How to update a table without changing the onscreen data displayed - javascript

I need to display a simple list of hundreds of items which are on a MySQL table, review the list onscreen and click on a link beside each unwanted item to delete it from the table. This is an internal management procedure; no outside user is involved. I do not need the item to disappear from the list immediately; I will refresh the list periodically so items deleted from the table are no longer listed. I do not need any message to confirm that the item has been deleted. The important thing is that I don't want to lose sight of the list each time I delete an item and have to click on a "go back" button to return to the list.
The table uses MySQL. All my coding to date has been in PHP. So I am using php to display the list of items, on a non-html screen. This is the code for each item:
echo $item." <a href='item_delete.php?id=".$item."'>Delete item</a><br />";
This is the code for item_delete.php:
<?php
require ('connect.php'); // To define connection $con
$id = $_POST['id'];
mysqli_query($con, "DELETE FROM `items_table` WHERE `id` = $id");
?>
The item is deleted correctly but a blank screen is (understandably) displayed.
I have done a lot of searching but most people needing help want to do more advanced things and - because I have so far managed to avoid learning JavaScript, jQuery and AJAX - I can't even work out which of those technologies I need to update a table without changing the screen.
I get the impression that each PHP script always takes "focus" with it, so maybe I need a little JavaScript script to do this ?
If so:
- can I just change item_delete.php to item_delete.js or do I have to define the non-html list as an html one ?
- what js code is needed in item_delete.js ?
I have read about using: header("location:javascript://history.go(-1)");
or: header('Location: ' . $_SERVER['HTTP_REFERER']);
but they don't go back to the onscreen list.
I don't think I want the js script to perform a virtual "go back" because the list is originally produced by using (about 20) $_POST parameters, so I still seem to have to refresh it each time.
So I'd like a solution to remain with the list - rather than leave it and return to it. Thanks.

It would make a lot of sense to do the deletion asynchronously using javascript. However, the simplest and messiest way to achieve what you want, is to add target="_blank" to the links, (which will leave you with a open blank tab for each delete request you do).
echo $item." <a href='item_delete.php?id=".$item."' target='_blank'>Delete item</a><br />";
Or you can solve it by adding checkboxes in front of every item, check the items you want to delete and submit them as form parameters to the delete script.

If you want to delete to row in the onscreen table after the actual PHP code has run you can use the following implementation:
The HTML structure for the link requires a unique class name, such as:
echo 'Delete item'
Note the item id is stored inside a HTML5 data attribute. I have also added an onclick event handler which returns false to avoid the link refreshing the page.
The javascript used to delete the item use the JQuery AJAX method and binds to the specified class, which is: item-delete. The implementation requires Jquery version >= 1.9.0
(function(){
$('.item-delete').click(function(event) {
var target = $(event.target);
var id = target.data('item-id');
$.ajax({
url: 'item_delete.php',
method: 'POST',
data: {
id: id
},
}).done(function() {
target.remove();
}).error(function(err) {
console.error('Could not delete item with ID: ' + id);
console.error(err);
});
});
}())
The event listener is defined inside a self-executing function, which is automatically executed when the page-load completes and avoids poluting the global namespace.

You can delete the item directly on the same page without moving to another
page by passing the id through a hyperlink and then get it to finally delete
the unwanted item. CHECK THIS OUT, and please let me whether or not is what you
wanted :-)
// connection
mysql_connect("host", "user", "password");
mysql_select_db("your database name");
// select all the items from table.
$selectQuery = mysql_query("SELECT * FROM table_name" );
// use while loop to list all the items...
while( $row = mysql_fetch_array($selectQuery) )
{
// list the items as a hyperlink, passing their id through the URL.
?>
<?php echo "delete " . $row["item_name"]; ?>
<?php
}
// Below is the code to delete the item.
if( isset( $_GET["id"] ) )
{
$itemId = $_GET["id"];
// query to delete item
$deleteQuery = mysql_query("DELETE FROM table_name WHERE id = '$itemId' ");
//-----------THE MOST IMPORTANT PART. >>>
// redirect if delete is successfull.
if( $deleteQuery )
{
// reload the page to get the items minus the deleted one...
// let's say your sript name is delete.php
header("Location:delete.php");
}
}
?>
</code>

