How to Retrieve Values Resulting From AJAX Live Search? - php

Question

Below is a jQuery function that retrieves 2 textbox values and posts them to another file ("Student Search Results.php"), where a live search is run using the values.
<script>
$(".search").keyup(function() {
var Team_Name = $('#TeamName').val();
var Teacher = $('#Teacher').val();
var Search_Data = Team_Name + '?????' + Teacher;
$.ajax({
type: "POST",
url: "Student Search Results.php",
data: {
query: Search_Data
},
cache: false,
success: function() {
alert('The values were sent');
}
});
});
</script>
Below is the PHP script on the search page ("Student Search Results.php") that makes use of these values.
<?php
include "Connection.php";
if(isset($_POST['query'])){
$searchData = explode('?????', $_POST['query']);
$teamName = $searchData[0];
$teacher = $searchData[1];
$query = "SELECT club_table.Club_Name, teacher_user_table.Teacher_Name
FROM club_table, teacher_user_table
WHERE club_table.Teacher_Email = teacher_user_table.Teacher_Email,
teacher_user_table.Teacher_Name LIKE '%" . $teacher . "%',
club_table.Club_Name LIKE '%" . $teamName . "%';";
}else{
$query = "SELECT club_table.Club_Name, teacher_user_table.Teacher_Name
FROM club_table, teacher_user_table
WHERE club_table.Teacher_Email = teacher_user_table.Teacher_Email;";
}
$result = mysqli_query($con, $query);
echo $query;
?>
How would I be able to take variables from the PHP script (such as $result) to the first page, so I can create a result table? Simply including the PHP file does not work, as the file is only included once.
Thank you for your time.

Answer 1

Best option is to serialize to JSON using json_encode

Answer 2

I think best you can do is,
success: function(result) {
alert(result);
}
and Student Search Results.php print result in tabular format.
P.S. : Please follow proper file naming convention

Answer 3

use a proper URL, and send the data (and stop using camelcase for everything) :
$(".search").on('keyup', function() {
var data = {
team_name : $('#TeamName').val(),
teacher : $('#Teacher').val()
}
$.ajax({
type: "POST",
url: "student_search_results.php",
data: data,
cache: false
}).done(function(result) {
console.log(result);
});
});
And in PHP, you have to actually get the result into an array and json_encode it :
<?php
include "Connection.php";
$team_name = !empty( $_POST['team_name'] ) ? $_POST['team_name'] : null;
$teacher = !empty( $_POST['teacher'] ) ? $_POST['teacher'] : null;
if ($team_name && $teacher) {
$query = "SELECT club_table.Club_Name, teacher_user_table.Teacher_Name
FROM club_table, teacher_user_table
WHERE club_table.Teacher_Email = teacher_user_table.Teacher_Email,
teacher_user_table.Teacher_Name LIKE '%" . $teacher . "%',
club_table.Club_Name LIKE '%" . $teamName . "%';";
}else{
$query = "SELECT club_table.Club_Name, teacher_user_table.Teacher_Name
FROM club_table, teacher_user_table
WHERE club_table.Teacher_Email = teacher_user_table.Teacher_Email;";
}
$result = mysqli_query($con, $query);
$data = $result->fetch_all( MYSQLI_ASSOC );
echo json_encode( $data );
?>

Answer 4

<script>
$(".search").keyup(function() {
var Team_Name = $('#TeamName').val();
var Teacher = $('#Teacher').val();
var Search_Data = "Team_Name="+'Team_Name'&Teacher='+Teacher;
$.ajax({
type: "POST",
url: "Student_Search_Results.php",
data: Search_Data,
cache: false,
success: function(result) {
$('$output').html(result);
}
});
});
</script>
Here is output div
<div id="output"></div>
On Student_Search_Results.php page get
$tname = $_POST['Team_Name'];
$teacher = $_POST['Teacher'];
//your search query & print data