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Php and ajax not returning any data from database, no errors

I have created a search using ajax and php, however when I click the search button nothing happens. There are no errors, however when I look at the network in the console it doesn't seem to be reaching searching.php as this file doesn't show up in the network. I want to be able to put the returned data from the search into seperate divs, although at the minute it isn't returning any data at all and I have no idea why.
This is my php
require_once('config1.php');
$output = '';
if(isset($_POST['search']) === true && empty($_POST['search']) === false){
$query = "SELECT a.attraction_name, a.lat, a.long, a.cost, a.image FROM zz_attractions a INNER JOIN zz_city c ON a.city_id = c.city_id WHERE c.city_name = '" . mysqli_real_escape_string(trim($_POST['search'])) . "' ";
$result = mysqli_query($conn, $query);
$count = mysqli_num_rows($result);
if ($count == 0) {
$output = 'there was no search results';
} else {
while ($row = mysqli_fetch_array($result)) {
$attraction_name = $row['attraction_name'];
$lat = $row['lat'];
$long = $row['long'];
$cost = $row['cost'];
$image = "<img src='{$row['image']}' height='100' width='100'>";
$output .= '<div>'.$attraction_name.' '.$lat.' '.$long.' '.$cost.' '.$image.'</div>';
}
}
}
This is my ajax
$('#submit').on('click', function() {
var search = $('#search').val();
if ($.trim(search) != '') { //if search is not equal to nothing - using trim allows users to type in blank space and it won't return anything
$.post('searching.php', {search: search}, function(data) {
$('#return').text(data);
});
}
});
Within the html the div id is #return. If someone could help me find out why no data is returning that would be great.

How to fetch questions ony by one on save and next buttons

I have a quiz management system and need to fetch the data from the database and displaying the data one by one on clicking next button.
I want to use ajax to insert data into database after clicking next button by using file read and write function.
$sql=mysqli_query($con,"SELECT * FROM tbl_question WHERE setid='$set' AND status=1 ORDER BY RAND()");
while($sha=mysqli_fetch_array($sql)) {
<h4><? echo $sha['id'];?></h4>

here's simple way to do it
first get question from database then put a button to the next id
<?php
$rows = 0;
$id = isset($_GET['id'] ? (int) $_GET['id'] : 0;
$query = 'SELECT * FROM `tbl_question` WHERE `tbl_question`.`status` = 1 AND `tbl_question`.`id` = ? LIMIT 1';
$mysqli = new mysqli('localhost','user','password','database');
$stmt = $mysqli->prepare($query);
$stmt->bind_param('i',$id);
$stmt->execute();
$stmt->bind_result($id,$something,$something_else,$status);
while($stmt->fetch())
$rows++;
$stmt->close();
$mysqli->close();
if($rows === 0) {
echo 'qustion not found';
} else {
// output qustion
// button
echo '<a type="buttton" href="?id=' . $id + 1 . '">next</a>';
}

