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JavaScript/PHP: Alert dialog after successfully saved

I new in programming. Currently, I develop a system that registration part. The registration part is successfully saved to the database. What I want to know is how to popup an alert dialog with one button e.g "Ok" after registration was successful and redirect to another page, such as home page. Now I only echo "successfully saved"
Below is my current code
<?php
require "DbConnect.php";
$name = $_POST['name'];
$badgeid = $_POST['badgeid'];
$position = $_POST['position'];
$department = $_POST['department'];
$factory = $_POST['factory'];
$reviewer = $_POST['reviewer'];
$title = $_POST['title'];
$year = $_POST['year'];
$month = $_POST['month'];
$suggestionwill = $_POST['suggestionwill'];
$present = $_POST['present'];
$details = $_POST['details'];
$benefit = $_POST['benefit'];
$sql_query = "INSERT INTO topsuggest (name,badgeid,position,department,factory,
reviewer,title,year,month,suggestionwill,present,details,benefit) VALUES('$name','$badgeid','$position','$department','$factory','$reviewer','$title','$year','$month','$suggestionwill','$present','$details','$benefit')";
if(mysqli_query($conn,$sql_query))
{
echo "<p id='msg'></p>";
}
else
{
echo "Error!! Not Saved".mysqli_error($con);
}
?>

Just use php header and use javascript to alert a message .
if(mysqli_query($conn,$sql_query))
{
echo "<script>alert('Successfuly Saved');</script>";
header('Location: PATH TO BE REDIRECTED');
}
For a example
if(mysqli_query($conn,$sql_query))
{
echo "<script>alert('Successfuly Saved');</script>";
header('Location: ../Insert/Index.php');
}
Please note that space between Location: is compulsory

After inserting data you can simply redirect to your interested page with a success message like:
header("location:page_of_interest.php?msg=Record Inserted");
and on page_of_interest.php you can simply check for msg and show if it is set like:
if(isset($_GET['msg'])){
echo $_GET['msg'];
}

PHP doesn't output error from mysql query for non-existent rows for autocomplete

I have a form that currently is able to auto complete base on user input, it queries the MySQL database and successfully lists all possible matches in the table and give suggestions. Now I want to handle rows that do not exist. I am having trouble to get my PHP file to echo the error. Here is what I have so far:
I'm guessing in my auto search function in my javascript in main.php I need to return the error message to the page?
search.php
<?php
//database configuration
$host = 'user';
$username = 'user';
$password = 'pwd';
$name = 'name';
//connect with the database
$dbConnection = new mysqli($host,$username,$password,$name);
if(isset($_GET['term'])){
//get search term
$searchTerm = '%'.$_GET['term'].'%';
//get matched data from skills table
if($query = $dbConnection->prepare("SELECT * FROM nametbl WHERE name LIKE ? ORDER BY name ASC")) {
$query->bind_param("s", $searchTerm);
$query->execute();
$result = $query->get_result();
//$row_cnt = $result->num_rows;
//echo $row_cnt;
if($result -> num_rows){
while ($row = $result->fetch_assoc()) {
$data[] = $row['name'];
}
//return json data
echo json_encode($data);
mysqli_close($dbConnection);
}
else { echo '<pre>' . "there are no rows." . '</pre>'; }
}
else {
echo '<pre>' . "something went wrong when trying to connect to the database." . '</pre>';
}
}
?>
main.php
<div id="gatewayInput">
<form method="post">
<input type="text" id="name" name="name" placeholder="Name..."><br><br>
<?php
include("search.php");
?>
</div>
...
...
...
<script src ="../../../jqueryDir/jquery-3.2.1.min.js"></script>
<script src ="../../../jqueryDir/jquery-ui.min.js"></script>
<script type="text/javascript">
//auto search function
$(function() {
$( "#name" ).autocomplete({
source: 'search.php'
});
});

1.your method type is post in the form
in main.php
and in the search.php, you have used "if(isset($_GET['term'])){"
this needs to be fixed I guess. either both needs to be POST or GET.
Again you are using include method which the whole code in search.php will be made in-line and treated as a one file main.php. so you need not use GET or Post method.
How does get and Post methods work is
3.1) you have a html or PHP which submits the data from browser(main.php), and this request is being served by an action class(search.php)
example :- in main.php
3.2) now in search.php you can use something like if(isset($_POST['term'])){

You can use num_rows (e.g. if ($result -> num_rows)) to see if the query returned anything.

