如何將select從select中傳遞給PHP？

-1

<script> 


    function firstStep(element) { 
var select_name = element.name; 
var option_value = element.value; 
var option_user_selection = element.options[ element.selectedIndex ].value; 
console.log(option_user_selection); 
element.id=option_user_selection; 



    }</script> 
</head> 
<body> 

    <div class="bck"> 

     <form name="login" action="" method="post" accept-charset="utf-8"> 

      <label for="usermail">Username</label> 
      <input type="text" name="nume" placeholder="username" required> 

      <label for="password">Password</label> 
      <input type="password" name="password" placeholder="password" required> 

      <input type="submit" value="Login"> 
     </form> 
     <?php 
      $l=0; 
      $nl=1; 
      $connection = mysqli_connect("127.0.0.1", "root", "", "agentie"); 
      if(isset($_POST['nume']) && isset($_POST['password'])) { 
       echo $_POST["nume"]; 
       echo $_POST["password"]; 
       $nl=0; 
       $query1 = mysqli_query($connection, "SELECT user, password FROM user u WHERE u.user='$_POST[nume]'") or die("Error in the consult.." . mysqli_error($connection)); 
       while ($row = mysqli_fetch_assoc($query1)) { 

        if (($row["user"] == $_POST["nume"]) && ($row["password"] == $_POST["password"])) 
        { 
        echo "Bine ati venit" . $row["user"]; 
        $l=1; 

     ?> 
     <span>Alegeti actiunea:</span> 

     <select class="nume" name="optiune" onchange="firstStep(this)" id="select"> 
      <option value="select">Afiseaza</option> 
      <option value="insert">Introdu</option> 
      <option value="delete">Sterge</option> 
      <option value="update">Modifica</option> 
     </select> 
      <script> 
      $(document).ready(function() { 
       $('.nume').on('change',function(){ 
       var id = $(this).attr('id'); 

       alert(id); 
       }); 
      }); 

      </script> 

     <?php 

     } 


     } 

     if($l==0) { 
     ?> 
     <div style="position:relative; top:300px;" class="col-md-offset-3 col-md-8"> 
      <span class="titlu">Jermaine Modelling</span> 
     </div> 
     <?php 
     } 

     } 
     if(($l==0) && ($nl==1)) { 
     ?> 
     <div style="position:relative; top:300px;" class="col-md-offset-3 col-md-8"> 
      <span class="titlu">Jermaine Modelling</span> 
     </div> 
     <?php 
     } 
     ?> 
    </div> 
</body>

我想，根據選定的選項，使用PHP和HTML做一些事情;所以我改變了使用Javascript的onChange事件的選擇ID，但我不知道如何將更改的ID傳遞給PHP;我該怎麼做（我知道還有另一個類似的問題，但我不知道如何處理這個功能）？或者，有沒有選擇一個選項，而沒有形式（因爲我已經有一個表格）與PHP的東西有另一種方式？如何將select從select中傳遞給PHP？

來源

2014-12-26 Marian Bratu

你爲什麼改變了ID？當您提交表單時，該值將被髮送到PHP。爲什麼不是'

'內的'