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Ajax call to php, get mysql data as array and use in JS function

I'm looking to make an ajax call to a PHP script to get data from MySQL, create a json array and pass it back to the success function of the ajax call, where i will then use it as parameters for a JavaScript function.
This is my ajax call,
$('button[name="message"]').click(function() {
var $row = $(this).closest("tr"); // Find the row
var $tenant_id = $row.find(".col-md-1 id").text(); // Find the tenants ID
var $landlord_id = "<?php echo $id; ?>"
$.ajax({
url : "./message.php",
type : "POST",
async : false,
data: {
landlord_id: $landlord_id,
tenant_id : $tenant_id
},
success: function(data){
console.log(data);
var messages = data;
insertChat(messages.sender_id, messages.body, messages.timestamp);
}
})
});
And this is my PHP file,
<?php
session_start();
require_once('../dbconnect.php');
// update tenants table to show deposit returned
if(isset($_POST['tenant_id'])){
$tenant_id = $_POST['tenant_id'];
$landlord_id = $_POST['landlord_id'];
$sql = "SELECT * from messages WHERE messages.sender_id OR messages.receiver_id = '$tenant_id' AND messages.sender_id OR messages.receiver_id = '$landlord_id'";
$result = mysqli_query($conn, $sql) or die("Error in Selecting " . mysqli_error($conn));
//create an array
$messages = array();
while($row =mysqli_fetch_assoc($result))
{
$messages[] = $row;
}
echo json_encode($messages);
}
?>
If anybody has a link to a tutorial or the individual parts that would be fantastic. I don't even know if the process i have outlined above is correct.
If anybody could tell me the correct way to go about this that would be of great help!
Thanks

Just a few things to adjust your javascript side (I won't explain the php sql injection issue you have... but please research prepare, bind_param and execute):
Since you are returning an ARRAY of $messages from php (json_encoded), you need to loop on those in your success handler.
Add dataType: 'JSON' to your options, so it explicitly expects json returned from php.
And you were missing a couple semicolons ;)
Adjustments added to your code:
$('button[name="message"]').click(function() {
var $row = $(this).closest("tr");
var tenant_id = $row.find(".col-md-1 id").text();
var landlord_id = "<?php echo $id; ?>";
$.ajax({
url : "./message.php",
type : "POST",
data: {
landlord_id: landlord_id,
tenant_id : tenant_id
},
dataType: 'JSON',
success: function(data){
console.log(data);
if (typeof data !== undefined) {
for(var i = 0; i < data.length; i++) {
insertChat(data[i].sender_id, data[i].body, data[i].timestamp);
}
}
}
});
});

Populating an array through jquery AJAX in php

I have a function in a compare.php that takes a parameter $data and uses that data to find certain things from web and extracts data and returns an array.
function populateTableA($data);
So to fill array I do this
$arrayTableA = populateTableA($name);
now this array is then used to iterate tables..
<table id="tableA">
<input type="text" name="search"/><input type="submit"/>
<?php foreach($arrayTableA as $row) { ?>
<tr>
<td><?php echo $row['name']?></td>
<td><?php echo $row['place']?></td>
</tr>
</table>
Now what I want to do is to enter some data on input and then through jquery ajax
function populateTableA($data);
should be called and $array should be refilled with new contents and then populated on tableA without refreshing the page.
I wrote this jquery but no results.
$(document).on('submit',function(e) {
e.preventDefault(); // Add it here
$.ajax({ url: 'compare.php',
var name = ('search').val();
data: {action: 'populateTableA(name)'},
type: 'post',
success: function(output) {
$array = output;
}
});
});
I have been doing web scraping and the above was to understand how to implement that strategy... original function in my php file is below
function homeshoppingExtractor($homeshoppingSearch)
{
$homeshoppinghtml = file_get_contents('https://homeshopping.pk/search.php?category%5B%5D=&search_query='.$homeshoppingSearch);
$homeshoppingDoc = new DOMDocument();
libxml_use_internal_errors(TRUE);
if(!empty($homeshoppinghtml)){
$homeshoppingDoc->loadHTML($homeshoppinghtml);
libxml_clear_errors();
$homeshoppingXPath = new DOMXPath($homeshoppingDoc);
//HomeShopping
$hsrow = $homeshoppingXPath->query('//a[#class=""]');
$hsrow2 = $homeshoppingXPath->query('//a[#class="price"]');
$hsrow3 = $homeshoppingXPath->query('(//a[#class="price"])//#href');
$hsrow4 = $homeshoppingXPath->query('(//img[#class="img-responsive imgcent"])//#src');
//HomeShopping
if($hsrow->length > 0){
$rowarray = array();
foreach($hsrow as $row){
$rowarray[]= $row->nodeValue;
// echo $row->nodeValue . "<br/>";
}
}
if($hsrow2->length > 0){
$row2array = array();
foreach($hsrow2 as $row2){
$row2array[]=$row2->nodeValue;
// echo $row2->nodeValue . "<br/>";
}
}
if($hsrow3->length > 0){
$row3array = array();
foreach($hsrow3 as $row3){
$row3array[]=$row3->nodeValue;
//echo $row3->nodeValue . "<br/>";
}
}
if($hsrow4->length > 0){
$row4array = array();
foreach($hsrow4 as $row4){
$row4array[]=$row4->nodeValue;
//echo $row3->nodeValue . "<br/>";
}
}
$hschecker = count($rowarray);
if($hschecker != 0) {
$homeshopping = array();
for($i=0; $i < count($rowarray); $i++){
$homeshopping[$i] = [
'name'=>$rowarray[$i],
'price'=>$row2array[$i],
'link'=>$row3array[$i],
'image'=>$row4array[$i]
];
}
}
else{
echo "no result found at homeshopping";
}
}
return $homeshopping;
}

