How to submit data from a modal? - javascript

These buttons are echo out from the database,
When the user clicks onto the button, the user will get to see a pop out before submitting it.
This is my code,
$con = getDbConnect();
$day = date("l");
if (mysqli_connect_errno($con)) {
"Failed to connect to MySQL: " . mysqli_connect_error();
} else {
$result = mysqli_query($con, "SELECT * FROM timetableschedule WHERE day='" . $day . "'");
while ($schedule = mysqli_fetch_array($result)) {
?>
<div class="col-md-3">
<button class="btn btn-warning" data-toggle="modal" data-target="#myModal" schedule="
<?php
echo "<br/>";
echo $schedule['academicInstitution'] . "<br />";
echo $schedule['startTime'] . "-" . $schedule['endTime'] . "hrs<br />";
echo "<br/>";
?>">
<?php
echo "<br/>";
echo $schedule['academicInstitution'] . "<br />";
echo $schedule['startTime'] . "-" . $schedule['endTime'] . "hrs<br />";
echo "<br/>";
?>
</button>
</div>
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">Insert Today's Activity</h4>
</div>
<div class="modal-body">
<p class="activity"></p>
<p>Click on the submit button if the above infomation is correct.</p>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Cancel</button>
<a class="btn btn-primary btn-ok">Submit</a>
</div>
</div>
</div>
</div>
<script>
$('#myModal').on('show.bs.modal', function(e) {
$(this).find('.btn-ok').attr('schedule', $(e.relatedTarget).attr('schedule'));
$('.activity').html('You have selected: <strong>' + $(this).find('.btn-ok').attr('schedule') + '</strong>');
});
</script>
<?php
}
mysqli_close($con);
}
?>
</div>
Once the user click onto the submit button, the "academicInstitution" and "duration" will be entered into a database called, "myRecord".But I do not know how to connect the submit button to the database.

If you need to stay on the same page, make an ajax post call to a server side php script: http://api.jquery.com/jquery.post/
This way you will be able to execute the query you want on the server
If you can accept to reload the page, you may simply use an html form
This question may be useful for you: jQuery Ajax POST example with PHP
Note that you don't necessarily need the form element if you choose the ajax post approach

Related

Pass Value From URL Link To Form Field

I have a link in a table from results of an SQL query. Each row has a unique ID. I define this in my script as $id
What I am trying to do is have a link that opens up a form in a modal box where a user can insert a note about that row and save it. This then gets inserted into DB with ID, rowID, Note, Date
echo "<td class='agentid'>" . $id. "<br><a href='#notes' data-toggle='modal' data-row-id='".$id."'>Add Note</a></td>" ;
The modal:
<div class="modal" id="notes">
<div class="modal-dialog">
<div class="modal-content">
<form action="postnote.php" method="post">
<div class="modal-body">
<div class="modal-header">
<h4 class="modal-title">Add a Note</h4>
<button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">×</span><span class="sr-only">Close</span></button>
</div>
<p>Note will show when page is refreshed</p>
<label for="fname">Note: </label>
<input type="hidden" name="rowID" value=""/>
<input class="form-control" type="text" name="notes" value="<?php echo date('m/d/Y h:i'); ?> - "/>
<div class="modal-footer">
<button type="submit" value="Submit" class="btn btn-primary">Submit</button>
</div>
</form>
</div>
</form>
</div>
</div>
</div>
<script type="text/javascript">
$('#notes').on('show.bs.modal', function(e) {
var row = $(e.relatedTarget).data('row-id');
$(e.currentTarget).find('input[name="rowID"]').val(row);
});
</script>
my postnote.php
$pdoOptions = array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
PDO::ATTR_EMULATE_PREPARES => false
);
$pdo = new PDO("mysql:host=" . MYSQL_HOST . ";dbname=" . MYSQL_DATABASE, MYSQL_USER,MYSQL_PASSWORD,$pdoOptions);
$data = [
'notes' => $_POST['notes'],
'id' => $_POST['rowID'],
];
$sql = "INSERT INTO act_recon_notes (note_id, note) Values(:id, :notes)";
$stmt= $pdo->prepare($sql);
$stmt->execute($data);
header('Location: ' . $_SERVER['HTTP_REFERER']);
I have this exact script on another page and it is working just fine. This does not post the $id into the form. So after submit it is a blank rowID value in the DB.
The issue was I had conflicting JS on the page. I removed the JS that was not needed for this page and the issue resolved.
<script type="text/javascript" src="https://code.jquery.com/jquery-3.3.1.js"></script>

