I need to display table header in my code - php

Question

The code that i made display the header after each row in the table.
I want just one time the header appear on the top.Any help?
print "<table border=1>\n";
while ($row = mysql_fetch_array($result)){
$files_field= $row['filename'];
$files_show= "Uploads/$files_field";
$descriptionvalue= $row['title'];
print "<tr>";
print "<th>";
echo "header1";
print "</th>";
print "<th>";
echo "header2";
print "</th>";
print "</tr>";
print "<tr>\n";
print "\t<td>\n";
echo "<font face=arial size=4/>$descriptionvalue</font>";
print "</td>\n";
print "\t<td>\n";
echo "<div align=center><a href='".$files_show."' target='_blank' title='CLICK TO OPEN'>$files_field</a></div>";
print "</td>\n";
print "</tr>\n";
}
print "</table>\n";

Answer 1

This program will:
get a row from db
print the header
print the data
get next row until the end
Sounds like you want to:
print the header
get a row from the db
print the data
get next row until the end

Answer 2

This should be adequate I think:
<?php
// print the beginning of the table, and the header
echo "<table border='1'>";
echo "<tr>";
echo "<th>";
echo "header1";
echo "</th>";
echo "<th>";
echo "header2";
echo "</th>";
echo "</tr>";
// than loop through the rows and print the rest of the table
while ($row = mysql_fetch_array($result)) {
$files_field = $row['filename'];
$files_show = "Uploads/{$files_field}";
$descriptionvalue = $row['title'];
echo "<tr>\n";
echo "\t<td>\n";
echo "<font face=arial size='4'/>$descriptionvalue</font>";
echo "</td>";
echo "<td>";
echo "<div align=center><a href='{$files_show}' target='_blank' title='CLICK TO OPEN'>{$files_field}</a></div>";
echo "</td>";
echo "</tr>";
}
echo "</table>";
Some additional notes / advises:
don't mix print / echo: they are almost the same, it just causes confusion
don't try to "prettify" the source with tabs and linebreaks: you are just bloating the output for no good reason, use the browsers inspector window instead
forget the mysql driver, use mysqli: the former is so out of dae, that it is removed from the new versions of PHP (from 7.0), so your code will stop working if the platform is updated; the mysqli variant is a bit different in how you have to use it (an extra parameter mostly), but the results are in the same format, and it behaves very similarly if not identically to the old one, so you will not have to rewrite many things