As mentioned in the comments PHP is a server side language so you will be unable to run your PHP function from javascript.
However if you want to update tableA (without refreshing the whole page) you could create a new PHP page that will only create tableA and nothing else. Then you could use this ajax call (or something similar) -
$(document).on('submit','#formReviews',function(e) {
e.preventDefault();
$.ajax({
url: 'getTableA.php', //or whatever you choose to call your new page
data: {
name: $('search').val()
},
type: 'post',
success: function(output) {
$('#tableA').replaceWith(output); //replace "tableA" with the id of the table
},
error: function() {
//report that an error occurred
}
});
});

Hi You are doing it in wrong way.You must change your response to html table and overwrite older one.
success: function(output) {
$("#tableA").html(output);
}
});
In your ajax page create a table with your result array

You are in a very wrong direction my friend.
First of all there are some syntax error in your JS code.
So use JavaScript Debugging
to find where you went wrong.
After that Basic PHP with AJAX
to get a reference how ajax and PHP work together
Then at your code
Create a PHP file where you have to print the table part which you want to refresh.
Write an AJAX which will hit that PHP file and get the table structure from the server. So all the processing of data will be done by server AJAX is only used for request for the data and get the response from the server.
Put the result in your html code using JS.
Hope this will help

Ajax call posting duplicates in DB

I am using to Fullcalendar jquery with php for event management. I using ajax call for adding events. The call works fine for the first event entry after refresh. But for the following event entries duplicate events are created for each entry. Not sure what causing this.
This is the error:
This is the jquery call:
Jquery
$('#evesav').bind('click',function(){
$('#evesav').attr('disabled','disabled');
var title = $('#evename').val();
var edes = $('#evedes').val();
var everegion = $('#everegion').val();
var eveserv = $('#eveserv').val();
$.ajax({
url: 'add_events.php',
data: 'title='+ title+'&start='+ start +'&end='+ end +'&edes='+ edes +'&everegion='+ everegion +'&eveserv='+ eveserv,
type: "POST",
success: function(json) {
$('#myModal').modal('hide');
$('#alertcon').html(json);
$('#alert').modal('show');
$('#evename').val("");
$('#evedes').val("");
$('#evesav').removeAttr('disabled');
$('#calendar').fullCalendar( 'refetchEvents' );
}
});
$('#calendar').fullCalendar( 'rerenderEvents' );
});
This is the PHP Code:
PHP
<?php
if(($_POST['title'] && $_POST['start'] && $_POST['end'] && $_POST['edes'] && $_POST['everegion'] && $_POST['eveserv'])!= NULL)
{
// Values received via ajax
$title = $_POST['title'];
$start = $_POST['start'];
$end = $_POST['end'];
$edes = $_POST['edes'];
$region = $_POST['everegion'];
$server = $_POST['eveserv'];
//echo $title."".$start."".$end."".$edes."".$region."".$server;
// connection to the database
include('includes/db.php');
// insert the records
$sql = "INSERT INTO evenement (title, start, end, edes, region, server) VALUES (:title, :start, :end, :edes, :region, :server)";
$q = $bdd->prepare($sql);
$q->execute(array(':title'=>$title, ':start'=>$start, ':end'=>$end, ':edes'=>$edes, ':region'=>$region, ':server'=>$server));
if($q->execute(array(':title'=>$title, ':start'=>$start, ':end'=>$end, ':edes'=>$edes, ':region'=>$region, ':server'=>$server))){
var_dump($q->execute(array(':title'=>$title, ':start'=>$start, ':end'=>$end, ':edes'=>$edes, ':region'=>$region, ':server'=>$server)));
}
$eveid=$bdd->lastInsertId();
// Get array of all source files
$files = scandir("uploads/");
// Identify directories
$source = "uploads/";
$destination = "evedata/".$eveid."/";
mkdir("evedata/".$eveid);
// Cycle through all source files
foreach ($files as $file) {
if (in_array($file, array(".",".."))) continue;
// If we copied this successfully, mark it for deletion
if (copy($source.$file, $destination.$file)) {
$delete[] = $source.$file;
}
}
// Delete all successfully-copied files
foreach ($delete as $file) {
unlink($file);
}
echo "Added Successfully";
}
else {
echo "Please Fill the data";
}
?>
Some one please help me with this.

