I don't know if it's possible, but I need to send some information across a form ou inside url come from checkbox value.
This code below is inside a products loop and create a checkbox on every products (product comparison approach).
In my case, it's impossible to make this code below across a form.
<?php
echo '<div><input type="checkbox" value="' . $products_id .'" id="productsCompare" title="Compare" onclick="showProductsCompare()" /> Compare</div>';
?>
To resolve this point, I started to use an ajax approach and put the result inside a $_SESSION
My script to for the checbox value
$(function() {
$('input[type=checkbox]').change(function() {
var chkArray = [];
$('#container').html('');
//put the selected checkboxes values in chkArray[]
$('input[type=checkbox]:checked').each(function() {
chkArray.push($(this).val());
});
//If chkArray is not empty create the list via ajax
if (chkArray.length !== 0) {
$.ajax({
method: 'POST',
url: 'http://localhost/ext/ajax/products_compare/compare.php',
data: { product_id: chkArray }
});
}
});
});
And at the end to send information on another page by this code. Like you can see there is no form in this case.
<div class="col-md-12" id="compare" style="display:none;">
<div class="separator"></div>
<div class="alert alert-info text-md-center">
<span class="text-md-center">
<button class="btn">Compare</button>
</span>
</div>
</div>
No problem, everything works fine except in my compare.php file, I have not the value of my ajax. I inserted a session_start in ajax file
But not value is inserted inside compare.php.
I tried different way, include session_start() inside compare.php not work.
My only solution is to include in my products file a hidden_field and include the value of ajax across an array dynamically, if it's possible.
In this case, values of hidden_fields must be under array and sent by a form.
This script must be rewritten to include under an array the chechbox value
without to use the ajax. How to insert the good code?
$(function() {
$('input[type=checkbox]').change(function() {
var chkArray = [];
$('#container').html('');
//put the selected checkboxes values in chkArray[]
$('input[type=checkbox]:checked').each(function() {
chkArray.push($(this).val());
});
//If chkArray is not empty show the <div> and create the list
if (chkArray.length !== 0) {
// Remove ajax
// some code here I suppose to create an array with the checkbox value when it is on true
}
});
});
and this code with a form
<?php
echo HTML::form('product_compare', $this->link(null, 'Compare&ProductsCompare'), 'post');
// Add all the js values inside an array dynamically
echo HTML::hidddenField('product_compare', $value_of_javascript);
?>
<div class="col-md-12" id="compare" style="display:none;">
<div class="separator"></div>
<div class="alert alert-info text-md-center">
<span class="text-md-center">
<button class="btn">Compare</button>
</span>
</div>
</div>
</form>
Note : this code below is not included inside the form (no change on that).
<?php
echo '<div><input type="checkbox" value="' . $products_id .'" id="productsCompare" title="Compare" onclick="showProductsCompare()" /> Compare</div>';
?>
My question is :
How to populate $value_of_javascript in function of the checkbox is set on true to send the information correctly inside compare.php
If my question has not enought information, I will edit this post and update in consequence.
Thank you.
You cannot pass JavaScript Objects to a server process. You need to pass your AJAX data as a String. You can use the JavaScript JSON.stringify() method for this...
$.ajax({
method: 'POST',
url : 'http://localhost/ext/ajax/products_compare/compare.php',
data : JSON.stringify({product_id: chkArray})
});
Once that has arrived at your PHP process you can turn it back into PHP-friendly data with PHP JSON methods...
<?
$myArray = json_decode($dataString, true);
// ... etc ... //
?>
See:
JSON # MDN
JSON # PHP Manual
Example: Form Submission Using Ajax, PHP and Javascript
Related
I am trying to display the data i retrieve from the database but it is not being displayed. I have a function in the file getComments.php called "getComments(page)" page is just a integer parameter to choose that database. and as you can see that i need to call this function to print the users comments. I am trying to use "load" but it is not being successful i just want to call this function to load the comments on the page. thank you in advance.
