How can I call multiple PHP scripts with AJAX/JQuery? - php

I am new with AJAX and JQuery. I am trying to use it to call two PHP scripts. I found some examples online but just to call functions. I am just trying to call the scripts so it will load everything on my main PHP file that will then be display on the screen the results withouth refreshing the page.
Here is the fiddle example, it works if I put all my PHP scripts in one file : http://jsfiddle.net/vw4w3ay5/
thanks in advance, your help is very much appreciated!
main_php file (where I want to call my other PHP scripts):
<div id="map_size" align="center">
<script type="text/javascript">
/*I WANT TO CALL THE TWO SCRIPTS BEFORE EXECUTE THE FUNCTION BELOW*/
$(".desk_box").click( function() {
$(".station_info").hide(); // to hide all the others.
$("#station_info"+ $(this).attr('data') ).show();
});
</script>
display_desk.php (Script I want to call):
<?php
include 'db_conn.php';
//query to get X,Y coordinates from DB for the DESKS
$desk_coord_sql = "SELECT coordinate_id, x_coord, y_coord FROM coordinates";
$desk_coord_result = mysqli_query($conn,$desk_coord_sql);
//see if query is good
if($desk_coord_result === false) {
die(mysqli_error());
}
//didsplay Desk stations in the map
while($row = mysqli_fetch_assoc($desk_coord_result)){
//naming X,Y values
$id = $row['coordinate_id'];
$x_pos = $row['x_coord'];
$y_pos = $row['y_coord'];
//draw a box with a DIV at its X,Y coord
echo "<div class='desk_box' data='".$id."' style='position:absolute;left:".$x_pos."px;top:".$y_pos."px;'>id:".$id."</div>";
} //end while loop for desk_coord_result
mysqli_close($conn); // <-- DO I NEED TO INCLUDE IT HERE OR IN MY db_conn.php SINCE IM INCLUDING IT AT THE TOP?
?>
display_stationinfo.php(second script I want to call):
<?php
include 'db_conn.php';
//query to show workstation/desks information from DB for the DESKS
$station_sql = "SELECT coordinate_id, x_coord, y_coord, section_name FROM coordinates";
$station_result = mysqli_query($conn,$station_sql);
//see if query is good
if($station_result === false) {
die(mysqli_error());
}
//Display workstations information in a hidden DIV that is toggled
while($row = mysqli_fetch_assoc($station_result)){
//naming values
$id = $row['coordinate_id'];
$x_pos = $row['x_coord'];
$y_pos = $row['y_coord'];
$sec_name = $row['section_name'];
//display DIV with the content inside
echo "<div class='station_info' id='station_info".$id."' style='position:absolute;left:".$x_pos."px;top:".$y_pos."px;'>Hello the id is:".$id."</br>Section:".$sec_name."</br></div>";
}//end while loop for station_result
mysqli_close($conn); // <-- DO I NEED TO INCLUDE IT HERE OR IN MY db_conn.php SINCE IM INCLUDING IT AT THE TOP?
?>

What about? :
<div id="map_size" align="center">
<?php
echo "<script>";
include "display_desk.php";
include "display_stationinfo.php";
echo "</script>";
?>
<script type="text/javascript">
/*I WANT TO CALL THE TWO SCRIPTS BEFORE EXECUTE THE FUNCTION BELOW*/
$(".desk_box").click( function() {
$(".station_info").hide(); // to hide all the others.
$("#station_info"+ $(this).attr('data') ).show();
});
</script>
To be sure add $(document).ready(function(){
});
/EDIT/
Hum you want to use Ajax . did you try with :
$.post("yourURL.php",function(html){
/*here what you want to do*/
/*return of your script in html*/
});

