For Future Readers, this was my first question and the answer has been found (read comments and replies below):
First of all, i've searched in Stackoverflow and i didn't found an answer for a similar problem.
i would like to link a html Button (among many buttons) with a JQuery function. The function shall execute AJAX method like so :
HTML Code in a separated file index.php:
<button id="submitbtn" type="button" class="btn btn-success">UPDATE</button>
JQuery Function :
$('#submitbtn').on('click', function(){
var id = $(this).data('id');
$.ajax({
url: 'includes/updatequery.php',
type: 'POST',
data: {id:id},
success: function(data){
if (data) {
console.log("updated");
} else {
$('#error').load("custom/static/error.html");
}
},
error: function(jqXHR, textStatus, errorThrown){
$('#error').html("oops" + errorThrown);
}
});
});
Here is the PHP file that should be called by AJAX Method :
<?php
include("src/db.php");
$query = "UPDATE mytable SET job='completed' WHERE id=id";
mysqli_query($conn, $query);
?>
The problem is that i CANNOT link the ID of the clicked button (because there are many buttons) to the ID of the Database Entry in order to update the Data in the Database according to this specific button.
Now i would like to have the results updated LIVE after updating the Database.
This is the PHP code that output menu items (items stored in the same Database table as before) and in front of every menu item, a badge should be displayed (with a value within it : "completed" or "not completed") :
<?php
foreach($data as $d) {
$id = $d['id'];
$mystatus = $d['status'];
?>
<li class="nav-item">
<a class="nav-link clickable blueMenuItem" id="nav-location" data-id="<?php echo $d['id']; ?>">
<i class="nav-icon fas <?php echo $d['icon']; ?>"></i>
<p><?php
echo $d["title"];
if ($d['type'] == "job") { ?>
<span id="updatedicon" class="right badge <?php if($mystatus == "completed"){echo "badge-success";} else {echo "badge-danger";}?>"><?php setJob($con, $id)?></span><?php
} ?>
</p>
</a>
</li><?php
}
?>
Here is the PHP file where the setJob method is defined :
<?php
function setJob($con, $idd) {
$sql = "SELECT status FROM mytable WHERE id=$id";
$result = mysqli_query($con, $sql);
while ($row = mysqli_fetch_assoc($result)) {
foreach ($row as $row => $value) {
echo $value;
}
}
}
?>
Any suggestions?
Thanks
Use the data-id attribute to add the id:
<button id="submitbtn" data-id="<id>" type="button" class="btn btn-success">UPDATE</button>
https://www.w3schools.com/tags/att_global_data.asp
By default, jQuery ajax uses a Content-Type of application/x-www-form-urlencoded; charset=UTF-8. This means in PHP the POST values can be accessed using $_POST. If using a Content-Type of application/json, you will need to do this.
include("src/db.php");
$id = $_POST['id']; // make sure to sanitize this value
$query = "UPDATE mytable SET job='completed' WHERE id=$id";
mysqli_query($conn, $query);
The above example only demonstrates how to reference the id value from the POST. However, this is not secure as-is. Make sure to sanitize the value as well as protect yourself from SQL Injection using prepared statements. Prepared Statements allow you to bind variables to SQL queries which are sent separately to the database server and can not interfere with the query itself.
Updated HTML - added data-id="" to button and replace with id
<button id="submitbtn" data-id="<id>" type="button" class="btn btn-success">UPDATE</button>
Updated jQuery - use attr to get the id of row/record by using data-id attribute
$('#submitbtn').on('click', function(){
var id = $(this).attr('data-id');
$.ajax({
url: 'includes/updatequery.php',
type: 'POST',
data: {id:id},
success: function(data){
if (data) {
console.log("updated");
} else {
$('#error').load("custom/static/error.html");
}
},
error: function(jqXHR, textStatus, errorThrown){
$('#error').html("oops" + errorThrown);
}
});
});
Related
I have been trying to delete a row in my mySQL database on the onclick of a delete button. But instead of the one mySQL row getting deleted, all rows in the database get deleted.
I am targeting just the specific ID, so I am unclear as to why all other ID's are getting deleted.