I'd give each event addition form a control, for instance a dynamic GUID, which then can be used to save to DB. This way you have a GUID to work with in dealing with CalDAV protocol, if you ever choose to do as such with your calendar, as well as have a way to make certain nothing is duplicated by chance in your database.
Now, do keep in mind this is simply a patch, not a fix. Therefore, you'll do yourself a lot of good to find a way to stop the multiple attempts to add an event to your DB. Regardless of your success in finding your bug, using a control mechanism or unique identifier is a good idea.

PHP is not reloaded automatically after processing

Hi I have a PHP file with data. The value is passed on to another php file which process it successfully. But the first php file does not refresh to update the new result. It have to do it manually. Can any one tell me where I'm wrong or what needs to be done. Please find my code below.
PHP code (1st page, index.php)
function display_tasks_from_table() //Displayes existing tasks from table
{
$conn = open_database_connection();
$sql = 'SELECT id, name FROM todolist';
mysql_select_db('todolist'); //Choosing the db is paramount
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}
echo "<form class='showexistingtasks' name='showexistingtasks' action='remove_task.php' method='post' >";
while($row = mysql_fetch_assoc($retval))
{
echo "<input class='checkbox' type='checkbox' name='checkboxes{$row['id']}' value='{$row['name']}' onclick='respToChkbox()' >{$row['name']} <img src='images/show_options.gif' /><br>";
}
echo "</form>";
echo "<label id='removeerrormsg'></label>";
close_database_connection($conn);
}
Javascript code which finds the selected value:
var selVal; //global variable
function respToChkbox()
{
var inputElements = document.getElementsByTagName('input'),
input_len = inputElements.length;
for (var i = 0; i<input_len; i++)
{
if (inputElements[i].checked === true)
{
selVal = inputElements[i].value;
}
}
}
jQuery code which passes value to another page (remove_Task.php):
$(document).ready(function() {
$(".checkbox").click(function(){
$.ajax({
type: "POST",
url: "remove_task.php", //This is the current doc
data: {sel:selVal, remsubmit:"1"},
success: function(data){
//alert(selVal);
//console.log(data);
}
});
});
});
PHP code (2nd page, remove_task.php);
session_start();
error_reporting(E_ALL);ini_set('display_errors', 'On');
$task_to_remove = $_POST['sel'];
function remove_from_list() //Removes a selected task from DB
{
$db_connection = open_database_connection();
global $task_to_remove;
mysql_select_db('todolist');
$sql = "DELETE FROM todolist WHERE name = "."'".$task_to_remove."'";
if($task_to_remove!='' || $task_to_remove!=null)
{
mysql_query($sql, $db_connection);
}
close_database_connection($db_connection);
header("Location: index.php");
}
if($task_to_remove != "") {
remove_from_list();
}
The selected value is getting deleted but the display on index.php is not updated automatically. I have to manually refresh to see the updated result. Any help would be appreciated.