Related

PHP drop down menu, once clicked

I have created a drop down menu in php that is displayed however, when a value has been clicked, I don't know how to collect this information.
<html>
<body>
<?php
$mydb = new mysqli('localhost','root','','TestTimeTableSolution');
$rows = $mydb->query("SELECT DISTINCT TeacherID FROM Teacher;")->fetch_all(MYSQLI_ASSOC);
$teachers = array();
foreach ($rows as $row) {
array_push($teachers, $row["TeacherID"]);
}
$dropdownMenu = "<select name='TeacherID' form='Teacher'><option value='Null' selected>All</option>";
foreach ($teachers as $topic) {
$dropdownMenu .= "<option value='" . $topic . "'>" . $topic . "</option>";
}
$dropdownMenu .= "</select>";
echo $dropdownMenu;
?>
</body>
</html>
Based on your last comment, "i want it to be dynamic so as soon as the user clicks on something the relevant information will pop up", it sounds like you will probably want to use Ajax/JavaScript (I will demonstrate a simple jQuery example, notating for clarity):
<?php
$mydb = new mysqli('localhost','root','','TestTimeTableSolution');
?>
<html>
<!-- Add the jQuery library -->
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script>
// Act when the document is ready
$(function(){
// listen for the select to change
$('select[name=TeacherID]').on('change',function(){
// Run the ajax – you can also use the shortcut $.post method found at:
// https://api.jquery.com/jquery.post/
$.ajax({
// This is the page that is going to do the data lookup/display action
url: '/lookup.php',
// This is how it's sending the data to that page
type: 'post',
// This is what is being sent ($_POST['submit'] in this case)
data: {
// Use $(this) to isolate the current selected element and get value (.val())
// The value is represented as $topic in your php
'submit': $(this).val()
},
// If all goes well and the page (lookup.php) returns a response
success: function(response) {
// Place the response into the div (see html snippet)
$('#loadspot').text(response);
}
});
});
});
</script>
<body>
<?php
$rows = $mydb->query("SELECT DISTINCT TeacherID FROM Teacher;")->fetch_all(MYSQLI_ASSOC);
$teachers = array();
foreach ($rows as $row) {
array_push($teachers, $row["TeacherID"]);
}
$dropdownMenu = "<select name='TeacherID' form='Teacher'><option value='Null' selected>All</option>";
foreach ($teachers as $topic) {
$dropdownMenu .= "<option value='" . $topic . "'>" . $topic . "</option>";
}
$dropdownMenu .= "</select>";
echo $dropdownMenu;
?>
<!---------------------------------------------->
<!-- THIS IS WHERE THE CONTENT WILL LOAD INTO -->
<!---------------------------------------------->
<div id="loadspot"></div>
</body>
</html>
In order for this to work, you need the page lookup.php in the domain root (you can make it whatever/where ever you want, but you need to match in the javascript url):
/lookup.php
<?php
# This is what will get placed into the parent page <div id="loadspot"></div>
# Do you your php here in place of this line and return whatever "relative information" you want
print_r($_POST);
You should review the jQuery page I have linked to to get more information and direction for that library and make sure you use your browser's developer tools to monitor javascript errors in the console. Ideally, you want to understand all this via the documentation instead of just copy and paste and move on...