<?php
use TastyRecipes\Controller\SessionManager;
use TastyRecipes\Util\Util;
require_once '../../classes/TastyRecipes/Util/Util.php';
Util::init();
function getComments($page){
echo "<br><br>";
$controller = SessionManager::getController();
$controller->getComments($page);
SessionManager::setController($controller);
}
and in my web page where i want to display it using java script, i tried the following
<div class="page" id="comments">
<p class="style">Comments</p>
<button class="btn" id="load-comments">See Previous Comments</button><br>
<br><br>
<?php
if(isset($_SESSION['u_id'])){
echo " <input type='hidden' id='uid' value = '".$_SESSION['u_uid']."'>
<input type='hidden' id='date' value = '".date('Y-m-d H:i:s')."'>
<textarea id='message'></textarea><br>
<button class = 'btn' type = 'submit' id = 'submitCom'>Comment</button>";
}
else{
echo "<p>Please log in to comment</p>";
}
?>
</div><br>
<script>
$(document).ready(function(){
$("#load-comments").click(function(){
document.getElementById('#comments').innerHTML =
$("#comments").load("../extras/getComments.php", getComments(1));
});
});
</script>
Just change your click handler to this:
$("#load-comments").click(function(){
$("#comments").load("../extras/getComments.php", { page: 1 }); //i also added where the elements are loaded
});
and in getComments.php (if practical, otherwise you might need to create a new PHP file which calls the getComments() function and call that from the click handler instead) add something like:
if (isset($_POST['page'])) {
getComments($_POST['page']);
// do any other necessary stuff
exit;
}
i have problem this code not working , i want display result of mysql without send no data
my simple code ajax
$('.myClass').on('click',function(){
$.post('resultAll.php');
});
code html
<li class="myClass" > click this </li>
code php / mysql for display result on page (name resultAll.php)
$stmt = $connect->prepare('SELECT * FROM table WHERE price = :aff');
$stmt->execute(array(':aff'=>'5'));
$res = $stmt->fetchAll();
foreach ($res as $row) {
?>
<div class="noon" style="width: 1000px;height: 500px">
<?php print_r($row['id'].' '); ?>
</div>
result these nothing
As per your code snippet, you want to fetch data from resultAll.php file into your HTML page. If I am correct then your code is wrong, you have to use below method instead of $.post() method
$('.myClass').on('click',function(){
$("#result").load('resultAll.php');
});
Your HTML code should be
<li class="myClass" > click this </li>
....
...
<div id="result"></div>
I mean in your page must have one div tag with id="result".
I hope this is works for you.
As you say that your Jquery is not working. Is your element .myClass is created with JS dynamically? Ensure this:
$(document.body).on('click', '.myClass', function(){
$.post('resultAll.php');
});
Edit
Add response from $.post to body:
$('.myClass').click(function(){
$.post('resultAll.php', function(response){
$(body).append(response);
});
});
At the moment, I try to create a survey in a webpage. At the end of the survey, users are able to fill two text fields with values. With these values, my plan is to calculate an output, displayed for them at the same page. So:
Input: a
Input: b
Result: ab+b-ab (do not concentrate this please, its just an example)
My plan is that the user is able to fill the two input fields and by a buttonclick, a php function is calculating the result field (by my own algorithm depending on input - this is already working) and fills this field. Do i have to link to another webpage for this purpose?
And how is it possible to grab the two input values and give it to my php function?
And as last thing, how is it possible to start a php function either embedded in html or in an own file?
I tried your solution and some others as well (fetching inputA and inputB from the DOM with document.getElementById does not work. Below is my code
<form>
InputA:<br>
<input type="text" id="inputA"/></br>
InputB:<br>
<input type="text" id="inputB"/></br>
Output:<br>
<input type="text" id="output"/>
</form>
<input name="go" type="button" value="Calculate" id="calculate" >
<script type="text/javascript" src="http://code.jquery.com/jquery-2.1.4.min.js" ></script>
<script type="text/javascript">
$("#calculate").click(function(){
$.get( "submit.php", { value1: $("#inputA").val(), value2: $("#inputB").val() } )
.done(function( data ) {
$("#output").val(data);
});
});
</script>
submit.php:
<?php
$value1 = $_POST['value1'];
$value2 = $_POST['value2'];
$output = $value1 + $value2;
echo($output);
}
?>
When I check with firebug the error, i get a: no element found exception in both (html and php) files. Seems like the problem is, that with: value1: $("#inputA").val(); no value is givent to the server or it can not be handled there.