HTML:
<?php foreach ($movies as $movie) : ?>
<div class="col-4">
<div class="card card-cascade">
<div class="view gradient-card-header purple-gradient">
<h2><?php echo $movie['name']; ?></h2>
<p><?php echo $movie['genre']; ?></p>
</div>
<div class="card-body text-center">
<!-- Delete -->
<a type="button" class="btn-floating btn-small btn-dribbble delbutton" data-toggle="tooltip" data-placement="top" title="Delete" id="<?php echo $movie['id']; ?>"><i class="fa fa-trash-o" aria-hidden="true"></i></a>
</div>
</div>
</div>
<?php endforeach; ?>
JS:
$(function () {
// Tooltips Initialization
$('[data-toggle="tooltip"]').tooltip();
// Delete Movie
$(".delbutton").click(function() {
console.log('watch me')
var del_id = $(this).attr("id");
var info = 'id=' + del_id;
if (confirm("Sure you want to delete this post? This cannot be undone later.")) {
$.ajax({
type : "POST",
url : "../movieApp/delete.php", //URL to the delete php script
data : {id:info},
success : function() {
console.log("success");
},
error: function () {
console.log("failed");
},
});
$(this).parents(".record").animate("fast").animate({
opacity : "hide"
}, "slow");
}
return false;
});
});
PHP:
require 'config/config.php';
require 'config/db.php';
if($_POST['id']){
$id=$_POST['id'];
$delete = "DELETE FROM movies WHERE id=$id";
$result = $conn->query($delete);
}
if (mysqli_query($conn, $sql)) {
mysqli_free_result($result);
mysqli_close($conn);
echo "Worked!";
exit;
} else {
echo "Error deleting record";
}
You set ajax method POST, But Post data format is not correct as per your requirement.
Change your ajax Data like as
//var info = 'id=' + del_id;
var info = {
id : del_id
}
And
$.ajax({
/*...*/
data : info,
/*.../
});
And also check if your id field is string, If integer then change the Query string to -
#$delete = "DELETE FROM movies WHERE id='$id'";
$delete = "DELETE FROM movies WHERE id=$id";
Also change -
#$_POST['info']
$_POST['id']
Because, You didn't set $_POST['info'] anywhere in your code.
Note : And don't forget to console your correct Ajax URL
In your HTML use data-id="<?php echo $movie['id']; ?>" for the tag. Then in your JS you can pick up the value like so: var del_id = $(this).data("id");. I would also inspect element in your browser to see if you are in fact sending an "id" to your PHP script. If you are then possibly you may want to enable error debugging in your PHP script like so: error_reporting(E_ALL);
ini_set('display_errors', 1);. Also wouldn't hurt to change your SQL statement to something like this: $delete = "DELETE FROM movies WHERE id='" . $id . "'";. Good luck with this one doesn't sound too hard.
I want to send id of element to php and create session for this.
This is piece from php file:
<?php
$sql = "SELECT id FROM products";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
?>
<tr class="table-manufacture-tr">
<td class="table-manufacture-td-statys">
<div class="warehouse-window-content-dropdown-plus2 plus">
<a class="open_item" data-id=<?php echo "\"".$row['p_id']."\"";?>
style="text-decoration: none; color: #D3D3D3;">Click</a>
</div>
</td>
</tr>
<?php
}
?>
And in this file javascript code:
$(document).on('click', '.open_item', function(event){
var data_item = this.getAttribute("data-id");
$.ajax({
url: 'get_id.php',
type: 'POST',
data-type: 'json',
data: { id: data_item },
contentType: 'application/x-www-form-urlencoded',
success: function(data){
console.log(data);
},
error: function(){
console.log("not working");
}
});
});
This is get_id.php:
<?php
session_start();
$_SESSION['item_id'] = json_encode($_POST);
header("Content-Type: application/json", true);
?>
I have tried also without content types and without json. "var data_item" prints id correct, but php doesn't create session and in console also clear(nothing).
The reason that you are not getting data in session is, you are not assigning proper value to session. Also it should be json_decode not json_encode.
replace
$_SESSION['item_id'] = json_encode($_POST);
with
if (!empty($_POST['id'])) {
$_SESSION['item_id'] = json_decode($_POST['id']); // use json_decode
}
It seems to me that you are making some small mistake in your code like you are echoing $row['p_id'] while your query should return id instead p_id also you are making mistake in ajax you are sending data-type JavaScript assuming your code is subtracting so try to use this code i code below.
// modify your php code
<?php
$sql = "SELECT id FROM products";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_assoc($result)) { ?>
<tr class="table-manufacture-tr">
<td class="table-manufacture-td-statys">
<div class="warehouse-window-content-dropdown-plus2 plus">
<a class="open_item" data-id=<?php echo "\"".$row['id']."\"";?>
style="text-decoration: none; color: #D3D3D3;">Click</a>
</div>
</td>
</tr>
<?php } ?>
// modify your jQuery
$(document).on('click', '.open_item', function(event){
var data_item = $(this).data("id");
$.ajax({
url: 'get_id.php',
type: 'POST',
dataType: 'json',
data: { id: data_item },
success: function(data){
console.log(data);
},
error: function(){
console.log("not working");
}
});
});
<?php
session_start();
header("Content-Type: application/json", true);
$_SESSION['item_id'] = json_encode($_POST["id"]);
echo json_encode(['data_id' => $_SESSION['item_id']]);
?>
You can use
$_SESSION['item_id'] = json_encode($_POST['id']);
instead of
$_SESSION['item_id'] = json_encode($_POST);
this will work fine.