Drop down list in PHP and MySQL

I need to make drop down list as a link to different pages. How do I do that using PHP, MySQL and HTML.
<?php
mysql_connect('localhost','root','');
mysql_select_db('test');
$sql="select first_name from users";
$result=mysql_query($sql);
echo "<select First_name=''>";
echo "<a href='index.html'>";
while($row=mysql_fetch_array($result)){
echo ":<option value='".$row['first_name']."'>".$row['first_name']."</option>";
}
echo"</a>";
echo"</select>";
?>
You can't use links on the option tag, in order to do that, you need to use javascript.
You can try to do something like this:
echo "<select name=\"First_name\" onchange=\"document.location='?'+this.value\">";
PHP is a server-side script and does not manipulate a page after a user has adjusted it. Like real time. Only javascript and others do that. PHP creates a page with what you want to see but if you need to change something bases on a dropdown use java. Here is a function that can do that. It unhides a div tag that can have your info you need.
<script type="text/javascript">
window.onload = function() {
var eSelect = document.getElementById('dropdown');
var divtag1 = document.getElementById('divtag1');
var divtag2 = document.getElementById('divtag2');
eSelect.onchange = function() {
if(eSelect.selectedIndex === 1) {
divtag1.style.display = 'block';
}
if(eSelect.selectedIndex === 2) {
divtag2.style.display = 'block';
}//or if you want it to open a url
if(eSelect.selectedIndex === 3) {
window.open("https://yourwebsite.com", "_NEW");
}
}
}
</script>
echo "<div id=\"divtag1\" style=\"display:none;\">/*your code*/
</div>";
echo "<div id=\"divtag2\" style=\"display:none;\">/*your code*/
</div>";

How to call a function with database data inside another function?

