在php中發送複選框值到數據庫

Question

我在頁面中有一個表格，它的前兩列來自數據庫中的表格，第三列是複選框，如果用戶知道表格中顯示的單詞的含義，檢查它們 我想以這種方式更新我在數據庫中的表格：對於選中的複選框，插入1到表中的檢查列，否則插入0 如何獲取複選框值並將其插入到數據庫的右側行中 我有此代碼直到如今

<?php 
     $con = mysql_connect("localhost", "root", "") 
     or die(mysql_error()); 
     if (!$con) { 
      die('Could not connect to MySQL: ' . mysql_error()); 
     } 
     mysql_select_db("project", $con) 
     or die(mysql_error()); 
     $result = mysql_query("select * from words"); 
     echo "<table border='1'> 
      <tr> 
       <th>word</th> 
       <th>meaning</th> 
       <th>checking</th> 
      </tr>"; 
      while($row = mysql_fetch_array($result)) { 
        echo "<tr>"; 
        echo "<td>" . $row['word'] . "</td>"; 
        echo "<td>". "<div class='hiding' style='display:none'>".$row['meaning']."</div>"."</td>"; 
        echo "<td>"; 
        echo "<input name=\"fahimeh\" type=\"checkbox\" value=\"\"> "; 
        echo "</td>"; 
        echo "</tr>"; 
        } 
     echo "</table>"; 
     mysql_close($con);       
     ?> 
     <button onclick="ShowMeanings()">ShowMeanings</button> 
     <button onclick="feedback()">sendfeedback</button> 

，這是我的javascript代碼：

<script type="text/javascript"> 
     function ShowMeanings(){ 
     var hidings = document.getElementsByClassName('hiding'); 
     for (var i=0; i<hidings.length; i++){ 
     hidings[i].style.display = 'block'; 
     } 

     } 
    </script> 

我不知道我怎麼能寫反饋（），我怎麼能得到複選框值，並將其插入到右排在數據庫

Answer 1

這應該工作:)

<?php 
     $con = mysql_connect("localhost", "root", "") 
     or die(mysql_error()); 
     if (!$con) { 
      die('Could not connect to MySQL: ' . mysql_error()); 
     } 
     mysql_select_db("project", $con) 
     or die(mysql_error()); 
     $result = mysql_query("SELECT * FROM words"); 
     echo "<table border='1'> 
      <tr> 
       <th>word</th> 
       <th>meaning</th> 
       <th>checking</th> 
      </tr>"; 
      while($row = mysql_fetch_array($result)) { 
        echo "<tr>"; 
        echo "<td>" . $row['word'] . "</td>"; 
        echo "<td>". "<div class='hiding' style='display:none'>".$row['meaning']."</div>"."</td>"; 
        echo "<td>"; 
        echo "<input name=\"fahimeh\" type=\"checkbox\" name='" . $row['id'] . "' value=\"\"> "; 
        echo "</td>"; 
        echo "</tr>"; 
        } 
     echo "</table>"; 
     mysql_close($con);       
     ?> 
     <button onclick="ShowMeanings()">ShowMeanings</button> 
     <button onclick="feedback()">sendfeedback</button>