I don't know what you are trying to do, but from your JS, it looks like that you are expecting that the PHP script --which you post some data to it-- to return a json with the data you have just posted in it. In that case, try this, change your get_id.php to be like:
<?php
session_start();
$_SESSION['item_id'] = json_encode($_POST);
header("Content-Type: application/json", true);
echo $_SESSION['item_id'];
?>
I'd troubleshoot this by making sure the click handler is actually going off. Put alert("clicked"); as the first thing in the in the click handler to make sure.
For the meantime, remove the contentType in the json call. Also remove the dataType (data-type) entirely. On the php side, replace the header() line so (as mentioned) the php is just:
session_start();
$_SESSION['item_id'] = $_POST["id"];
echo $_SESSION['item_id'];
Do not use json_encode/decode right now. From your code, it is not needed.
I'm trying my best to get this to work. But AJAX is pretty new to me. So hang in there...
Ok, I've asked a couple of questions here about getting this issue that I'm having to work. I (We)'ve come a long way. But now the next issue is here.
I'm trying to echo a session in a div using AJAX.
The AJAX code is working, I can echo plain text to the div I want it to go. The only problem I have is it does not display the title of the item.
I have some items (lets say 3 for this example) and I would like to have the custom save the Items in a Session. So when the customer clicks on save. The ajax div displays the title. And if the custom clicks the 3rd item it show the 1st and 3rd item, etc...
My HTML:
<button type="submit" class="btn btn-primary text-right" data-toggle="modal" data-target=".post-<?php the_ID(); ?>" data-attribute="<?php the_title(); ?>" data-whatever="<?php the_title(); ?>">Sla deze boot op <span class="glyphicon glyphicon-heart" aria-hidden="true"></span></button>
My AJAX code:
$(".ajaxform").submit(function(event){
event.preventDefault();
$.ajax({
type: "POST",
url: "example.com/reload.php",
success: function(data) {
$(".txtHint").html(data);
},
error: function() {
alert('Not OKay');
}
});
return false;
});
My PHP reload.php:
<h4>Saved Items</h4>
<p>Blaat...</p>
<?php echo "Product Name = ". $_SESSION['item'];?>
I saw this code on here: I'm not using this code. Only wondering if I can use it for my code, and then how?
Change session.php to this:
<?php
session_start();
// store session data
$_SESSION['productName'] = $_POST['productName'];
//retrieve session data
echo "Product Name = ". $_SESSION['productName'];
?>
And in your HTML code:
function addCart(){
var brandName = $('iframe').contents().find('.section01a h2').text();
$.post("sessions.php", {"name": brandName}, function(results) {
$('#SOME-ELEMENT').html(results);
});
}
How I'm getting my title();:
<?php
// Set session variables
$_SESSION["item"][] = get_the_title();
?>
Is this some thing I can use? And could someone help me with the code?
Thanks in advance!
I'm not too sure on what exactly you're trying to accomplish, but here's a quick and dirty example of making an HTTP request (POST) with a name of a product, storing it in a PHP session, and outputting all product names in the session:
HTML
<p>Product A <button class="add-product" data-product="Product A">Add Product</button></p>
<p>Product B <button class="add-product" data-product="Product B">Add Product</button></p>
<p>Product C <button class="add-product" data-product="Product C">Add Product</button></p>
<div id="response">
</div>
JavaScript
$('.add-product').click(function() {
var productName = $(this).data('product');
$.post('addProduct.php', {productName: productName}, function(data) {
$('#response').html(data);
})
});
PHP (addProduct.php)
<?php
session_start();
if (!array_key_exists('products', $_SESSION) || !is_array($_SESSION['products'])) {
$_SESSION['products'] = [];
}
$productName = array_key_exists('productName', $_POST) ? (string) $_POST['productName'] : '';
if ($productName) {
$_SESSION['products'][] = $productName;
}
?>
<h4>Your added products:</h4>
<ul>
<?php foreach ($_SESSION['products'] as $product): ?>
<li><?php echo htmlspecialchars($product); ?></li>
<?php endforeach;?>
</ul>
I'm performing CRUD oprations using JQuery/Ajax and php/MySQL
i'm able to insert/select and delete data but i gotta stuck in edit/update. im pulling data into text box when i click on edit button but after editing when i click on save button unable to update in mysql db!!