Sending value of PHP variable to Javascript document

So I am trying to reach into a MySQL table and draw out a value. I have the following PHP that does so:
<!DOCTYPE html>
<html>
<body>
<?php
$username = strval($_GET['userName']);
$con = mysqli_connect('localhost','PRIVATE','PRIVATE','PRIVATE');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
} else {
$sql="SELECT * FROM users WHERE username = '".$username."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
$wealth = $row['wealth'];
echo $wealth;
}
}
mysqli_close($con);
//return $wealth;
?></body>
</html>
I've put PRIVATE where the database ID and password go for security reasons. Essentially, this PHP takes the value out of the 'wealth' column in according to the logged on user. I have this AJAX function that triggers this PHP (the ajax is located inside of the document I would like to send the PHP variable to). Note that this function sends the username of the current logged on user and their 'score' (the var clicks) to the PHP.
function sendScore() {
$.post("sendScore.php",{username:localStorage.getItem('userName'),wealth:clicks},function(response){
console.log("The service replied"+response);
});
}
Now, I know the value I retrieved is equal to the PHP variable $wealth. I also understand that PHP is server based and Javascript/html are client based, so you can't simply reach into another document and find the value of the variable. I'd like to assign the value of $wealth to a javascript variable named: userWealth
Thanks for reading!
EDIT: I WROTE THE Q WRONG ...
#Ilan Kleiman Small problem... I have two separate PHP files, sendScore and getScore. I had mistakenly pasted the wrong ones in the question. So, I have the sendScore AJAX which you can see in the question, but this triggers a different PHP code which isn't shown above, which essentially writes into the 'wealth' column. I have a separate piece of PHP, shown above, which is used to RETRIEVE written info from the wealth column (like how cookie clicker saves the number of clicks you have when you close the tab, this code activates when you open the website back up again, and it loads the last written value in 'wealth'). I am looking into how to create a piece of AJAX that can turn the $wealth PHP variable generated by the code into a javascript variable. Sorry for the confusion.
EDIT #2: CODE
AJAX FOR SENDSCORE
$.post("sendScore.php",{username:localStorage.getItem('userName'),wealth:clicks},function(response){
console.log("The service replied"+response);
});
PHP FOR SENDSCORE
<!DOCTYPE html>
<html>
<body>
<?php
$username = strval($_POST['username']);
$wealth = strval($_POST['wealth']);
$con = mysqli_connect('localhost','PRIV','PRIV','PRIV');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
} else {
$sql = "UPDATE users SET wealth=".$wealth." WHERE username='".$username."'";
$result = mysqli_query($con,$sql);
}
mysqli_close($con);
?></body>
</html>
AJAX FOR GETSCORE IS WHAT I AM TRYING TO FIND
PHP FOR GETSCORE
<!DOCTYPE html>
<html>
<body>
<?php
$username = strval($_GET['userName']);
$con = mysqli_connect('localhost','PRIV','PRIV','PRIV');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
} else {
$sql="SELECT * FROM users WHERE username = '".$username."'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
$wealth = $row['wealth'];
echo $wealth;
}
}
mysqli_close($con);
//return $wealth;
?></body>
</html>

In your PHP file:
change
echo $wealth;
to
echo "<div id='wealth'>" . $wealth . "</div>";
AJAX Request:
var userWealth;
function getScore() {
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
userWealth = document.getElementById("wealth").innerHTML;
alert(userWealth);
}
};
xhttp.open("GET", "getScore.php", true);
xhttp.send();
}
This will set "userWealth" to $wealth from the PHP.
Keep in mind, the Javascript should be on the same "server/website" as the PHP otherwise the AJAX request won't work.

Insert an paragraph with jQuery in an PHP 'If' statement

I've written a simple login script that connects to a db, and now I want to insert a paragraph with jQuery in my #loginbox which says 'Login failed' when
if (!$row = mysqli_fetch_assoc($result))
is true.
My thought was:
[function.js]
function loginfailed() {
$('#loginbox').html("<p>Login failed.</p>");
}
[login.php]
<head>
<script type="text/javascript" src="functions.js"></script>
</head>
<?php
include '../config.php';
include 'dbh.php';
session_start();
$uid = $_POST['uid'];
$pw = $_POST['pw'];
$sql = "SELECT * FROM user WHERE uid='$uid' AND pw='$pw'";
$result = mysqli_query($conn, $sql);
if (!$row = mysqli_fetch_assoc($result))
{
header('Location: ../index.php');
echo '<script> loginfailed(); </script>';
}
else
{
header('Location: ../index.php');
}
?>
But it doesn't work.

DON'T EVER STORE PASSWORDS IN PLAIN TEXT!!
Regarding your question.
The header function redirects to index.php and does not execute the echo. One solution can be to add a $_GET parameter and after the redirect check if it exists and echo the message or append it with JS.
if (!$row = mysqli_fetch_assoc($result))
{
header('Location: ../index.php?status=fail');
}
In the index.php file at the bottom (if you want to use JS/jQuery to show message)
<script>
var status = "<?php echo (!empty($_GET['status']) && $_GET['status'] === 'fail') ? 0 : 1; ?>";
if(!status) loginfailed();
</script>

Thanks guys, but i've found my own solution with the help of Allkin.
My header now redirects to
header('Location: ../index.php?status=fail');
and my #loginbox checks if status is set and then executes my loginfailed() function.
if(isset($_GET['status'])) {
echo '<script> loginfailed(); </script>';
}
Nothing easy like that!
Thanks for your help everyone.