EDITED: with new code after help from Sgt AJ.
Ok, so I am learning all the time, but since my coder stopped coding for our website, I am now having to learn PHP fully myself.
And I see all the time the coding where my coder made function calls inside other function calls.
So first of all the setup, we have a file for pretty much 95% of all functions in our site. That functions file basically has about 40-50 functions in it.
So I'm asking if someone can explain to me how is this possible to call a function inside another which works in the below instance, but when I try replicate it, it doesn't work? displays no data when I try to echo out the $user_info?
Like for example this function below: So Sgt AJ helped me solve the user avatar issue, so that will be removed from this question!
function showComments($v)
{
$mysqli = db_connect();
$v = mysqli_real_escape_string($mysqli,$v);
$sql = "SELECT * FROM `cl45-tbn_dir`.`comments` WHERE `v` = ? ORDER BY `id` ASC";
$stmt = $mysqli->prepare($sql);
$stmt->bind_param("s",$v);
$stmt->execute();
$result = $stmt->get_result();
while ($myrow = $result->fetch_assoc()) {
if ($myrow['post_approved']==1){
$user_info = getUserInfo($myrow['poster_id']);
if ($user_info['user_avatar_type']==1) {
$avatar = "https://www.tubenations.com/forum/download/file.php?avatar=".$user_info['user_avatar'];
} else {
$avatar = "https://www.tubenations.com/forum/styles/Flato%20-%20LightBlue%20-%20Main%20Style/theme/images/no_avatar.gif";
}
echo '<div class="comment">
<div class="avatar">
<a href="https://www.tubenations.com/users.php?id='.$myrow['poster_id'].'">
<img src="'.$avatar.'" />
</div>
<div class="name"><a class ="myaccount'.$user_info['group_id'].'" href="https://www.tubenations.com/users.php?id='.$myrow['poster_id'].'">'.$user_info['username'].'</a></div>
<div class="date" title="report this post">'.date("d M Y",$myrow['post_time']).'<form action="" class="flag" method="post"><button type="submit" value="'.$myrow['id'].'" name="vote" id="votebutton" alt="vote"><img src="/images/flag.png" alt="report this post!" /></button></form></div>
<p>'.stripslashesFull(clean($myrow['post_text'])).'</p>
</div>';
}
}
$stmt->close();
$mysqli->close();
}
As you can see, there is a line where it calls another function getUserInfo, $user_info = getUserInfo($myrow['poster_id']); that is another function inside this file, and that basically connects to our forum database and gets data.
But when I try to replicate this method by using this type of call within another, it doesn't work.
So basically what I was trying to play with was trying to make a function for displaying X users data with this below function
function getYouTubeInfo($page)
{
#$id = $_GET['id'];
print_r ($userdata['user_id']);
echo $myrow['user_id'];
echo $userdata['user_id'];
$db_link = mysqli_connect ('localhost', 'HIDDEN', 'HIDDEN', 'HIDDEN');
if ( !$db_link )
{
die('following error occured: '.mysqli_error());
}
$query = "SELECT user_id, yt_channelTitle, channel_id FROM points WHERE channel_id IS NOT NULL AND yt_channelTitle IS NOT NULL ORDER BY channel_id DESC;";
if($result = mysqli_query($db_link, $query)){
echo "";
$i = -1;
$objectsPerPage = 14;
$show_records = FALSE;
while ($row = $result->fetch_assoc())
{
if (!isset($_SESSION['last_slide'])) { $_SESSION['last_slide'] = $row['channel_id']; }
if ($row['channel_id'] == $_SESSION['last_slide']) { $show_records = TRUE; }
if ($show_records)
{
$i = $i+1;
if ($i > $objectsPerPage) { $_SESSION['last_slide'] = $row['channel_id']; echo 'BREAK: ', $row['channel_id']; break; }
$page = abs(floor($i/$objectsPerPage));
$youtube_info = $row;
$userdata = getUserInfo($row['user_id']);
if ($userdata['user_avatar_type']==1) {
$avatar = "/forum/download/file.php?avatar=".$userdata['user_avatar'];
} else {
$avatar = "/images/no_image.png";
}
if (($i/$objectsPerPage)==$page)
{
if ($page !=0) {
echo "</div></div>";
}
echo '<div class="cslide-slide">
<div class="slideTitles">Youtube Users Slide '.$page.'</div>
<div class="sections grouped">';
}
echo '
<div class="cols span_1_of_2">
<div class="memberTitles">'.$youtube_info['yt_channelTitle'].''.$i.';</div>
<div class="memberPicture"><img src="'.$avatar.'" title="Tube Nations Profile Picture" alt="Tube Nations Profile Picture"/></div>
<div class="memberTwitter"><div class="g-ytsubscribe" data-channelid="'.$youtube_info['channel_id'].'" data-layout="full" data-count="default" data-onytevent="onYtEvent"></div></div>
</div> ';
}
}
echo '</div></div>';
}
mysqli_free_result($result);
echo $_SESSION['last_slide'];
session_destroy();
mysqli_close($db_link);
}
So basically in the page in question, youtube.php, I just echo this getYouTubeInfo function.
This function I need to try get the users profile pictures that are in the forum database which is from the getUserInfo($id).