If i grab the value from the DOM, I can "bring" the value inside the .click function but there is still a "no element found exception" by calling the submit.php.
I have no idea what i am doing wrong, any suggestions? Do i need to install/bind anything in for using JQuery?
After some additional changes, it finally worked (one thing was the header line in the submit.php file):
<form>
WorkerID:<br>
<input type="text" id="workerId"/></br>
CampaignId:<br>
<input type="text" id="campaignId"/></br>
Payment Code:<br>
<input type="text" id="payCode"/>
</form>
<input name="go" type="button" value="Calculate" id="calculate" >
<script type="text/javascript">
$("#calculate").click(function(){
$.get( 'submit.php', { wId: $('#workerId').val(), cId: $('#campaignId').val()} )
.done(function( data ) {
$('#payCode').val(data.payCode);
});
});
and submit.php:
<?php
header('Content-Type: text/json');
$workerId = $_GET['wId'];
$campaignId = $_GET['cId'];
$payCode = $campaignId . $workerId;
$result = array("status" => "success",
"payCode" => $payCode);
echo json_encode($result);
?>
To simplify, i am using jQuery, doing this in vanilla JS is a real pain in the a** in my opinion.
You can use .get(), which is the GET shorthand for .ajax().
With that code, you bind a handler on your submit button and make a AJAX request to your PHP and fill the result your PHP gives into your result field asynchronously.
$("#calculate").click(function(){
$.get( "path/to/your_php.php", { value1: $("#inputA").val(), value2: $("#inputB").val() } )
.done(function( data ) {
$("#output").val(data);
});
});
Also change your submit to something like this:
<input name="go" type="button" value="Calculate" id="calculate" >
Like that, your button won't submit a form and therefore synchronously load your PHP.
Since you seem new to JavaScript and you had this comment
my button, but here i got redirected to submit, no idea how i can go back to page before with filled textfield
in your question, i'll tell you, JavaScript works while the DOM (Document Object Model) is loaded, means you can access your elements when already loaded and alter them.
Getting the value of a input is as easy as that in jQuery:
$("#inputA").val();
With the AJAX you get what your php will return in data.
// the { value1: $("#inputA").val(), value2: $("#inputB").val() } object
// is what you send to your PHP and process it
$.get( "path/to/your_php.php", { value1: $("#inputA").val(), value2: $("#inputB").val() } )
.done(function( data ) {
// data is what your php function returned
});
Using JS you can now change your elements as just said, effectively meaning to change the value of your output here:
$("#output").val(data);
"Working" Example: JSFiddle (There is no PHP to access to, so it will not do anything actively)
A grid table is displayed via PHP/MySQL that has a column for a checkbox that the user will check. The name is "checkMr[]", shown here:
echo "<tr><td>
<input type=\"checkbox\" id=\"{$Row[CONTAINER_NUMBER]}\"
data-info=\"{$Row[BOL_NUMBER]}\" data-to=\"{$Row[TO_NUMBER]}\"
name=\"checkMr[]\" />
</td>";
As you will notice, there is are attributes for id, data-info, and data-to that are sent to a modal window. Here is the JavaScript that sends the attributes to the modal window:
<script type="text/javascript">
$(function()
{
$('a').click(function()
{
var selectedID = [];
var selectedBL = [];
var selectedTO = [];
$(':checkbox[name="checkMr[]"]:checked').each(function()
{
selectedID.push($(this).attr('id'))
selectedBL.push($(this).attr('data-info'))
selectedTO.push($(this).attr('data-to'))
});
$(".modal-body .containerNumber").val( selectedID );
$(".modal-body .bolNumber").val( selectedBL );
$(".modal-body .toNumber").val( selectedTO );
});
});
</script>
So far so good. The modal retrieves the attributes via javascript. I can choose to display them or not. Here is how the modal retrieves the attributes:
<div id="myModal">
<div class="modal-body">
<form action="" method="POST" name="modalForm">
<input type="hidden" name="containerNumber" class="containerNumber" id="containerNumber" />
<input type="hidden" name="bolNumber" class="bolNumber" id="bolNumber" />
<input type="hidden" name="toNumber" class="toNumber" id="toNumber" />
</form>
</div>
</div>
There are additional fields within the form that the user will enter data, I just chose not to display the code. But so far, everything works. There is a submit button that then sends the form data to PHP variables. There is a mysql INSERT statement that then updates the necessary table.