Any help is Appreciated Thanks
html code
<span class="noedit name" idl='<?php echo $row->id;?>'>
<?php echo $row->url;?>
</span>
<input id="url1" name="url1" class="form-control edit name url1" value="<?php echo $row->id;?>"/>
<a ide='<?php echo $row->id;?>' id="edit" class='editOrder' href="#" style="display:block-inline;">EDIT</a>
<a idu='<?php echo $row->id;?>' id="update" class='update saveEdit' href='#' style='display:none;'>SAVE</a>
<a idc='<?php echo $row->id;?>' id="cancel" class='cancelEdit edit' href='#' style='display:none;'>CANCEL</a>
Jquery code
$('body').delegate('.edit','click',function(){
var IdEdit = $(this).attr('ide');
alert(IdEdit);
$.ajax({
url:"pages/feeds.php",
type:"post",
data:{
editvalue:1,
id:IdEdit
},
success:function(show)
{
$('#id').val(show.id);
$('#url1').val(show.url);
}
});
});
$('.update').click(function(){
var id = $('#id').val()-0;
var urls = $('#url1').val();
$.ajax({
url:"pages/feeds.php",
type:"post",
async:false,
data:{
update:1,
id:id,
upurls:urls
},
success:function(up)
{
$('input[type=text]').val('');
showdata();
},
error:function(){
alert('error in updating');
}
});
});
PHP Code
if(isset($_POST['editvalue']))
{
$sql = "select * from test where id='{$_POST['id']}'";
$row = mysql_query($sql);
$rows = mysql_fetch_object($row);
header("Content-type:text/x-json");
echo json_encode($rows);
exit();
}
if(isset($_POST['update']))
{
$sql = "
update test
set
url='{$_POST['upurls']}'
where id='{$_POST['id']}'
";
$result = mysql_query($sql);
if($result)
{
//alert('success');
echo 'updated successfully';
}
else
{
//alert('failed');
echo 'failed to update';
}
}
I don't see an #id input in your code. is it there? I think the problem is here.
If this input exists, use the following tips:
Check if all values (id, url) are sended to your PHP script.
You can use console.log in Javascript or print_r, var_dump functions in PHP.
Change
$('.update').click(function(){
to
$('.saveEdit').click(function(){
I am trying to add some data to a relational database, and would like the session_user_id to be the foreign key for that database. When a user clicks a button, I want to make a database entry with the session_user_id and some other information I have POSTed to the page. My ajax posts to the php webpage page which it is run on (meaning all my scripts are on the same page)
I am currently getting a Uncaught ReferenceError: $sess_user_id1 is not defined. The jquery is firing. While I would love to get the undefined variable fixed, overall this does not seem like a very direct way to to this, and has added a bunch of confusing variables, when all the variables I need were already in my PHP statement. Is there any way to trigger the PHP entry without going through ajax and having to define the variables again?
Here is my php, which is at the header which is on the same page as my JS and HTML:
<?php
$markerid = $_POST["id"];
$name = $_POST["name"];
$type = $_POST["type"];
$point = $_POST["point"];
$lat2 = $_POST["lat"];
$lng2 = $_POST["lng"];
$locationdescription = $_POST["locationdescription"];
$locationsdirections = $_POST["locationdirections"];
session_start();
if (!isset($_SESSION['sess_user_id']) || empty($_SESSION['sess_user_id'])) {
// redirect to your login page
exit();
}
$sess_user_id1 = $_SESSION['sess_user_id'];
if ((isset($_POST['usid'])) && (isset($_POST['usid']))) {
$user_id_follow = strip_tags($_POST['usid']);
echo $user_id_follow;
$query = "INSERT INTO markerfollowing ( userID, markerID, type )
VALUES ('$user_id_follow', '$markerid', '$type');";
$result = mysql_query($query);
if (!$result) {
die('Invalid query: ' . mysql_error());
}
mysql_close();
}
?>
Here is the HTML button:
<div class="btn pull-right">
<button class="btn btn-large btn-followmarker" type="submit"id="followmarker">Add me to the list</button>
</div>
Here is the jquery/ajax post:
<script/javascript>
$(document).ready(function () {
$("#followmarker").click(function(){
$.ajax({
type: "POST",
url: "", //
data: { usid: <?php echo '$sess_user_id1'; ?>},
success: function(msg){
alert("success");
$("#thanks").html(msg);
},
error: function(){
alert("failure");
}
});
});
});
</script
A sincere thanks for any and all help. I haven't worked with relational databases before.
<?php echo '$sess_user_id1'; ?>
is wrong. If you wont to get
data: { usid: 123} at $sess_user_id1 is 123, you should write
data: { usid: <?php echo "$sess_user_id1"; ?>}
See your html source code in your brawser. I think there is data: { usid: $sess_user_id1}, and javascript is not understand what is the $sess_user_id1
This is the only one problem that I can see now, but I don't understand your current task whole to say more.