Following is the code I used to print the delete button for each data echoed.
<?php
$sno=1;
while($row=mysqli_fetch_assoc($datas)){
echo('<b><tr><td>'.$sno.'</td><td>' .$row['id'].'</td>
<td>'.$row['name'].'</td>
<td>'.$row['address'].'</td>
<td>'.$row['faculty'].
"</td><td>
<button onclick=deletee(); name='del_user' value='<?= $row['id']; ?>'> Delete </button><br>
</td></tr>");
++$row;
$sno++;
}
?>
In script;
function deletee() {
<?php
$datas="";
$sql="";
$idd=$row['id'];
$datas=mysqli_query($database,"SELECT*FROM data");
if(isset($_POST['delete_user'])){
$newrow=mysqli_fetch_assoc($datas);
if($newrow['id']==$idd)
{
$sql = "DELETE FROM data WHERE id='$idd' ";
mysqli_query($database,$sql);
header("Location: main.php");
}
}
?>
}
Yet I'm not able to delete a specified tuple. Whats the problem?
I'm trying to delete a row.
"After every data row is a delete button which deletes a specific row."
A better option to your above would be to create your button as
<button onclick=deletee(<?=Row['id'] ?>)
From there your function would be
function deletee(int id){
var url = "/Url/Where/PHP/is/run"
$.ajax({
url: url,
type: "post",
data: { id = id },
success: function (response) {
// Redirect to a page, or update whats displayed on the page here
},
error: function(jqXHR, textStatus, errorThrown) {
// do your error logging / display to page here
}
});
Now you just create a page to run the delete via your php and make sure the url in the ajax function points to that page, and it accepts a field called "id" (Note, this is using as a "POST", you can change to get etc. just make sure you do some checking on the data sent through.
If you don't want to use jquery at all, an alternative would be to make every lineitem a form pointing to the same url with a different value i.e.
//line item details Here
// where the submit button is use the code below instead
<form action="url/with/php/code" method="post">
<input type="hidden" name="id" value="<?= $row['id']?>" />
<input type="submit" value="delete">
</form>
Then have the php form it directs to check for post values, and redirect on success / do what it needs to do on failure
Related
I am building a budget application with HTML, Javascript, and PHP. My goal is to have the user be able to add data into a database from a form provided. I already have a ton of php at the top of my 'dashboard.php' (which contains the form) so I didn't want to run dashboard.php on submit, so instead I created a button that preforms an AJAX call to a different php file 'addIncome.php'.
I have two different files...
dashboard.php &
addincome.php
dashboard.php contains my form, as well as my javascript to run an AJAX call.
addincome.php is using $_POST to grab the values from the form in dashboard.php and make a mysqli_query. However, at first nothing was happening so I decided to echo the value of one of the return values from my $_POST. And ended up getting this error...
undefined index iName in addIncome.php
undefined index iAmount in addIncome.php
So from there I though that maybe I didn't have access to the dashboard.php by default so I included...
include('dashboard.php');
Still no difference...
I'm really at a stand still here. Any thoughts?
Thanks
The form...
<form>
<input type="text" name="iName" placeholder="income name">
<input type="number" step="0.01" min="0" name="iAmount" placeholder="amount">
<input type="date" name="iDate">
</form>
The javascript...
<script>
$('.in-btn').click(function() {
$.ajax({
url: "addIncome.php",
type: "POST",
data: 'show=content',
success: function(data) {
$('.in-btn').html(data);
}
});
setTimeout(() => {
// location.reload();
}, 2000);
});
</script>
The php...
<?php
echo "adding...";
require_once('connection.php');
include('dashboard.php');
$iUser = $_SESSION["username"];
$iName = $_POST["iName"];
$iAmount = $_POST["iAmount"];
echo $iName;
$sql = "INSERT INTO income (user, name, amount, date) VALUE ('pmanke', '$iName', '$iAmount','1/16/19')";
mysqli_query($dbCon, $sql);
?>
You are not sending any post data with your AJAX call except for:
show=content. You want to send your form data. You can retrieve your form data with:
$("#id-of-form").serialize()
That way your PHP code is able to retrieve the correct values from your POST data.
An even more general way to do this is to just create a normal form with a submit button and an action and use javascript to catch the submit event and make an AJAX call instead:
HTML:
<form id="idForm" action="addIncome.php">
<input type="text" name="iName" placeholder="income name">
<input type="number" step="0.01" min="0" name="iAmount" placeholder="amount">
<input type="date" name="iDate">
<input type="submit" />
</form>
Javascript:
$("#idForm").submit(function(e) {
var form = $(this);
var url = form.attr('action');
$.ajax({
type: "POST",
url: url,
data: form.serialize(), // serializes the form's elements.
success: function(data) {
alert(data); // show response from the php script.
}
});
e.preventDefault(); // avoid to execute the actual submit of the form.
