How to pull out from the table all records in which the field "user"=$search?

Question

PHP

It is necessary to pull out from the table all records in which the field "user"=$search and further to deduce in the table.
Here is an example, but it displays only one record - the last one, and all records are needed in which the field "user" = $search.

$result = mysql_query("SELECT * FROM table WHERE user='$search'",$db);
while ($myrow = mysql_fetch_array($result))
   {
     $th1 = '<th>'.$myrow['php'].'</th>';
     $th2 = '<th>'.$myrow['code'].'</th>';     
   }

Answer 1

$result = mysql_query("SELECT * FROM table WHERE user='$search'",$db);
while ($myrow = mysql_fetch_array($result))
   {

     $th1 .= '<th>'.$myrow['php'].'</th>';
     $th2 .= '<th>'.$myrow['code'].'</th>';

       
   }

Answer 2

$result = mysql_query("SELECT * FROM table WHERE user='$search'",$db);

1. This feature is deprecated. It is deprecated in 5.6 and missing in 7.0.
2. You need to use prepared statements, you have SQL injection.