Related

Bootstrap Modal - How to provide data from php (database)

This is probably very simple, but am learning PHP, Javascript as I go. I find it easier to learn using real examples than the contrived examples given online.
I am creating an attendance register page, based on selecting a class, then all members of that class ordered by Surname and Firstname.
The table row has it's id set, by PHP, as the record's mem_id, and contains just forename+" "+surname, and some checkboxes.
All this is working fine, but now I have been asked to add a link so that clicking on it brings up a modal containing related data for the person selected. The extra data is already in the $a_fetch array.
Have added a glyphicon link for every row and clicking it displays a modal alright, and by having a javascript function I know I can get the row index and row id
<tbody>
<?php
while($g_fetch = $a_query->fetch_array()) {
$checked = array();
$memid = $g_fetch['mem_id'];
$name = $g_fetch['firstname'].' '.$g_fetch['lastname'];
$attendences = explode(",",$g_fetch['attend']);
for ($x = 0; $x <= 12; $x++) {
if ($attendences[$x]!="0") {
$checked[$x] = 'checked = "checked"';
}
else $checked[$x] = '';
}
echo "<tr id='".$memid."'>";
echo "<td>".$name."</td>";
echo "<td align='center'><div id='".$memid."' class='glyphicon glyphicon-info-sign' onclick='getId(this.id)' style='cursor:pointer' data-toggle='modal' data-target='#ModalCentre'></div>";
for ($y = 0; $y <= 12; $y++) {
echo '<td align="center"><input type="checkbox" value = "" '.$checked[$y].'></td>';
}
}
unset($checked);
unset($attendences);
?>
</tbody>
</table>
I am at a loss as how to proceed - is it even possible to pass data to the modal to display related data?
If it is would I need to run a new query (SELECT), or as the row is the same index as the data in the $A_fetch, and the row id has the correct mem_id is it possible to get the data from the existing $a_fetch array using either of those, or would I need to run a new SELECT?
Many thanks
There are multiple ways to provide data to the modal - and (in my opinion) it depends on how much data you need to pass to your modal and how many rows you have.
I want to describe you two ways:
Light+Easier Solution
If you don't want to display a lot of data and you have just a few rows.
The idea is to add the data directly to each div.glyphicon (as data attributes) and then use it in the modal
In your foreach add it to your model like that:
<div id='".$memid."' class='glyphicon glyphicon-info-sign' onclick='getId(this.id)' style='cursor:pointer' data-toggle='modal' data-target='#ModalCentre' data-link='".$g_fetch['additional_link'] ."' data-moreInfo='".$g_fetch['moreInfo']."'></div>
You haven't posted the modal's HTML or your JS code, but you wrote you are using bootstrap, so stick to
https://getbootstrap.com/docs/4.0/components/modal/#varying-modal-content
and fetch/set the relevant data (related clicked glyphicon) as it's described.
More complex solution
For more data / more rows. The additional data is not provided in the inital loaded HTML page - Therefore not all data needs to be loaded in the beginning.
Instead the additional data is loaded via ajax when clicking on one row.
For that you need to provide an additional endpoint (php) which provides the modal content for one row.
Check out second answer in Bootstrap 3 - How to load content in modal body via AJAX?
Basically you have a php file (e.g. getAdditionalData.php)
In this file you access the mem_id via GET
$mem_id = $_GET['mem_id'];
fetch the additional data from database
and print/render out the modal content (full html like in the second answer)
And in JS (inital page) you load the modal content onClick (fetched from php with provided mem_id as parameter)
var clicked = $(e.relatedTarget);
$(this).find(".modal-body").load("%PATH%/getAdditionalData.php?mem_id="+clicked.attr("id"));
I hope it will help you solving your problem and if you need additional infos just let me know. There are more ways to archive your goal but I think this 2 possibilities are enough in the beginning :-)

jQuery responds with html table containing a (form in each row plus jquery code to edit each row) newly created forms ignore submit()