Bootstrap modal popup upon failed login

I have a login page (login.html) that calls a php with POST method to valid login credentials, which works just fine. If the email an password are incorrect, then in my current implementation, an alert pops up and the page reloads. I would like to know how to activate a modal instead of said popup.
<div id="myModal" class="modal fade">
<div class="modal-dialog modal-confirm">
<div class="modal-content">
<div class="modal-header">
<div class="icon-box">
<i class="material-icons"></i>
</div>
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
</div>
<div class="modal-body text-center">
<h4>Ooops!</h4>
<p>Something went wrong. File was not uploaded.</p>
<button class="btn btn-success" data-dismiss="modal">Try Again</button>
</div>
</div>
</div>
</div>
Complete PHP File:
<?php
$email = $_POST['email'];
$pwd = $_POST['pwd'];
$conn = new mysqli('localhost','root','','rm');
if($conn->connect_error){
die('Connection Error: '.$conn->connect_error);
}
else{
$sql = "select * from superfan where email='".$email."' and password='".$pwd."' ";
$result = mysqli_query($conn,$sql);
$records = mysqli_num_rows($result);
$row = mysqli_fetch_array($result);
if ($records==0)
{
echo '<script type="text/javascript">';
echo '$(document).ready(function() {';
echo '$("#myModal").modal("show");';
echo '});';
echo '</script>';
}
else
{
header('location:index.html');
}
mysqli_close($conn);
}
?>

