='$id'");
$down = mysql_query("UPDATE uploads SET down=down-1 WHERE id='$id'");
$vote = mysql_query("UPDATE votes SET type=up WHERE img_id='$id' AND ip='$ip'");
}
}
$result = mysql_query("select up from uploads where id='$id'");
$row = mysql_fetch_array($result);
$up_value = $row['up'];
echo $up_value;
}
?>
Not an answer but too long for a comment. This script:
while ($row = mysql_fetch_array($result)) {
$vote_type = $row['type'];
}
if ($vote_type == 'down') {
/* ... */
}
I don't think you need it like it is now. You are actually only using the last fetched row, not all of them. If that's your intention (because there's only one), then you don't need the while() at all. You could change it for:
$row = mysql_fetch_row($result);
$vote_type = $row['type'];
if ($vote_type == 'down') {
/* ... */
}
Furthermore, I'd recommend to change your code to PDO. It's more secure and mysql_* is deprecated.
I apologise if this comes across as really stupid. I have searched but can't seem to find an answer. I hope I can explain what it is I am trying to do.
I want to be able to query a database and if there is a record in it to show the record in the span/div or show a not found error message if there isn't.
I have a jquery check up and running to check if a username is in the database, what I want to know is how easy it would be to ammend this to pull all the data and show it in the span/div on the original page.
This is the jquery I have:
$(document).ready(function () {
$('#username').keyup(username_check);
});
function username_check() {
var username = $('#username').val();
if (username == "" || username.length < 2) {
$('#username').css('border', '1px #D5D5D5');
$('#cross').hide();
$('#tick').hide();
} else {
jQuery.ajax({
type: "POST",
url: "check.php",
data: 'username=' + username,
cache: false,
success: function (response) {
if (response == 1) {
$('#username').css('border', '2px #C33 solid');
$('#tick').hide();
$('#cross').fadeIn();
} else {
$('#username').css('border', '2px #090 solid');
$('#cross').hide();
$('#tick').fadeIn();
}
}
});
}
}
Can I do all this on the one page and query the db from the same page, instead of posting it to another page as I don't know how to get the results back to the calling page?
I hope I have explained what I want to do. Apologies if I haven't
Here is the PHP code:
$username = trim(strtolower($_POST['username'])); $username = mysql_escape_string($username); $query = "SELECT adbkid FROM person WHERE adbkid = '$username' LIMIT 1"; $result = mysql_query($query); $num = mysql_num_rows($result); echo $num; mysql_close()
You will normally send ajax requests to pages hosted on your server. So you can't directly access your database without going through your server. You'll need to write a function on your server that queries the database, and then call that function from javascript using ajax.
You can output a string in PHP and then set that text value to an element with jQuery ( $('#element').val(responseFromServer);
or $('#element').html(responseFromServer);
Instead of sending back "1" send back a json response something like:
/* record exists */
{status:1, html:'server generated message about record'}
/* doesn't exist */
{status:0}
This will allow you to still change css based on response data status value
Can use $.post ajax shorthand method:
$.post('check.php',{username: username}, function(response){
var upDateElement=$('#spanID');
if(response.status && response.status== 1){
$('#username').css('border', '2px #C33 solid');
$('#tick').hide();
$('#cross').fadeIn();
upDateElement.html( response.html)
}else{
$('#username').css('border', '2px #090 solid');
$('#cross').hide();
$('#tick').fadeIn();
upDateElement.html('Message html for no record found')
}
},'json')
check.php
$data = array();
$data['exists'] = false;
if(!isset($_POST['username'])) {
echo json_encode($data);
exit();
}
$username = mysql_escape_string($_POST['username']);
$query = "SELECT adbkid FROM person WHERE adbkid = '$username' LIMIT 1";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
if(count($row) == 1) {
$data = $row;
$data['exists'] = true;
}
return json_encode($data);
from your jQuery:
success: function(response) {
/**
* For instance, for a table with id, username, password and email you have
* data.exists = true/false;
* data.id = 1;
* data.username = 'foo';
* data.password = 'sample data password';
* data.email = 'foo#bar.com';
*/
if(response.exists === true) {
$('#username').val(response.username);
$('#email').val(response.email);
}
}
My PHP script generates a table with rows which can optionally be edited or deleted. There is also a possibility to create a new Row. The PHP is activated through jQuery Events.