Three days and I cannot find an answer to this or a solution. I am far from being a jQuery guy.
User arrives at a php page that shows a form to choose the language code and submit. Form gets submitted and the jQuery response builds a table in a div container in the original page. So far so good. The resulting table contains hundreds of rows where the language variables can be edited. On clicking the edit button, I get nothing, no errors in console, nothing. If I use on click, I can fire an alert. On submit, bind, and many others do not work.
I am attempting to load the table, perform basic editing, submit the edits to the db, and refresh the div.table.row with the edited results. I have not gotten to the part where I refresh the rows with the edited data, still stuck trying to submit the dynamic form in each row.
One interesting thing that happens when I use on click, it works with just the first button I click on, it somehow disables all the other rows. I need to be able to submit a row for changes, have that row refresh and move on to the next row all without redirecting. Is this too much to ask? Or should I just move them from page to page?
The ajax php page returns the table using the following code:
if(!empty($_POST['edit_language']) && $_POST['edit_language'] == 1){
edit_language($_POST['lang']); //call function to edit language
} else {
echo "You got here but no variables<br>"; //testing
print_r($_POST); //testing
}
function edit_language($lang){
//query table to get language vars list
$sql = "SELECT lang_site.lid, lang_codes.iso_code, lang_codes.`language`, lang_varnames.varid, lang_varnames.varname, lang_site.varval FROM lang_codes LEFT JOIN lang_site ON lang_site.langid = lang_codes.langid LEFT JOIN lang_varnames ON lang_site.varid = lang_varnames.varid where lang_codes.iso_code = '" . $lang . "'";
$result = db_query($sql);
//generate report table
echo "<table cellspacing='0' border='1'><tr><th>Count</th><th>Language</th><th>Variable Id</th><th>Variable Name</th><th>Variable Value</th><th>Edit</th></tr>";
$a=1; //count the rows to number the rows for easy viewing
while($data = db_fetch_assoc($result)){
//create form in each tr
echo "<form name='edit_" . $data['lid'] . "' id='edit_" . $data['lid'] . "' action=''><tr><td>" . $a++ . "</td><td>" . $data['language'] . "</td><td>" . $data['varid'] . "</td><td>" . $data['varname'] . "</td><td><textarea class='form-control' name='varval' id='varval' cols='100' wrap='virtual'>" . $data['varval'] . "</textarea></td><td id='editresponse'><button type='submit' class='button' type='submit' id='but_" . $data['lid'] . "'>Edit</button></td></tr></form>";
?>
//jquery to post edits
<script language="javascript">
$(document).ready(function()
{ //using $.ajax() function
//alert("document ready"); //this alerts each row
//this function will not work, no errors, simply nothing
$(document).on("submit", "#edit_<?php echo $data["lid"]; ?>", function(e)
{
//alert("button clicked"); //this does not work
var data = $("#edit_<?php echo $data["lid"]; ?>").serialize();
$.ajax({
type : "POST",
url : "/lang/ajax_langs.php",
data : data,
success : function(response)
{
{
$("#editresponse").html(response);
};
}
});
e.preventDefault();
});
});
</script>
<?php
}
echo "</table>";
}
?>
How do I get the jQuery to submit the individual form in each row to effect an edit and then refresh each row individually as languages are edited? I need to add another form to each row to delete an entry then remove the row dynamically, but I am stuck here without being able to submit any of these forms.
I should add that this generated form shows well in the website, but is not shown in view source. Using chrome browser. I'm open to any suggestions. Thanks in advance for reading. I seriously hope I can find a solution here. Otherwise, I'm back to the 90's on this software.
It's not working because you are only submitting one id. You need to make each of your forms be of the same class. Therefore, any form clicked will be processed. Like
<form class="myForm" .....
Then, the Jquery would look like this
$(".myForm").submit(function(e){
var inputToChange = this.varval; // saw input varVal, but use whatev
var dataToChange = inputToChange.value;
// then send the data to php for processing using ajax or post
Then, use inputToChange.value = whatever new value you want to put into the form input or other div/td etc. you want to change. If you want to change data that isn't in an input, just make that td a text input and disable it from being edited by the client
<input type="text" name="lname" disabled>
then on callback send new data to this.lname.value = (many other ways to do this as well)
Don't forget, if you send through ajax you want to stop the form itself from submitting so use e.preventDefault()