how to get specific id to show specific detail from modal

i created an article and using modal i can view the article but the problem is when i tried to view the article only the first article is being shown. not specifically the one that i clicked.. i noticed that the variable that i declared only gets the first id.
<?php
include_once('../model/articles.php');
$database = new Database();
$conn= $database->getConnection();
$db= new articles($conn);
$output = $db->viewAllarticles();
$articleid ='';
$artname = '';
$artpublished = '';
$arttitle= '';
$artbody = '';
foreach ($output as $key) {
$articleid = $key['art_id'];
$artpublished = $key['date_published'];
$arttitle = $key['art_title'];
$artbody =$key['art_body'];
if($key['art_isSaved'] == '0'){
?>
<style>
.ellipsis {
overflow: hidden;
white-space: nowrap;
text-overflow: ellipsis;
}
</style>
<div id="article" class="col-lg-4" style="background:#f9f9f9;width:218px;margin-right:10px;">
 <strong class="font-1"><?php echo $key['firstname'].' '.$key['lastname'] ?></strong><br>
<?php echo $key['art_body']?><br>
<?php
<i class="fa fa-eye"></i> <?php echo $key['art_seen']?>
<span class="pull-right"><span style="color:#59960b;" data-toggle="modal" data-target="#myModal">Read more..</span><input type="hidden" value="<?php echo $articleid; ?>"> </span>
</div>
<?php
}
}
?>
what i wanted to do is when i clicked read more the id of the one that i clicked will be sent to the modal.
<div class="modal fade margin-top-70" id="myModal" role="dialog" style="margin-left:-50px;">
<div class="modal-dialog">
<?php
$output = $db->viewArticle($articleid);
foreach ($output as $key):
if($key['art_id'] == $articleid){
?>
<!--view Modal content-->
<div id="articlepost" class="modal-content-article">
<div class="modal-header-article">
<input type="hidden" name="aid" id="aid" value="<?php echo $articleid ?>"/>
<button type="button" style="padding-left:1155px;position:fixed;" class="close" data-dismiss="modal">×</button>
</div>
<img src="./<?= $key['art_cphoto'];?>" style="margin-left:100px;"/>
<div class="modal-body">
<div align="center" style="color:#222;">
<strong class="font-2" id="title"><?php echo $arttitle ?></strong>
<br>
</div>
<?php }?>
</div>
</div>
i tried to use javascript to get the id of the article and it works but the problem is i dont know how to use the variable from javascript to declared it to my php if condition...do you have any idea on how can i get the id of the specific article so when i click the article the one that i click will be the one to shown???
This is happen because you are using repeated modal id. You have to make Modal Id unique like this:
<span class="pull-right" id="artid"><span style="color:#59960b;" class="read" data-id="myModal_<?php echo $articleid; ?>">Read more..</span><input type="hidden" value="<?php echo $articleid; ?>"> </span>
<div class="modal fade margin-top-70" id="myModal_<?php echo $articleid; ?>" role="dialog" style="margin-left:-50px;">
<div class="modal-dialog">
<?php
$output = $db->viewArticle($articleid);
foreach ($output as $key):
if($key['art_id'] == $articleid){
?>
<!--view Modal content-->
<div id="articlepost" class="modal-content-article">
<div class="modal-header-article">
<input type="hidden" name="aid" id="aid" value="<?php echo $articleid ?>"/>
<button type="button" style="padding-left:1155px;position:fixed;" class="close" data-dismiss="modal">×</button>
<img src="./<?= $key['art_cphoto'];?>" style="margin-left:100px;"/>
</div>
<?php }?>
</div>
</div>
jQuery for Modal Open:
$("body").on("click",".read",function(){
var id = $(this).data('id');
('#'+id).modal('toggle');
});
You have to use AJAX with PHP. Because each click you have to fetch data from database depend on you article id.kindly read through about ajax.
Now i just try to solve you problem .
First add jquery library in your file
<script src='https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.js'></script>
Replace you read more button with this
<span class='pull-right'><span style='color:#59960b;' data-toggle='modal' data-id='<?php echo $articleid; ?>' data-target='#myModal'>Read more..</span><span>
Change your modal like below
<div class="modal fade margin-top-70" id="myModal" role="dialog" style="margin-left:-50px;">
<div class="modal-dialog">
</div>
</div>
Add this script below to your page
<script>
$(document).ready(function(){
$(document).on('click','span[data-toggle=modal]',function(){
var articleID=$(this).data('id');
$.ajax({
'url':'../model/read_more_articles.php',//create a page and give the url
'data':{id:articleID},//pass id into read_more_articles.php page using method post
'method':'POST',
success: function (data) { //data contain return of the read_more_articles.php
$('.modal-dialog').html(data);//put generate html of the modal-dialog div
}
})
});
})
</script>
Create a page like read_more_articles.php. Here get details of crrosponding article id and populate html .
<?php
$output="";
$articleid=$_POST['id'];//receive article id which is sent by ajax.
$output = $db->viewArticle($articleid);
foreach ($output as $key){
if($key['art_id'] == $articleid){
$output .="<div id='articlepost' class='modal-content-article'>
<div class='modal-header-article'>
<input type='hidden' name='aid' id='aid' value='".$articleid."'/>
<button type='button' style='padding-left:1155px;position:fixed;' class='close' data-dismiss='modal'>×</button>
</div><img src='".$key['art_cphoto']."' style='margin-left:100px;'/>
<div class='modal-body'>
<div align='center' style='color:#222;'>
<strong class='font-2' id='title'>".$arttitle."</strong><br></div>"
}
}
echo $output;
?>
i finally found a way how to do it.. so here is what i did.. first i change my span to button
<button type="button" class="btn openModal" data-toggle="modal" data-target="#myModal" data-id="<?php echo $articleid;?>">Read More....</button>
and then
my modal div
<div class="modal fade margin-top-70" id="myModal" role="dialog" style="margin-left:-50px;">
<div class="modal-dialog">
</div>
</div>
then created a jquery script to get the id from the button and send it to another page where my html content is and return it and display it in modal-dialog.
<script type="text/javascript">
$('.openModal').click(function(){
var id = $(this).attr('data-id');
alert(id);
$.ajax({url:"../controller/displayarticle.php?id="+id,cache:false,success:function(result){
$(".modal-dialog").html(result);
}});
});
</script>
after sending the id to another page i use $_GET to get the id that is being sent.
$artid = '';
if ( !empty($_GET['id'])) {
$artid = $_GET['id'];
}else{
header("Location: home.php");
}
after sending the id i use for loop to display the data from database in a modal based on the id that is being sent.
$output = $db->viewArticle($artid);
foreach ($output as $key):
<div id="articlepost" class="modal-content-article">
<div class="modal-header-article">
<input type="hidden" name="aid" id="aid" value="<?php echo $artid ?>"/>
<img src="./<?= $key['art_cphoto'];?>" style="margin-left:100px;"/>
</div>
<div class="modal-body">
<img src="./<?= $key['image'];?>" style="border-radius:50%;margin-top:10px;" width="5%" height="5%" />
</div>
<?php
foreach ($db->countarticlecomment($artid) as $value) {
?>
<i class="fa fa-comment" style="color:#59960b;"></i> <span style="color:gray;"><b><?php echo $value['comments']?></b></span> 
</div
<?php endforeach ?>