Also on a side note, I also can not work out how to re arrange the $i and $objectsPerPage variables and if statements so I can then use the $page inside the query LIMIT $page; because at the moment the page crashes with no limit, so I have had to put limit to 16 for now.
I use a jQuery slide script for displaying X per slide, so if I can just somehow work out how to make the query further down after the variables and if statements for the page stuff or get any help, I would appreciate it.
EDIT UPDATED REPLY: So now the problem is it now displays X per slide/page, but it now displays a gap after 8 results are displayed when it shows 10, but with a gap, and then on the the next slide button isn't showing up? so Sgt AJ said we need to somehow connect it to the jquery?, so I now will add a tag for jquery. (But can i say a big thanks to Sgt AJ for his help, really appreciate it) :)
Wow, you've got several things going on here.
First, your query right now says LIMIT 0;, which means you should get zero rows returned. Are you getting data back from this query??
Second, to get the page and items per page working right, you could go with something like this:
In your loop, keep your i=i+1 line
Add this if:
if ($i == $objectsPerPage)
{
++$page;
i = 1;
}
This will increment the page counter once the page is full, then reset the item count for the next page.
I just wanted to add more to my question, I think now the answer is with AJAX, so I think I need to somehow make the cslide jquery code recall the getYoutubeInfo($page) function
the jquery code for the cslide is this:
(function($) {
$.fn.cslide = function() {
this.each(function() {
var slidesContainerId = "#"+($(this).attr("id"));
var len = $(slidesContainerId+" .cslide-slide").size(); // get number of slides
var slidesContainerWidth = len*100+"%"; // get width of the slide container
var slideWidth = (100/len)+"%"; // get width of the slides
// set slide container width
$(slidesContainerId+" .cslide-slides-container").css({
width : slidesContainerWidth,
visibility : "visible"
});
// set slide width
$(".cslide-slide").css({
width : slideWidth
});
// add correct classes to first and last slide
$(slidesContainerId+" .cslide-slides-container .cslide-slide").last().addClass("cslide-last");
$(slidesContainerId+" .cslide-slides-container .cslide-slide").first().addClass("cslide-first cslide-active");
// initially disable the previous arrow cuz we start on the first slide
$(slidesContainerId+" .cslide-prev").addClass("cslide-disabled");
// if first slide is last slide, hide the prev-next navigation
if (!$(slidesContainerId+" .cslide-slide.cslide-active.cslide-first").hasClass("cslide-last")) {
$(slidesContainerId+" .cslide-prev-next").css({
display : "block"
});
}
// handle the next clicking functionality
$(slidesContainerId+" .cslide-next").click(function(){
var i = $(slidesContainerId+" .cslide-slide.cslide-active").index();
var n = i+1;
var slideLeft = "-"+n*100+"%";
if (!$(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-last")) {
$(slidesContainerId+" .cslide-slide.cslide-active").removeClass("cslide-active").next(".cslide-slide").addClass("cslide-active");
$(slidesContainerId+" .cslide-slides-container").animate({
marginLeft : slideLeft
},250);
if ($(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-last")) {
$(slidesContainerId+" .cslide-next").addClass("cslide-disabled");
}
}
if ((!$(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-first")) && $(".cslide-prev").hasClass("cslide-disabled")) {
$(slidesContainerId+" .cslide-prev").removeClass("cslide-disabled");
}
});
// handle the prev clicking functionality
$(slidesContainerId+" .cslide-prev").click(function(){
var i = $(slidesContainerId+" .cslide-slide.cslide-active").index();
var n = i-1;
var slideRight = "-"+n*100+"%";
if (!$(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-first")) {
$(slidesContainerId+" .cslide-slide.cslide-active").removeClass("cslide-active").prev(".cslide-slide").addClass("cslide-active");
$(slidesContainerId+" .cslide-slides-container").animate({
marginLeft : slideRight
},250);
if ($(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-first")) {
$(slidesContainerId+" .cslide-prev").addClass("cslide-disabled");
}
}
if ((!$(slidesContainerId+" .cslide-slide.cslide-active").hasClass("cslide-last")) && $(".cslide-next").hasClass("cslide-disabled")) {
$(slidesContainerId+" .cslide-next").removeClass("cslide-disabled");
}
});
});
// return this for chainability
return this;
}
}(jQuery));
I also tweaked the code that Sgt AJ helped me with again, By adding a session_destroy() just before the closing brace. And also a few other bits, because I noticed that when you refreshed the page over and over, it just loaded the next 14 results, instead of the same first 14 results, so the actual code and logic seems to be working. so it is basically now down to we need to find a way to use AJAX and recall the function from the onclick event of the next/previous buttons.

