需要幫助來修改php腳本與開關案例

Question

我需要修改這個源代碼php腳本代碼。當我與開關的情況下修改我有一個錯誤的text1沒有價值，當我給值的text1

$query = "INSERT INTO mytable(text_name, text_image, text_detail) VALUES (:name,  :image, :detail) "; 
$query_params = array(
    ':name' => $_POST['text_name'], 
    ':image' => $_POST['text_image'], 
    ':detail' => $_POST['text_detail '] 
); 

向（像）這

switch("text1" || "tex2" || "text3"){ 
      case "text1": 
      $query = "INSERT INTO comments (text_name, text_image, text1_detail) 
    VALUES (:name, :image, :detail) "; 
     //Update query 
    $query_params = array(
     ':name' => $_POST['text_name'], 
    ':image' => $_POST['text_image'], 
    ':detail' => $_POST['text1_detail']    
      ); 
      break; 
      case "tex2": 
      $query = "INSERT INTO comments (text_name,text_image, text2_detail) 
    VALUES (:text_id, :image, :detail) "; 
     //Update query 
    $query_params = array(
     ':name' => $_POST['text_name'], 
     ':image' => $_POST['text_image'], 
     ':detail' => $_POST['text2_detail '] 
      ); 
      break; 
      case "text3": 
      $query = "INSERT INTO comments (text_name, text_image, text3_detail) 
    VALUES (:text, :image, :detail) "; 
     //Update query 
    $query_params = array(
     ':name' => $_POST['text_name'], 
    ':image' => $_POST['text_image'], 
    ':detail' => $_POST['text3_detail'],   
      ); 
      break; 
      default: echo "Something is wrong fuckingshit!! Give me at least text1 or text2 or text3 !!!"; 
    } 

，但我得到錯誤的text1沒有價值當我給那裏text1，有什麼不對？有什麼建議麼？

Answer 1

一個switch語句接受一個變量，並評估其對在case小號的每一個值：

switch ($yourVariable) { 
    case "text1": # Equivilent to if ($yourVariable == "text1") 
     # do stuff 
     break; 
    case "text2": 
     # do different stuff 
     break; 
    case "text3": 
     # etc. 
     break; 
}