});
i want to make like/unlike system with PHP and jQuery/AJAX..
Here is my form in PHP foreach... Here i have own id's for every form;
<?php foreach ($vid as $var) { ?>
<form class="classform" action="functions/videolike.php" method="post">
<input type="text" name="id" value="<?php echo $var['video_id'];?>">
<button class="submitbuttonclass"type="submit">Like</button>
</form>
<?php } ?>
Here is my Ajax script;
<script>
// this is the id of the submit button
$(".submitbuttonclass").click(function() {
$.ajax({
type: 'post',
url: "functions/videolike.php",
data: $(".classform").serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
return false; // avoid to execute the actual submit of the form.
});
</script>
the codes are working but not correctly;
When i click "like" button is working good, i cheked the database, caunting, inserting, deleting , working good...
But I want to make this with AJAX becouse, refreshing page is stopping the video when user watching video if he click the like button. Video is preloading becouse page refresh...
After i add my ajax script its working. But when i am clicking the like button, AJAX is posting to PHP, only the last id in the foreach loop,
THE Question?
How to make AJAX to get all of the id's in PHP foreach loop ??
And this is my videolike.php if you want to check;
<?php
session_start();
if($_POST['id'] && #$_SESSION["userid"]){
require_once "connectdb.php";
$id = $_POST["id"];
$VLcheck = "SELECT count(*) FROM `videolikes` WHERE user_id = ? AND liked_vid_id=?";
$reslike = $conn->prepare($VLcheck);
$reslike->execute(array($_SESSION["userid"],$id));
$VLrow = $reslike->fetchColumn();
echo $VLrow;
if ($VLrow > 0){
$VLcheck = "DELETE FROM `videolikes` WHERE user_id = ? AND liked_vid_id=?";
$reslike = $conn->prepare($VLcheck);
$reslike->execute(array($_SESSION["userid"],$id));
} else {
$curentsess= $_SESSION["userid"];
$INSlike = $conn->prepare("INSERT INTO videolikes(user_id, liked_vid_id)
VALUES('$curentsess','$id')");
$INSlike->execute();
}} else {die;}
?>
As you have a lot forms with class .classform, so how do you think your script should select the proper one?
The asnwer is - script can't, you should help it). Use .closest function to find closest <form> for a clicked button:
$(".submitbuttonclass").click(function() {
var form = $(this).closest("form");
// or find closest element with class `classform`
//var form = $(this).closest(".classform");
$.ajax({
type: 'post',
url: "functions/videolike.php",
data: form.serialize(), // serializes the form's elements.
success: function(data)
{
alert(data); // show response from the php script.
}
});
return false; // avoid to execute the actual submit of the form.
});
I have the following problem:
What i'm trying to accomplish is:
User clicks on a submit/image type button
Ajax call handles the submit, calls another PHP script to update a record in a MySQL table, all without reloading the page ( obviously )
My PHP code is working fine without the AJAX, as it will reload and update the record. but somehow the ajax call is not working and/or returning any error.
My code:
$(function() {
$('#like_form').submit(function(event) {
event.preventDefault(); // Preventing default submit button
var formEl = $('#like_form');
var submitButton = $('input[type=submit]', formEl);
$.ajax({
async: true,
type: 'POST',
url: formEl.prop('action'),
accept: {
javascript: 'application/javascript'
},
beforeSend: function() {
submitButton.prop('disabled', 'disabled');
}
}).done(function(data) {
submitButton.prop('disabled', false);
$("#like").fadeOut();
$("#like").fadeIn();
});
});
});
<!-- LIKE een gebruiker -->
<form action="" id="like_form" method='POST' enctype="multipart/form-data">
<input onmouseover="this.src='img/heart2.png'" onmouseout="this.src='img/heart.png'" name='like' id="like" src='img/heart.png' type="image" />
</form>
my PHP (just in case):
<?php
include_once "dbconnection.php";
//if like button (submit button) clicked
if ($_POST){
$conn = DatabaseConnection::getConnection();
$sql = "UPDATE dating_members
SET likes = likes + 1
WHERE member_id = 3";
$stmt = $conn->prepare($sql);
$stmt->execute();
}
?>
I figured out what the problem is, Kind of silly I didn't notice but better late than never.
I had to 'include' the JS script at the end of my PHP page in order to catch my onsubmit function and therefore disable the default event a.k.a submit button submitting my form and page reload / POSTs to the other PHP file.
Everything is working fine now
I am using codeigniter 3.1
Ajax post not working and i am getting 403 (Forbidden) in console.
