当我运行此代码时，它不会显示查询的任何结果

The query does not produce any results. The variable response is supposed to hold the query result but when I test it contain using var_dump it shows that it is empty.

 try
 {
    $bdd = new PDO('mysql:host=localhost;dbname=ems', 'root', '');
 } catch(Exception $e)
 {
    die('Erreur : '.$e->getMessage());
 }
 $code=123456;

 $query = $bdd->prepare('SELECT Code,sender FROM info WHERE Code = ?');
 //,reciever,origin,destination,transit
 $query->execute(array($code));
 if($query->fetchColumn())
 {
     echo"good";
     while($reponse = $query->fetch()){
         var_dump($reponse);

         echo"good";
         echo $reponse['sender'];
     }

   } else{
     echo" The Code entered  was not  found please verify and  enter again";
 }

I'm not sure if this was simply a typo in your code, but I noticed this block of code:

if($query->fetchColumn())
{
    echo"good";
    while($reponse = $query->fetch()){
     var_dump($reponse);

     echo"good";
     echo $reponse['sender'];
    }

else{
 echo" The Code entered  was not  found please verify and  enter again";
}
}

Looking at your code (and your comment), you have an open brace { for the if statement, followed by an open brace for the while loop. Then there is a closing brace. The closing brace } is for the while loop. Then you have an pair of braces for the else block. You also have another } after the else block. Try adding a } before your else block and remove one of the } after the else block. The else block should not be nested in the if block (if I understand you completely). This is a syntax error.

How the code should look:

if($query->fetchColumn())
{
    echo"good";
    while($reponse = $query->fetch()){
     var_dump($reponse);

     echo"good";
     echo $reponse['sender'];
    }
}
else{
 echo" The Code entered  was not  found please verify and  enter again";

}