Set_value fallback in code igniter to obtain integer from URL address

I'm using the codeigniter framework in order to update records from a database, where the user first selects the ID of the item from a drop down list on the form
This ID gets added onto the URL where all of the relevant information for that item is obtained from the database.
I used a small bit of javascript in order to achieve the ID being added onto the URL.
My problem is that even though I have the set_value() field on the form to the ItemID, it still doesn't retrain the ID within the drop down box after the page reloads?
I need the ID to stay inside of the drop down list, as well as being on the URL at the top of the page
here is my code for the form
<?php echo form_dropdown('ItemID', $ItemIDListFromDatabase,
set_value('ItemID', 1), 'id="ItemID"'); ?>
<script type="text/javascript">
$( '#ItemID' ).on( 'change', function( e ){
document.location.href = "<?php echo site_url('site/Myform) ?>" + "/" + $( this ).val();
});
</script>
As you can see above I've tried adding a random fallback number to this set_value call, which works but only to update the item with ID of 1, I need it to be able to update the Item corresponding to the ID on the URL
I hope this makes sense, all help appreciated thanks.
You can do this by very simply using
$this->uri->segment()
Without knowing your full URL address, I don't know the exact number to put into the segment, but going by your code supplied just change the following to this:
<?php echo form_dropdown('ItemID', $ItemIDListFromDatabase,
set_value('ItemID', $this->uri->segment(3)), 'id="ItemID"'); ?>
This will obtain the value within segment 3 of your URL.
A quick tip if you can't figure out what the correct URI segment should be, you can test it out by echoing different values until you find the correct one
E.g.
<?php echo $this->uri->segment(3); ?>
This will display what ever is displayed on your third segment of the URL

Populating drop downs with MySQL entries

I have a MySQL database that contains, amongst other things, 2 tables. One is the events table, containing event names and other details. The other is the instance table. This table links the events table to a venue table and adds a date, so each row is an instance of the linked event.
I am making an event booking form for internal use for these events. I want to allow selection of the event to be booked via a dropdown list. So, I have populated one dropdown with the event names:
$qEvent = "SELECT event_name, event_id FROM events";
$rEvent = mysqli_query($dbc,$qEvent);
echo '<select>';
while ($row = mysqli_fetch_assoc($rEvent)) {
echo '<option value="'.$row['event_id'].'">'.$row['event_name'].'</option>';
}
echo '</select>';
What I now want to do is, for the selected event, grab all the instances associated with that event, and populate another dropdown with the dates.
Can I do this with PHP, or do I need to dip into Javascript? I think I just need some way to grab the event_id value of the dropdown selection and then query based on that, but I don't know how without Javascript.
You should be looking at Javascript or jQuery for achieving your goal. I've used jQuery based on my question to you earlier. It's also simpler and less code.
Your PHP:
Add an ID attribute event_menu to your select menu
echo '<select id="event_menu">';
while ($row = mysqli_fetch_assoc($rEvent)) {
echo '<option value="'.$row['event_id'].'">'.$row['event_name'].'</option>';
}
echo '</select>';
<div id="container_for_new_menu"></div>
Using jQuery:
$('#event_menu').on('change', function() {
// get selected value and build data string for AJAX
var event_selected = "event_selected="+$(this).val();
// send the selected data to a PHP page to build the populated menu
$.ajax({
url : 'populate-menu.php',
type: 'POST',
data : event_selected,
dataType : 'html',
success : function(data) {
$('#container_for_new_menu').html(data);
}, error : function() {
alert("Something went wrong!");
}
});
});
On populate-menu.php, have something like:
$event_selected = isset($_POST['event_selected']) ? $_POST['event_selected'] : null;
// do SQL query here based on user's selection
// making sure you validate the data in the POST request for malicious BS
// or use parameterized queries
// then build a new menu to send back
echo '<select>';
// loop through results and build options
echo '</select>';
This new menu will then be posted back to your original page into the container_for_new_menu element.
By the looks of it, you want to populate the "instances" dropdown based on the selection the user makes on the "event" dropdown. You cannot do this without Javascript.
My suggested way of doing this is to use AJAX to pull the instance data and populate the "instances" dropdown on change of the "event" dropdown. Useful resources below for simple AJAX get with jQuery:
http://api.jquery.com/jQuery.get/
http://remysharp.com/2007/01/20/auto-populating-select-boxes-using-jquery-ajax/
You need some kind of Javascript to accomplish this. Either:
Basic- submit the form on select and let php populate the instance drop-down.
More elegant- use Javascript to make an Ajax call on select which will dynamically replace the instance drop-down's div.
You will need JavaScript to populate the second drop down box. I suggest you load all the values into JSON on the page and then you can just use a jQuery on change event to populate the second select box.