Now all works well, I can edit delete and create an Item. After each action which makes use of the PHP script the HTML table gets updated.
But when I try after an Event to do an action again the HTML Table doesn't get updated though in the background the PHP script makes an entry into the database.
Does someone of you know why my HTML Table doesn't update itself when I trigger a second event?
Here is the Script:
PHP
<?php
require_once "../../includes/constants.php";
// Connect to the database as necessary
$dbh = mysql_connect(DB_SERVER,DB_USER,DB_PASSWORD)
or die ("Unaable to connnect to MySQL");
$selected = mysql_select_db(DB_NAME,$dbh)
or die("Could not select printerweb");
$action = $_POST['action'];
$name = $_POST['name'];
$id = $_POST['id'];
if($action == "new")
{
mysql_query("INSERT INTO `place` (`id`, `name`) VALUES (NULL, '$name')");
}
elseif($action == "edit")
{
mysql_query("UPDATE `place` SET `name` = '$name' WHERE `id` = '$id'");
}
elseif($action == "delete")
{
mysql_query("DELETE FROM place WHERE id = '$id'");
}
echo "<table><tbody>";
$result = mysql_query("SELECT * FROM place");
while ($row = mysql_fetch_array($result)) {
echo "<tr><td id=".$row["id"]." class=inputfield_td><input class=inputfield_place type=text value=".$row["name"]." /></td><td class=place_name>".$row["name"]."</td><td class=edit>edit</td><td class=cancel>cancel</td><td class=delete>delete</td><td class=save>SAVE</td></tr> \n";
}
echo "</tbody>";
echo "</table>";
echo "<input type=text class=inputfield_visible />";
echo "<button class=new>Neu</button>";
?>
JS
$(function() {
$.ajax({
url: "place/place_list.php",
cache: false,
success: function (html){
$("#place_container").append(html);
}
});
$(".edit").live("click", function() {
$(this).css("display","none").prevAll(".place_name").css("display","none").prevAll(".inputfield_td").css("display","block").nextAll(".cancel").css("display","block").nextAll(".save").css("display","block").prevAll(".inputfield_td").css("display","block");
});
$(".cancel").live("click", function() {
myvariable5 = $(this).prevAll(".place_name").html();
$(this).css("display","none").prevAll(".edit").css("display","block").prevAll(".place_name").css("display","block").prevAll(".inputfield_td").css("display","none").nextAll(".save").css("display","none").siblings().find("input[type=text]").val(myvariable5);
});
$(".save").live("click", function() {
var myvariable1 = $(this).siblings().find("input[type=text]").val();
var myvariable2 = $(this).prevAll("td:last").attr("id");
$(this).css("display","none").prevAll(".cancel").css("display","none").prevAll(".edit").css("display","block").prevAll(".place_name").css("display","block").prevAll(".inputfield_td").css("display","none");
$.post("place/place_list.php", {action: "edit", name: ""+myvariable1+"", id: ""+myvariable2+""}, function (html){$("#place_container").replaceWith(html);});
});
$(".delete").live("click", function() {
var myvariable3 = $(this).prevAll("td:last").attr("id");
$.post("place/place_list.php", {action: "delete", id: ""+myvariable3+""}, function (html){$("#place_container").replaceWith(html);});
});
$(".new").live("click", function() {
var myvariable4 = $(this).prevAll("input[type=text]").val();
$.post("place/place_list.php", {action: "new", name: ""+myvariable4+""}, function (html){$("#place_container").replaceWith(html);});
});
});
I think I know. You do replaceWith instead of append, so your DIV with ID #place_container disappears after the first operation (you are left with only a table in your page), and of course jQuery does not find it and is unable to refresh it with new content from the second operation.
Just use append or, better yet, html methods.
Shouldnt you replace the complete table ?
$("#place_container").html(html);