How to compare and output variables in psql?

Question

PHP

there is a php script that displays a certain word, tried to compare by assigning additional variables, nothing helped. If the value is true, you need to display another column with this id, but did not even display the column itself

if($_POST['country']){
  $result = $pdo->pg_query("SELECT `name_doc` FROM `test_table` WHERE `number` = ".$_POST['name']."");
  while ($row = $result->fetch_row()) {
        printf ("%s (%s)\n", $row[0], $row[1]);
    }
}
$pdo->close();
?>

Answer 1

Long story short, I solved this problem myself.

$query = 'SELECT name FROM database';
$result = pg_query($query);
$myrow = pg_fetch_assoc($result, [$myVar = array()] );
$value = $myrow[name];
//echo "$value";
//echo "$word";
if ($value == "$word") {
    $query2 = 'SELECT path FROM documents';
    $result2 = pg_query($query2);
    $myrow2 = pg_fetch_assoc($result2, [$myVar = array()]);
    $value2 = $myrow2[path];
    echo "$value2";
} elseif ($value <> $word) {
    echo "pozshral";
}