PHP multiple records insert

I am attempting to reword my issue.
I have a datatable that can return thousands of records, each with multiple columns. There is a checkbox in the first column that, once the user checks it, they then click a button, and the CONTAINER_NUMBER that is associated with the row is sent to a modal window to be used in a form.
Here is the code for the checkbox:
echo "<tr><td><input type=\"checkbox\" id=\"{$Row[CONTAINER_NUMBER]}\" name=\"checkMr[]\" /></td>";
This is the javascript that retrieves the CONTAINER_NUMBER and sends it to the modal window:
<script type="text/javascript">
$(function()
{
$('a').click(function()
{
var selectedID = [];
$(':checkbox[name="checkMr[]"]:checked').each(function()
{
selectedID.push($(this).attr('id'))
});
$(".modal-body .containerNumber").val( selectedID );
});
});
</script>
This is the section of the modal window that displays the CONTAINER_NUMBER:
<div class="modal-body">
<form action="" method="POST" id="serviceModalForm" name="serviceModalForm">
<input type="text" name="containerNumber" id="containerNumber" class="containerNumber">
Here is the section of PHP that takes the id="containerNumber" and converts it to a PHP variable. After that, there is an INSERT statement that inserts the containerNumber into a database table:
<?php
$container = $_POST['containerNumber'];
if(isset($_POST['submit'])){
$container = mysql_real_escapse_string(stripslashes($container));
$sql = "INSERT INTO myTable (container_num) VALUES ('$container')";
if(mysql_query($sql)){
echo "Insert complete";
}
else {
echo "Insert was not completed";
}
?>
This code is fine. It works good. It does what it's supposed to do...for when the user checks ONE checkbox. It DOES NOT work when the user checks multiple checkboxes.
Basically, from what I've been researching is that I need to separate the records from the variable $container, as there can be multiple containers in that variable, which is why the query does not work when there are more than one container numbers selected.
I need to be able to separate the container numbers and store them in an array or something. The query will read each record separately and generate multiple INSERT statements for each record.
I've tried several times to create an array and get the sql statement to recognize it, but have been unsuccessful. I'm not sure if I'm placing the array in the right place. I'm not sure if this has to be done in the javascript before the container gets sent to the modal window.
I know I need to utilize a FOREACH loop to go through the array, but like I said, I'm not sure where the array needs to go in my code.
Please help. I know I need to learn PDO or MYSQLI. I will be sure to utilize PDO or MYSQLI on my next application. Until then, please help me with this issue.
Thank you, and sorry for so much wording.
Your containerNumber will be posted as a converted string from a js array. Something like id1, id2, id3[...]
In your php code, convert the $container back to an array ($containerArray = explode(",", $container)) and construct the sql dynamically to add all the rows in a single query so that the statment becomes something like
INSERT INTO myTable (container_num) VALUES ('$containerArray[0]'), ('$containerArray[1]')[...]

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