Pass array from javascript to page in wordpress

I have a page in wordpress, that uses an jquery and ajax to get information from an external api. The form sends the array generated in javascript back to the same page with another variable that the php in the page uses to determine which page to display. Outside of wordpress, the code works fine. Inside Wordpress the first portion runs, but then instead of loading the same page again it goes to a search page and says nothing found.
The url on the output is:
http://kltools.net/?s=&post_type%5B%5D=portfolio&post_type%5B%5D=post&post_type%5B%5D=page
Which seems odd considering I'm using post not get.
The javascript that generates the array and submits the form:
function submitchans(){
for (var i=0;i<chans.length;i++)
{ var newHidInp = document.createElement('input');
newHidInp.type = 'hidden';
newHidInp.name = 'chans[]';
newHidInp.value = chans[i];
form.appendChild(newHidInp);
}
}
function livearray(input){
if (input != null) {
chans.push(input);
}
if (Y === cSize){
submitchans();
document.forms[0].submit();
}
}
Previously the array was outArray[] instead of chans[], I changed it thinking that may be triggering the result, but no luck.
This is the PHP portion of the code:
<?php
$page_to_load = $_POST[view];
switch($page_to_load) {
case '':
echo "<script src=\"../scripts/jquery-3.2.0.min.js\"></script>";
echo "<script type=\"application/javascript\" src=\"../scripts/raid.js\"></script>";
echo "<font size=\"+3\" color=\"#FFFFFF\">Who should I host?<br>Please wait while channel is selected<br></font>";
echo "<font size=\"+2\">";
echo "<br><br>";
echo "<img src=\"../_images/ajax_loader_blue_350.gif\">";
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "database";
$chanarray[] = null;
$offline = 0;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT `TwitchNames` FROM TK_Members WHERE Validated='1' AND RaidMe='1'";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
array_push($chanarray, $row['TwitchNames']);
}
} else {
echo "0 results";
}
array_splice($chanarray, 0, 1);
$conn->close();
echo "<script type=\"application/javascript\">";
echo "var channels = ". json_encode($chanarray);
echo "</script>";
echo "</font>";
echo "<form id=\"form\" method=\"post\">";
echo "<input type=\"hidden\" name=\"view\" value=\"page2\">";
echo "</form>";
break;
case 'page2':
echo "<font size=\"+3\" color=\"#FFFFFF\">Who should I host?<br>";
echo "Your channel to host is:<br></font>";
echo "<font size=\"+2\">";
echo "<br><br>";
$chans[] = null;
$test = $_POST['chans'];
foreach ($test as $chan) {
$temparray = array(rand(),$chan);
array_push($chans, $temparray);
}
array_splice($chans,0,1);
sort($chans);
echo "".$chans[0][1]."";
echo "<br><br><br>";
echo "See All Live Channels";
echo "</font>";
break;
}
?>
After working with what blokeish has suggested I've modified the javascript file working out where the problem is.
The new javascript file is:
// JavaScript Document
var chans = ["test1","test2","test3"];
function submitchans(){
for (var i=0;i<chans.length;i++)
{ var newHidInp = document.createElement('input');
newHidInp.type = 'hidden';
newHidInp.name = 'chans[]';
newHidInp.value = chans[i];
document.getElementById('chansform').appendChild(newHidInp);
}
}
jQuery(function ($) {
submitchans();
document.getElementById('chansform').submit();
});
Using only the javascript clicking submit, it passes to the next page. When adding in the array pass is when it fails. This is the page log that is returning during the execution. !!--CORRECTION--!! there was a typo in the code, after correcting ID to Id the code is working as intended.

document.forms[0].submit() is likely submitting the wp search form as that could be the first form in the DOM. I see "http://kltools.net/?s=" in the URL where "s" is the search term.
Using document.getElementById('idOfForm').submit() should get you around that problem if there are multiple forms in a page and you cant be sure of its index.

javascript load php on div doesn't work with Firefox

I have several php pages that create dynamically links with javascript scripts which load different content in the same div. The function that creates the links is the following:
$variablephp = $_GET['id'];
$result = mysql_query( "SELECT `BASIC` FROM `roleflow_training` WHERE `ROLE_ID`=".$variablephp.""); //Once the text has been saved in the BBDD:
$result = mysql_result ($result, 0, "BASIC");
$result = preg_replace("/\n+/", "\n",$result);
$lineas = explode("\n", $result);
echo "<ul>\n";
foreach ($lineas as $bullet)
{
$test = preg_replace("/\n+/", "", $bullet);
$prueba = chop($test, "\n");
$prueba = chop($prueba, "\b");
$prueba = chop($prueba, "\br");
$prueba = "'".$prueba."'";
$link = mysql_query("SELECT * FROM `cursos_training` WHERE `COURSE_NAME`=".$prueba."");
$existe = mysql_num_rows($link);
if ($existe < 1)
{
echo "<li>".$test."</li>\n";
}
else
{
$fila = mysql_fetch_array($link);
$enlace = $fila['IDLINK'];
$direccion = "<a href=\"javascript:$('#floatFlowCursos').css('display','none');$('#contentFlow').load('curso1_info.php?id=".$enlace."')\">";
echo "<li>".$direccion.$bullet."</a>"."</li>\n";
}
}
echo "</ul>";
This sentence works perfectly with Chrome but not with Firefox or I Explorer, do you have any idea?
Thank you so much for your support!! Cheers!!

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";
}
?>