Here is the PHP code (within the modal window):
<?php
$bol = $_POST['bolNumber'];
$container = $_POST['containerNumber'];
$to = $_POST['toNumber'];
if(isset($_POST['submit'])){
$bol = mysql_real_escape_string(stripslashes($bol));
$container = mysql_real_escape_string(stripslashes($container));
$to = mysql_real_escape_string(stripslashes($to));
$sql_query_string =
"INSERT INTO myTable (bol, container_num, to_num)
VALUES ('$bol', '$container', '$to')
}
if(mysql_query($sql_query_string)){
echo ("<script language='javascript'>
window.alert('Saved')
</script>");
}
else{
echo ("<script language='javascript'>
window.alert('Not Saved')
</script>");
}
?>
All of this works. The user checks a checkbox, the modal window opens, the user fills out additional form fields, hits save, and as long as there are no issues, the appropriate window will pop and say "Saved."
Here is the issue: when the user checks MULTIPLE checkboxes, the modal does indeed retrieve multiple container numbers and I can display it. They seem to be already separated by a comma.
The problem comes when the PHP variables are holding multiple container numbers (or bol numbers). The container numbers need to be separated, and I guess there has to be a way the PHP can automatically create multiple INSERT statements for each container number.
I know the variables need to be placed in an array somehow. And then there has to be a FOR loop that will read each container and separate them if there is a comma.
I just don't know how to do this.
When you send array values over HTTP as with [], they will already be arrays in PHP, so you can already iterate over them:
foreach ($_POST['bol'] as $bol) {
"INSERT INTO bol VALUES ('$bol')";
}
Your queries are vulnerable to injection. You should be using properly parameterized queries with PDO/mysqli
Assuming the *_NUMBER variables as keys directly below are integers, use:
echo '<tr><td><input type="checkbox" value="'.json_encode(array('CONTAINER_NUMBER' => $Row[CONTAINER_NUMBER], 'BOL_NUMBER' => $Row[BOL_NUMBER], 'TO_NUMBER' => $Row[TO_NUMBER])).'" name="checkMr[]" /></td>';
Then...
$('a#specifyAnchor').click(function() {
var selectedCollection = [];
$(':checkbox[name="checkMr[]"]:checked').each(function() {
selectedCollection.push($(this).val());
});
$(".modal-body #checkboxCollections").val( selectedCollection );
});
Then...
<form action="" method="POST" name="modalForm">
<input type="hidden" name="checkboxCollections" id="checkboxCollections" />
Then...
<?php
$cc = $_POST['checkboxCollections'];
if (isset($_POST['submit'])) {
foreach ($cc as $v) {
$arr = json_decode($v);
$query = sprintf("INSERT INTO myTable (bol, container_num, to_num) VALUES ('%s', '%s', '%s')", $arr['BOL_NUMBER'], $arr['CONTAINER_NUMBER'], $arr['TO_NUMBER']);
// If query fails, do this...
// Else...
}
}
?>
Some caveats:
Notice the selector I used for your previous $('a').click() function. Do this so your form updates only when a specific link is clicked.
I removed your mysql_real_escape_string functions due to laziness. Make sure your data can be inserted into the table correctly.
Make sure you protect yourself against SQL injection vulnerabilities.
Be sure to test my code. You may have to change some things but understand the big picture here.
I have an application (drag and drop using JqueryUI.GridSort) that allows the user to upload photos, and then sort the photos in the order that they would like using drag and drop.
