Show only columns with values as result of a dynamic sql-query - php

Question

I have an html form to read out data from an sql-database. According to the selected criteria, the results are shown in a dynamically generated table. This works well with the following code. The SELECT query is composed according to the selected criteria.
$sql = "SELECT DISTINCT $selection FROM $tabelle WHERE $whereclause";
$result = mysqli_query($db, $sql) or die("<b>No result</b>"); //Running
the query and storing it in result
$numrows = mysqli_num_rows($result); // gets number of rows in result
table
$numcols = mysqli_num_fields($result); // gets number of columns in
result table
$field = mysqli_fetch_fields($result); // gets the column names from the
result table
if ($numrows > 0) {
echo "<table id='myTable' >";
echo "<thead>";
echo "<tr>";
echo "<th>" . 'Nr' . "</th>";
for($x=0;$x<$numcols;$x++){
$key = array_search($field[$x]->name, $custom_column_arr);
if($key !== false){
echo "<th>" . $key . "</th>";
}else{
echo "<th>" . $field[$x]->name . "</th>";
}
}
echo "</tr></thead>";
echo "<tbody>";
$nr = 1;
while ($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $nr . "</td>";
for ($k=0; $k<$numcols; $k++) { // goes around until there are no
columns left
echo "<td>" . $row[$field[$k]->name] . "</td>"; //Prints the data
}
echo "</tr>";
$nr = $nr + 1;
} // Ende der while-Schleife
echo "</tbody></table>";
}
}
mysqli_close($db);
Now, I want to display only those columns that contain at least one value and not those with NULL, 0, empty string or empty. My problem is, that $numcols counts the empty columns, too.