Data not required is being displayed - php

Question

I have a form where the user enters data e.g. AXZAA QS1QS. This data is Posted to my PHP script. The PHP Script connects to a MYSQL database which contains at the moment 2 records.
The idea is that the PHP script will take the input and compare it to the records in the database. If the records exist they are displayed in a table on a web page otherwise, an error message is displayed.
I am having a number of problems with my PHP script and have modified my script a number of times. However, the thing I am having the biggest problem with is this:
When the form appears for the first time, the message record doesn't exist appears twice, this is before the user has entered any data and is seeing the form for the first time. See picture below.
After entering data (when the PHP script was partially working correctly), if there is a match i.e. records existed, along with the records in the table I would receive an error message telling me that records were not found. To see if I could resolve the problem I added code to tell me what records could not be found, the records that couldn't be found were the ones that were found and the other records from the database which I wasn't looking for. I know the SQL query in my PHP script tells the script to get everything from the database however, I would have thought the if statement would have fixed the problem.
Sorry about writing such a long problem and I hope it's not confusing.
enter code here
<?php
//Connect to the database connection file
require 'databaseconnection.php';
$searchBar=(isset($_POST['searchBar']) ? $_POST['searchBar'] :null);
$userdata = trim($searchBar);
$cleaned_data = preg_split('/[\s]+/', $userdata);
$sql = "SELECT DISTINCT * FROM atable_2";
$result = mysqli_query($database_connection, $sql);
echo "<table border>
<tr>
<th>Allocation</th>
<th>Codes</th>
<th>Names</th>
</tr>";
while($putdatabaseanswer_intoarray = mysqli_fetch_array($result)) {
$allocation_id = $putdatabaseanswer_intoarray["allocation"];
$codes_id = $putdatabaseanswer_intoarray["codes"];
$names_id = $putdatabaseanswer_intoarray["names"];
foreach($cleaned_data as $value) {
if($value==$codes_id) {
echo "<tr>";
echo "<td>" . $allocation_id. "</td>";
echo "<td>" . $codes_id . "</td>";
echo "<td>" . $names_id . "</td>";
echo "</tr>";
}
else
{
echo "<br />";
echo "One or more of the records have not been found: $codes_id";
echo"<br />";
}
}
}
echo "</table>";
?>

Answer 1

Wouldn't it be better to assign $searchbar after an if statement like
`<?php
//Connect to the database connection file
require 'databaseconnection.php';
if(isset($_POST['searchBar']))
{
$searchbar = $_POST['searchBar'];
$userdata = trim($searchBar);
$cleaned_data = preg_split('/[\s]+/', $userdata);
$sql = "SELECT DISTINCT * FROM atable_2";
$result = mysqli_query($database_connection, $sql);
echo "<table border>
<tr>
<th>Allocation</th>
<th>Codes</th>
<th>Names</th>
</tr>";
while($putdatabaseanswer_intoarray = mysqli_fetch_array($result)) {
$allocation_id = $putdatabaseanswer_intoarray["allocation"];
$codes_id = $putdatabaseanswer_intoarray["codes"];
$names_id = $putdatabaseanswer_intoarray["names"];
foreach($cleaned_data as $value) {
if($value==$codes_id) {
echo "<tr>";
echo "<td>" . $allocation_id. "</td>";
echo "<td>" . $codes_id . "</td>";
echo "<td>" . $names_id . "</td>";
echo "</tr>";
}
else
{
echo "<br />";
echo "One or more of the records have not been found: $codes_id";
echo"<br />";
}
}
}
echo "</table>";
}
else{
echo "<p>Please enter a search term</p>";
}
?>
You could then execute the MySQL query within that "if" statement rather than having it execute assuming there is a value