By calling header("Location: index.php"); you don't redirect main page. You sent an ajax request - you can think about it as of opening a new page at the background, so this code redirects that page to index.php.
The better way to solve your task is to return status to your success function and remove items which were deleted from the database.
success: function(data){
if(data.success){
//remove deleted items
}
}

How to pass row name from php to ajax using jquery

I have a table in which the details are fetched from the DB.
if(mysql_num_rows($sql) > 0)
{
$row_count_n=1;
while($rows=mysql_fetch_assoc($sql))
{
extract($rows);
$options1 = select_data_as_options("project_resources", "name", $resource_allocated);
$options2 = select_data_as_options("project_roles", "name", $role);
echo "<tr>";
echo "<td><select name='ra_$row_count_n'><option value=''>-- Select --$options1</option></select></td>";
echo "<td><select name='role_$row_count_n'><option value=''>-- Select --$options2</option></select></td>";
echo "<td><input type='text' name='start_date_tentative_$row_count_n' class='date_one' value=$tentatively_starts_on /></td>";
echo "</tr>";
$row_count_n++;
}
}
I wanted to update the table when required, am doing this using Ajax by collecting data from the form using Jquery and saving it on button click.
$("#save_changes_id").click(function()
{
// To retrieve the current TAB and assign it to a variable ...
var curTab = $('.ui-tabs-active'); // in NEWER jQueryUI, this is now ui-tabs-active
var curTabPanelId = curTab.find("a").attr("href");
if(curTabPanelId == "#tab_dia")
{
var curTab = $('#sub_tabs .ui-tabs-active');
var curTabPanelId = curTab.find("a").attr("href");
}
responseData = doAjaxCall($(curTabPanelId + " form"));
if(responseData == 1)
showMessage('status_msg', 'Project details updated successfully', 'green');
else
showMessage('status_msg', 'Error: Please check all the fields', 'red');
});
function doAjaxCall(objForm)
{
var values = objForm.serialize();
$.ajax({
url: ajaxURL,
type: "post",
data: values,
async: false,
success: function(data)
{
responseData = data;
},
error:function()
{
alert('Connection error. Please contact administrator. Thanks.');
}
});
return responseData;
}
Ajax code is as below:
case "allocate_ba_details":
for($i=1; $i<=$row_count; $i++)
{
$resource = $_REQUEST["ra_$i"];
$role = $_REQUEST["role_$i"];
$start_date_tentative = $_REQUEST["start_date_tentative_$i"];
$already_available_check = mysql_num_rows(mysql_query("select * from project_allocate_ba where project_id = $pdid"));
if($already_available_check > 0)
{
$sql = ("UPDATE project_allocate_ba SET resource_allocated='$resource', role='$role', tentatively_starts_on='$start_date_tentative' WHERE project_id=$pdid");
}
}
echo $sql;
break;
As I am new to this am not sure how to pass the row name in order to update a particular row.
Please suggest a solution. Thanks in advance.

firstly use PDO or some php framework that has nice API to work with mysql. Second don't use success/error callback in jquery is too deprecated. Instanted use done/fail.always.
I understand that you want update row in html table data from the server ?
In success callback simply update the table using jquery text method for jquery object. You don't paste all code so i write example:
in server.php
<?php
[...]
$already_available_check = mysql_num_rows(mysql_query("select * from project_allocate_ba where project_id =" . intval($pdid)));
[...]
echo $already_available_check;
?>
This code return the integer, so in doAjaxCall:
function doAjaxCall(objForm)
{
var values = objForm.serialize();
$.ajax({
url: ajaxURL,
type: "post",
data: values,
async: false,
success: function(data)
{
if(typeof data !== 'undefined' && $.isNumeric(data)) {//check that server send correct anserw
$('whereIsData').text(data);
}
},
error:function()
{
alert('Connection error. Please contact administrator. Thanks.');
}
});
}
Now in success method you populate some DOM element using text method. You cannot simply return data from ajaxCall method because $.ajax is asynchronized method and responseData has value only when ajax request ends, so always return undefined in you example. You must present responseData to the user in success callback method.

For one thing...
$sql = ("UPDATE project_allocate_ba SET resource_allocated='$resource', role='$role', tentatively_starts_on='$start_date_tentative' WHERE project_id=$pdid")
needs single quotes around $pdid
Also don't echo the $sql. Instead do your inspection and form a response.
$response = array();
if(EVERYTHING_IS_GOOD){
$response['status'] = 'good to go';
}else{
$response['status'] = 'it went horribly wrong';
}
echo json_encode($response);