単一の条件が満たされない場合、このコードの実行を停止する方法PHP

Question

このコードは、最後まで実行され、条件が満たされなくても送信します。私はそれが実行されるのを止めて、どんな条件も見逃してしまうとエラーをエコーし​​たい。親切に助けてください。注：exit（）を試してみたところ、html全体の実行に影響がありました。あなたがしなければならない場合

$verifiedPhone = phoneNumberValidator($phone); 
    $correctsurname = nameValidator($surname); 
    $correctlastname = nameValidator($lastname); 
    $correctusername = nameValidator($username); 
     if(!$verifiedPhone){ 
      echo "<p class='alert alert-warning'>Please use a Valid Phone Number</p>"; 

          } 
     if(!$correctsurname){ 
      echo "<p class='alert alert-warning'>Surname can only contain alphabets</p>"; 
          } 
     if(!$correctlastname) { 
      echo "<p class='alert alert-warning'>Last Name can only contain alphabets</p>"; 
          } 
     if (!$correctusername) { 
      echo "<p class='alert alert-warning'>Username can only contain alphabets</p>"; 
          } 
     $checkuser = " SELECT * FROM staff 
         WHERE username = '$correctusername'"; 
     $checkuserresult = mysqli_query($connection, $checkuser); 
     $checkuserrow = mysqli_num_rows($checkuserresult); 
    if($checkuserrow > 0){ 
     echo "<p class='alert alert-danger'>Username \"".$username."\" already exist! Try another</p>"; 
          } 
    else{ 
     $harsedpassword = md5("$password"); 
    $datainsert = " INSERT INTO staff (surname, lastname, phone, username, password) VALUES ('$correctsurname', '$correctlastname', '$verifiedPhone', '$username','$harsedpassword')"; 
$datainsertresult = mysqli_query($connection, $datainsert); 
    if($datainsertresult){ 
    echo "<p class='alert alert-success'><b>Staff Added Successfully</b></p>"; 
           } 

Answer 1

（ "メッセージ"）

if(!$correctsurname){ 
         echo "<p class='alert alert-warning'>Surname can only contain alphabets</p>"; 
} 
else if(!$correctlastname) { 
         echo "<p class='alert alert-warning'>Last Name can only contain alphabets</p>"; 
} 
else if (!$correctusername) { 
         echo "<p class='alert alert-warning'>Username can only contain alphabets</p>"; 
} 
else { 
    //the rest 

IFS他に使用するか、死ぬ次のいずれかの

if(!$correctsurname){ 
         die("<p class='alert alert-warning'>Surname can only contain alphabets</p>");  

}