On page load, the user is prompted to upload photos which are posted to the next page. When they arrive on the next page my php script creates a <ul id="sortable"> containing <li> for each of the files they uploaded. For each picture that they have uploaded to the site, a new <li> is created. Inside of that <li> is a <img> that sets the picture for <li> with the image they have uploaded.
My goal is to be able to "save" the order of the pictures after they have arranged them in the drag and drop interface. For example, once they have finished arranging and sorting the pictures in the order they want them in, I would like to be able to send them another page that creates an xml file ( I don't need help with the XML, only saving the order) with using the list that they created in the correct order.
After hours of tinkering with PHP, I have come to realization that because PHP is a serverside language, it cannot see what is sorted post render. So my question is, is there a way to have JavaScript or Ajax read the current order of the list, and post it to the next page? If you do know how, could you please provide an example of both the POST from one page, and the post receiving on the other? I am not very familiar with Ajax.
Thank you greatly for any assistance you could provide.
Sample Code (The contents of the foreach statement that creates a LI for each file uploaded)
$imgID++;
echo '<li class="ui-state-default"><img id="'.$imgID.'"'.' src="user_files/'.$file_name.'" draggable="true" height="90" width="95"></li>';
EDIT
main page :
<script>
$('#my_form').on('submit', function() {
var ordered_list = [];
$("#sortable li img").each(function() {
ordered_list.push($(this).attr('id'));
});
$("#ordered_list_data").val(JSON.stringify(ordered_list));
});
</script>
<div id="tesT">
<form id="my_form" action="update_data.php">
<!-- other fields -->
<input type="hidden" id="ordered_list_data"></input>
<input type="submit" value="Proceed to Step 2"></input>
</form>
</div>
update_data.php:
<?php
// process other fields as normal
if(isset($_POST['ordered_list_data'])) {
$img_ordering = json_decode($_POST['ordered_list_data']);
echo "1";
} else {
echo "nodata";
}
// do things with the data
?>
I built a JSFiddle doing basically the same thing that David posted.
I added a piece to write out the result to a div on the page, so you can see what's going on:
<input type="button" id="savebutton" value="save"/>
<div id="output"></div>
<form id="listsaveform" method="POST" action="script.php">
<input type="hidden" name="list" id="hiddenListInput" />
</form>
Javascript:
$(function() {
$( "#sortable" ).sortable();
$( "#sortable" ).disableSelection();
$( "#savebutton" ).click(function() { LISTOBJ.saveList(); });
});
var LISTOBJ = {
saveList: function() {
var listCSV = "";
$( "#sortable li" ).each(function() {
if (listCSV === "") {
listCSV = $(this).text();
} else {
listCSV += "," + $(this).text();
}
});
$("#output").text(listCSV);
$("#hiddenListInput").val(listCSV);
//$("#listsaveform").submit();
}
}
If you're using a <form> you can do something like this (assuming jQuery is being used):
$('#my_form').on('submit', function() {
var ordered_list = [];
$("#sortable li img").each(function() {
ordered_list.push($(this).attr('id'));
});
$("#ordered_list_data").val(JSON.stringify(ordered_list));
});
In essence, what you're doing is looping over the <ul>, fetching each <img> and appending the ids (in order of appearance) to an array. Arrays preserve ordering in JavaScript and JSON, so one can turn it into a JSON string using the JSON.stringify function, set it as the value of a <input type="hidden"> field and then submit the form.
If you want to use AJAX, the functionality is very similar. However, instead of using an onsubmit (or onclick) you'd use $.post.
Let's go with the <form> option since it's simpler. All told you'll have something similar to the above JS along with HTML like this:
<form id="my_form" method="post" action="./update_data.php">
<!-- other fields -->
<input type="hidden" name="ordered_list_data" id="ordered_list_data"></input>
<input type="submit" value="Submit"></input>
</form>
Then, in update_data.php (or whatever your script is named):
<?php
// process other fields as normal
if(isset($_POST['ordered_list_data'])) {
$img_ordering = json_decode($_POST['ordered_list_data']);
} else {
// handle case where there is no data
}
// do things with the data
?>