How to use php fetched data into html using jquery post

Hi i have a function in jquery and using $.Post to send data on a php file where my query is working fine and sending data back
js
function send_agenda_data(cidade_data){
var data = {'cidade_data':cidade_data};
$.post('includes/agenda_data.php',data,function(info){
});
}
This function works fine and when i alert the data coming back that also works fine
here is php
<?php
include_once("connection.php");
$cidade_data = $_POST['cidade_data'];
if (isset($cidade_data)) {
$sql = mysql_query("select * from agenda where cidade = '$cidade_data'", $con) or die(mysql_error());
if (mysql_num_rows($sql) > 0) {
while ($data = mysql_fetch_object($sql))
{
$date = $data->data;
$cidade = htmlentities($data->cidade);
$estado = htmlentities($data->estado);
$local = htmlentities($data->local);
$endereco = htmlentities($data->endereco);
$site_local = htmlentities($data->site_local);
$site_ingresso = htmlentities($data->site_ingresso);
$endereco = htmlentities($data->endereco);
}
}
else{
echo "No Data";
}
}
?>
this works fine if i use directly using php and echo the variables in tags now
$.post('includes/agenda_data.php',data,function(info){
//need data here
});
i want to know how i can get the php returned data in js here and how i assign that data into tags. also want to know there is while loop in php is here will be also loop to populate all rows ?
if i use direct php in my webpage then i can use like this in while loop
<div><?php echo $date; ?></div>
<div><?php echo $cidade; ?></div>
<div><?php echo $estado; ?></div>
<div><?php echo $local; ?></div>
.
.
.
how can i get in $.Post case
Your loop is somewhat incorrect. You're simply fetching each row's data, and then overwriting the previous row's data in all those variables. You should be building an array of results, which you can then send over to the client-side JS code. e.g. something like
$data
while($row = mysql_fetch_assoc($result)) {
$data[] = $row;
}
echo json_encode($data);
On the client side, you'll receive an array of objects you can iterate over:
$.post('includes/agenda_data.php',data,function(data){
$.each(data, function(idx, row) {
$('#date').innerHTML = row['date'];
...
});
});
exactly what you do in that JS loop is up to you.
As well, note that your PHP code is vulnerable to SQL injection attacks.
$.post is an AJAX call, thus you need to echo the result in PHP, to make sure you can use the results.
Try using
echo json_encode($your_data)
And to use the data in client side (browser), you can use
http://api.jquery.com/jQuery.parseJSON/
That parseJSON is used so you can use the data easier.
Why not use jQuery to fill the contents of the div by assigning a class as in this example:
<div class="date"><?php echo $date; ?></ div>
<div class="cidade"><?php echo $cidade; ?></ div>
<div class="estado"><?php echo $estado; ?></ div>
<div class="local"><?php echo $local; ?></ div>
and with jQuery, the success of $.post retrieve the information returned and integrate like this.
$('.date').html(info[0]['date']);

PHP Show/hide link

I have a function that prints out articles from my database and three links Edit , Add , Show/hide.
In the show/hide link i want to be able to hide/show that particular article.
How can i do that?
EDIT: I need to be able to hide/show articles in my backend page and it needs to stay hidden in the frontend page
function displaynews()
{
$data = mysql_query("SELECT * FROM news") // query
or die(mysql_error());
while ($info = mysql_fetch_array($data))
{
$id = $info['id'];
echo "<br>
<a href=Edit.php?id=$id>Edit</a></a>
<a href='addnews.php'> Add </a>
<a href='#'>Show/Hide</a><br><strong>" .
$info['date'] .
"</strong><br>" .
$info['news_content'] .
"<hr><br>"; // Print Articles and Date
}
}
You could use some Javascript and set the style attribute to display:none to hide, then display:block to show it again. Or use jQuery.
Use jquery.
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
</head>
show/hide
<div id="whatever">
Content
</div>
<script>
//Try these too
$('#whatever').hide();
$('#whatever').show();
$('#whatever').toggle();
</script>
Use following code:
PHP Code:
function displaynews()
{
$data = mysql_query("SELECT * FROM news") // query
or die(mysql_error());
while ($info = mysql_fetch_array($data))
{
$id = $info['id'];
echo "<div class="news"><br><a href=Edit.php?id=$id>Edit</a></a><a href='addnews.php'> Add </a><a href=hide.php>Show/Hide</a><br><strong>". $info['date']."</strong><br>". $info['news_content'] . "<hr><br></div>"; // Print Articles and Date
}
}
Javascript/jQuery Code (Don't forget to add jQuery in your page)
<script type="text/javascript">
$(document).ready(function(){
$(".news").click(function(){
$(this).toggle();
});
});
</script>

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