[POST http://localhost/test/post 403 (Forbidden)]
HTML
<div class="post">
<input type="text" id="data1" name="data1" value="">
<input type="text" id="data2" name="data2" value="">
</div>
<button id="post">Submit</button>
JAVASCRIPT
$('#post').on('click', function () {
var value1=$("#data1").val();
var value2=$("#data2").val();
$.ajax({
url: window.location.href+'/post',
type: "POST",
data:"{'data1':'"+value1+"','data2':'"+value2+"'}"
});
CONTROLLERS
public function post()
{
$data1 = $this->common->nohtml($this->input->post("data1", true));
$data2 = $this->common->nohtml($this->input->post("data2", true));
$this->data_models->update($this->data->INFO, array(
"data1" => $data1,
"data2" => $data2,
)
);
}
If you want CSRF protection on (a good idea) then you must pass the CSRF token when posting form data - via AJAX or not. Consider this approach.
The easiest way to put the token in your form is to use Codeigniter's "Form Helper" (Documented here) You can load the function your controller or use autoloading. This view code assumes you have the helper loaded.
HTML
<div class="post">
<?= form_open('controller_name/post'); //makes form opening HTML tag ?>
<input type="text" id="data1" name="data1" value="">
<input type="text" id="data2" name="data2" value="">
<?php
echo form_submit('submit','Submit', ['id'=>'post']); //makes standard "submit" button html
echo form_close(); // outputs </form>
?>
</div>
The form_open() function also automatically adds a hidden field containing the CSRF token to the HTML.
Javascript
$('#post').submit(function( event ) {
//the next line will capture your form's fields to a format
//perfect for posting to the server
var postingData = $( this ).serializeArray();
event.preventDefault();
$.ajax({
url: window.location.href + '/post',
type: "POST",
data: postingData,
dataType: 'json',
success: function(data){
console.log(data);
}
});
});
controller
By the time $_POST gets to your controller the CSRF token has been striped away so you don't have to worry about it "polluting" your incoming data.
public function post()
{
//get all the posted data in one gulp and NO, you do not want to use xss_clean
$posted = $this->input->post();
//With the above the var $posted has this value (showing made up values)
// array("data1" => "whatever was in the field", "data2" => "whatever was in the field");
//sanitize the field data (?)
//just stick the clean data back where it came from
$posted['data1'] = $this->common->nohtml($posted["data1"]);
$posted['data2'] = $this->common->nohtml($posted["data2"]);
$this->data_models->update($this->data->INFO, $posted);
//you must respond to the ajax in some fashion
//this could be one way to indicate success
$response['status'] = 'success';
echo json_encode($response);
}
You could also send back some other status if, for instance, the model function reported a problem. You then need to react to that status in you javascript. But if you don't respond it will likely result in problems down the road.
I've never used Ajax before, but from researching and other posts here it looks like it should be able to run a form submit code without having to reload the page, but it doesn't seem to work.
It just redirects to ajax_submit.php as if the js file isn't there. I was trying to use Ajax to get to ajax_submit without reloading anything.
Is what i'm trying to do even possible?
HTML form:
<form class="ajax_form" action="ajax_submit.php" method="post">
<input class="input" id="license" type="text" name="license" placeholder="License" value="<?php echo htmlentities($person['license1']); ?>" />
<input class="input" id="license_number" type="text" name="license_number" placeholder="License number" value="<?php echo htmlentities($person['license_number1']); ?>" />
<input type="submit" class="form_button" name="submit_license1" value="Save"/>
<input type="submit" class="form_button" name="clear1" value="Clear"/>
</form>
in scripts.js file:
$(document).ready(function(){
$('.ajax_form').submit(function (event) {
alert('ok');
event.preventDefault();
var form = $(this);
$.ajax({
type: "POST",
url: "ajax_submit.php",//form.attr('action'),
data: form.serialize(),
success: function (data) {alert('ok');}
});
});
});
in ajax_submit.php:
require_once("functions.php");
require_once("session.php");
include("open_db.php");
if(isset($_POST["submit_license1"])){
//query to insert
}elseif(isset($_POST['clear1'])) {
//query to delete
}
I have "<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>"
in the html head
form.serialize() doesn't know which button was used to submit the form, so it can't include any buttons in the result. So when the PHP script checks which submit button is set in $_POST, neither of them will match.
Instead of using a handler on the submit event, use a click handler on the buttons, and add the button's name and value to the data parameter.
$(":submit").click(function(event) {
alert('ok');
event.preventDefault();
var form = $(this.form);
$.ajax({
type: "POST",
url: "ajax_submit.php",//form.attr('action'),
data: form.serialize() + '&' + this.name + '=' + this.value,
success: function (data) {alert('ok');}
});
});
Your ajax call is working perfectly. You have few conceptual error with your code -
form.serialize() will not attach submit button's info.
If you want to clear your form, you can do it using something like this
$('#resetForm').click(function(){
$('.ajax_form')[0].reset();
});
Lastly complete your task & return success or failed value to ajax call using echo like echo 'successful' or echo failed etc. Use an else condition with your code. It will be more clearer to you.
Remove the "action" and "method" attributes from the form. You shouldn't need them.