j query updating div of drop-down after inserting data

I have a drop down which i am filling from database (ss and source attached)
<div id="divmedium">
<label>Medium:</label> <a data-toggle="modal" role="button" href="#medium_m">[Add New Medium]</a>
<select data-placeholder="Select Medium" class="select-full" tabindex="2" id="media" name="media">
<option value=""></option>
<?php
$quee = 'SELECT `media_id` , `mediatype` FROM `media` WHERE `bus_id_fk` = "'. $_SESSION['bus_id_fk'].'" order by `mediatype` asc';
$rs=$DBH->query($quee);
while($row = $rs->fetch_assoc()){
echo "<option value=$row[media_id]>$row[mediatype]</option>";
}
?>
</select>
</div>
if i click on [Add New Medium] , a model appears in which i can add value.
<div id="medium_m" class="modal fade" tabindex="-1" role="dialog">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title"><i class="icon-paragraph-justify2"></i> Add New Medium </h4>
</div>
<!-- Form inside modal -->
<form action="#" role="form" id="med1">
<div class="modal-body with-padding">
<div class="form-group">
<div class="row">
<div class="col-sm-9">
<label>First Medium</label>
<input type="text" name="fname" placeholder="Eugene" class="form-control">
</div>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-warning" data-dismiss="modal">Close</button>
<button type="submit" id = "m_btn" class="btn btn-primary">Submit form</button>
</div>
</form>
</div>
</div>
click on submit form (AJAX)
$(document).ready(function(){
$('#med1').submit(function(){
var formData = new FormData(this);
$.ajax({
type: 'POST',
url: 'addmedium.php',
//data: formData.serialize()
data: $(this).serialize()
})
.done(function(data){
$('#response').html(data);
if(data.status == 'success'){
$("#divmedium").html(data);
}else if(data.status == 'error'){
alert("Error on query!");
}
})
.fail(function() {
// just in case posting your form failed
alert( "Posting failed." );
});
// to prevent refreshing the whole page page
return false;
});
});
addmedium.php
<?php
session_start();
INCLUDE './config/databases.php';
header('Content-type: application/json');
$loc= $_POST['fname'];
$busid=$_SESSION['bus_id_fk'];
$sql = "INSERT INTO media (mediatype,bus_id_fk)VALUES ( '$loc','$busid' )";
//echo $sql;
if ($DBH->query($sql) === TRUE) {
// echo "New record created successfully";
} else {
// echo "Error: " . $sql . "<br>" . $DBH->error;
}
$response_array['status'] = 'success';
echo json_encode($response_array);
exit;
?>
Now the problem is
data getting inserted in the database but i am unable to refresh the
div,
After clicking on submit form model is not disappearing. i need to click on close or some where else besides model.
after clicking on Submit form , the div divmedium is disappearing.
Let me know what i am doing wrong.
I'll start from the end, if you won't mind.
after clicking on Submit form , the div divmedium is disappearing.
In your AJAX code $("#divmedium").html(data); you are replacing divmedium content with json_encode($response_array);
After clicking on submit form model is not disappearing. i need to click on close or some where else besides model.
I don't see any code that should close it. Try to add data-dismiss="modal"
data getting inserted in the database but i am unable to refresh the div
Same stuff as in 3rd question, check this and edit your AJAX success callback.

codeIgniter Javascript alert with success message on click ok page refresh

I have a code like this:
<div class="alert alert-success">
<a class="close" data-dismiss="alert">×</a>
<?php
$message = "Your Upload was successful";
if((isset($message))&&($message!='')){
echo '<script> alert("'.str_replace(array("\r","\n"), '', $message).'"); </script>';
}
redirect($this->uri->uri_string()); //refresh page
?>
I want to show this success alert message and then if the user click on OK it will refresh the browser. In my case it is just refreshing the browser.
What will be the best way to do it.
Thanks a lot in advance.
To make your code work as expected, you have to write the refresh function in Javascript instead of using PHP redirect function like the below:
<?php
$message = "Your Upload was successful";
if ((isset($message)) && ($message != '')) {
echo '<script>
alert("'.str_replace(array("\r","\n"), '', $message).'");
location.reload(true);
</script>';
}
?>
If you want to use Bootstrap modal, try this:
<?php
$message = "Your Upload was successful";
if ((isset($message)) && ($message != '')):
?>
<div class="modal" id="alert-dialog">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Alert</h4>
</div>
<div class="modal-body">
<?php echo $message; ?>
</div>
<div class="modal-footer">
<button data-dismiss="modal" type="button" class="btn btn-primary">OK</button>
</div>
</div>
</div>
</div>
<script>
$(function() {
$('#alert-dialog').modal('show').on('hidden.bs.modal', function () {
location.reload(true);
});
});
</script>
<?php endif; ?>
I'm not sure this is possible with a standard alert box, you can do it with a confirm.
e.g.
var con = confirm('Are you sure');
if (con == true) {
//means the user clicked on `OK`
//refresh the page
} else {
//means the user clicked `Cancel`
}
Alternatively you could use a customized alert box, to find a suitable one just search on google.

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