Two questions about possible mysql and php functionalities (maybe javascript?), and where I should look to learn more

First, is it possible for when I insert a record onto my mysql table, a page is automatically generated using the new record in some way. EXAMPLE: My column "image" is on autoincrement, so my image names are always numbers. Furthermore, is it possible for when I insert a record, I automatically generate a page with my image name. So basically, I submit record 367, the image name is 367, and my site will automatically generate mysite.com/367? I want to go in more details but you get the point. Is it possible? If not, what's the closest thing possible?
Also, is there someway to automatically update my page periodically. Such as I set it so at 5pm, it'll automatically insert a code. 5:30pm, it'll insert a different code, which I preprogrammed to do. This is useful, for say I'm on vacation but I still want to update my site regularly.
Can you guys point me to any specific tutorial/terminology/methods/programs/codes/anything? All help would be appreciated!
EDIT: Code I have so far (just want to show to Nick)
<html>
<head>
<title>tgh</title>
</head>
<body>
<?php
$objConnect = mysql_connect("localhost","root","") or die(mysql_error());
$objDB = mysql_select_db("thegoodhumor");
$strSQL = "SELECT * FROM gallery";
if (!isset($_GET['Page'])) $_GET['Page']='0';
$objQuery = mysql_query($strSQL);
$Num_Rows = mysql_num_rows($objQuery);
$Per_Page = 16; // Per Page
$Page = $_GET["Page"];
if(!$_GET["Page"])
{
$Page=1;
}
$Prev_Page = $Page-1;
$Next_Page = $Page+1;
$Page_Start = (($Per_Page*$Page)-$Per_Page);
if($Num_Rows<=$Per_Page)
{
$Num_Pages =1;
}
else if(($Num_Rows % $Per_Page)==0)
{
$Num_Pages =($Num_Rows/$Per_Page) ;
}
else
{
$Num_Pages =($Num_Rows/$Per_Page)+1;
$Num_Pages = (int)$Num_Pages;
}
$strSQL .=" order by GalleryID ASC LIMIT $Page_Start , $Per_Page";
$objQuery = mysql_query($strSQL);
$cell = 0;
echo '<table border="1" cellpadding="2" cellspacing="1"><tr>';
while($objResult = mysql_fetch_array($objQuery))
{
if($cell % 4 == 0) {
echo '</tr><tr>';
}
if($cell == 2) {
echo '<td>RESERVED</td>';
} elseif ($cell == 3) {
echo '<td>The other cell</td>';
} else {
echo '<td><img src="https://s3.amazonaws.com/imagetitle/' . $objResult["Picture"] . '" />' .
$objResult["GalleryName"] . '</td>'; }
$cell++;
}
echo '</tr></table>';
?>
<br>
view more:
<?php
if($Prev_Page)
{
echo " <a href='$_SERVER[SCRIPT_NAME]?Page=$Prev_Page'>prev</a> ";
}
{
echo "|";
}
if($Page!=$Num_Pages)
{
echo " <a href ='$_SERVER[SCRIPT_NAME]?Page=$Next_Page'>next</a> ";
}
?>
</body>
</html>
<?php
mysql_close($objConnect);
?>

It sounds like you want a dynamic web page. To make a dymaic webpage I'd suggest using PHP which would interact with the mysql server.
For example, a user would visit 'mysite.com/info.php?image=367' and the php script would get the information 'image=367'. Your PHP script could do a select query against the mysql database 'SELECT paragraph FROM table WHERE image_id = 367' and then write that data out to the user's web browser.
As far as the user is concerned they just visited 'mysite.com/info.php?image=367', but in the background, PHP dynamically created the webpage content after it got that request.
More basic info about dynamic webpages: http://way.clicktracks.com/help/en/pr650/index.html?dynamicwebsiteshowtheywork.htm
Simple Intro to PHP:
http://www.tizag.com/phpT/
http://www.w3schools.com/php/php_intro.asp
Here is a head start I wrote for you, feel free to use it.
<?php
if (!isset($_GET['imageNumber']))
die("You must specify an image number");
$image_requested = mysql_real_escape_string($_GET['imageNumber']); //sanitizes input
$dbhost = 'localhost'; //TODO: Set this to the ip address of your mysql server if it is not on the same machine
$dbuser = 'root'; //TODO: Set the username you use to access your mysql db here
$dbpass = 'password'; //TODO: Set the password you use to access your mysql db here
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql');
$dbname = 'database_name_here'; //TODO: Set the database name here
mysql_select_db($dbname);
$query = "SELECT paragraph FROM table_name WHERE image_id = " . $image_requested; //TODO: Set table_name, column to get, and image_id to the correct column name
$result = mysql_query($query);
$row = mysql_fetch_array($result) or die(mysql_error());
echo "Here is the paragraph of text" . $row['paragraph']; //TODO: Set paragraph to the same column you retrieved 3 lines above.
mysql_close($conn);
?>
As for the second part of your question, it can also be done with PHP
<?php
$specifictime = strtotime("tuesday 3pm");
if (time("now") > $specifictime)
{
echo " its after 3pm on tuesday";
}
else {
echo " not 3pm on tuesday yet";
}
?>