Really Easy PHP query. Displaying from separate PHP file - php

Question

I'm a PHP n00b. I have this code, however it is on my main page. I assume like jQuery there must be a way to display this without having the actual script on the main document that the users access?
Or is there a way I can do this tidier without echo'ing everything?
include('talentsearch.php');
while($row = mysqli_fetch_array($result2))
{
echo "<div class='talent_result_wrapper' data-experience='" . $row['divTagExp'] . "' data-salary='" . $row['divTagExp'] . "'>";
echo "<ul>";
echo "<li><strong>Talent ID: </strong>" . $row['CandidateID'] . "</li>";
echo "<li><strong>Resides: </strong>" . $row['Town'] . "</li>";
echo "<li><strong>Salary Required: </strong>£" . $row['SalaryMin'] . "</li>";
echo "<li><strong>Experience: </strong>" . $row['CandidateExperience'] . " Years </li>";
echo "<li><strong>Industy: </strong>" . $row['PrimarySector'] . "</li>";
echo "<li><strong>Specialism: </strong>" . $row['PrimarySector'] . "</li>";
echo "</div>";
}
I have tried putting it in a function like this:
function getTalent() {
}
And then calling the function like this
<?php
include('talentsearch.php');
getTalent()
?>
But I get an error whereas I don't if I just run the code normally.
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in
So I assume I'm not doing it right.

Answer 1

is this what you are trying to achive?
talentsearch.php file
function getTalent($row) {
return "<div class='talent_result_wrapper' data-experience='" . $row['divTagExp'] . "' data-salary='" . $row['divTagExp'] . "'>
<ul>
<li><strong>Talent ID: </strong>" . $row['CandidateID'] . "</li>
<li><strong>Resides: </strong>" . $row['Town'] . "</li>
<li><strong>Salary Required: </strong>£" . $row['SalaryMin'] . "</li>
<li><strong>Experience: </strong>" . $row['CandidateExperience'] . " Years </li>
<li><strong>Industy: </strong>" . $row['PrimarySector'] . "</li>
<li><strong>Specialism: </strong>" . $row['PrimarySector'] . "</li>
</div>";
}
///////
main.php file
include('talentsearch.php');
while($row = mysqli_fetch_array($result2)) {
echo getTalent($row);
}