如何訪問多個複選框名稱，同時複選框名稱來自數據庫在php

Question

下面的代碼顯示來自數據庫的一些數據，這裏在每一行中都有一個複選框，名稱爲來自數據庫的primaryid ..並且當用戶單擊時小於5複選框出現一個警告顯示所有的代碼正在工作。但

在這裏我想，當用戶點擊5複選框我想單獨的名稱5複選框選擇和提交按鈕單擊，我想做一些PHP代碼與這5個獨特的ID。

$sql="some sql"; 
$result = mysqli_query($mysql,$sql); 

while($res1=mysqli_fetch_array($result,MYSQLI_ASSOC)) { 


        ?> 
           <tr> 

           <td> 

           <input type="checkbox" name="studentid" value="<?php echo $res1['leadID']?>" ></td> 


       <td>some stuff here </td> 
           </tr> 
        <?php } 
           ?>  

<input type="submit" name="submit" value="Submit" class="btn newcol1" id="submit2" onClick="checkboxes();"> 


       <script> 
    function checkboxes() 
     { 

     var inputElems = document.getElementsByTagName("input"), 
     count = 0; 

     for (var i=0; i<inputElems.length; i++) {  
      if (inputElems[i].type == "checkbox" && inputElems[i].checked == true){ 
       count++; 

      } 

     } 
     if(count < 5){ 
        alert ("Select at least 5 leads"); 
        } 




    } </script> 

Answer 1

不知道這是否是你想要什麼，但在這裏我已經作出如此的複選框（）函數將所有的唯一ID到一個數組，然後如果超過5個複選框已被點擊，張貼在到當前頁面，您可以在PHP中訪問ID。

function checkboxes(){ 
    var inputElems = document.getElementsByTagName("input"), 
    count = 0; 

    var ids = [] 
    for (var i=0; i<inputElems.length; i++) {  
     if (inputElems[i].type == "checkbox" && inputElems[i].checked == true){ 
      ids.push(inputElems[i].getAttribute("value")); 
      count++; 
     } 
    } 

    if(count < 5){ 
     alert ("Select at least 5 leads"); 
    } 
    else{ 
     //more than 5 checkboxes selected 
     post(window.location.href, {ids: ids}); 
    } 
} 

function post(path, params, method) { 
    method = method || "post"; // Set method to post by default if not specified. 

    // The rest of this code assumes you are not using a library. 
    // It can be made less wordy if you use one. 
    var form = document.createElement("form"); 
    form.setAttribute("method", method); 
    form.setAttribute("action", path); 

    for(var key in params) { 
     if(params.hasOwnProperty(key)) { 
      var hiddenField = document.createElement("input"); 
      hiddenField.setAttribute("type", "hidden"); 
      hiddenField.setAttribute("name", key); 
      hiddenField.setAttribute("value", params[key]); 

      form.appendChild(hiddenField); 
     } 
    } 

    document.body.appendChild(form); 
    form.submit(); 
}