如何將選定的複選框值插入到數據庫中

Question

我想將選中的複選框值插入數據庫表 ，該表從數據庫表中動態傳遞。你能建議我該怎麼做

我只是想插入所有數據在另一個表中。這樣

while($row = mysql_fetch_assoc($result)) 
     { 
     ?> 

      <tr> 
       <td><input type="checkbox" name="check[]" id="check" value="<?php echo $row['id'] ?>"></td> 

      </tr> 

在for循環

<form name="form1" id="show_sub_categories"> 
      <table width="90%" border="0" cellpadding="1" cellspacing="1" bgcolor="#c0c0c0"> 
       <tr> 
       <td width="9%" align="center" bgcolor="#FFFFFF" class="head01">Select</td> 
       <td width="7%" align="center" bgcolor="#FFFFFF" class="head01">Sr. No. </td> 
       <td width="25%" align="center" bgcolor="#FFFFFF" class="head01">Leader name</td> 
       <td width="13%" align="center" bgcolor="#FFFFFF" class="head01">Facebook Like</td> 
       <td width="11%" align="center" bgcolor="#FFFFFF" class="head01">Twitter Like</td> 
       <td width="10%" align="center" bgcolor="#FFFFFF" class="head01"> Value</td> 
       <td width="12%" align="center" bgcolor="#FFFFFF" class="head01">View Profile</td> 
       <td width="15%" align="center" bgcolor="#FFFFFF" class="head01">Select PM</td> 
       </tr> 

<?php 
    $election=$_REQUEST['election']; 
    $result = mysql_query("SELECT * FROM leader_info"); 
    $i=1; 
    while($row = mysql_fetch_assoc($result)) 
     {     
?> 
      <tr> 
       <td bgcolor="#FFFFFF" align="center"><input type="hidden" value="0" id="chkids" name="chkid[0]"> 
       <input type="checkbox" id="ChkIds_<?php echo $row['leader_id']; ?>" value="" name="selected[]" onChange="myFunction(this)" class="<?php echo $row['state_point'];?>"> 
       </td> 
       <td bgcolor="#FFFFFF" align="center"><span class="subdetails"><?php echo $i++; ?></span></td> 
       <td bgcolor="#FFFFFF" align="left"><span class="subdetails"><?php echo $row['leader_name']; ?></span></td> 
       <td bgcolor="#FFFFFF" align="left"><?php echo $row['facebook']; ?></td> 
       <td bgcolor="#FFFFFF" align="left"><?php echo $row['twitter']; ?></td> 
       <td bgcolor="#FFFFFF" align="left" class="<?php echo $row['state_point'];?>"><span id="state_point_<?php echo $row['leader_id'];?>"></span></td> 
       <td align="center" bgcolor="#FFFFFF"><a href='#'><img style='vertical-align:bottom;' src='images/view profile.png' width='10' height='10' border='0'></a></td> 
       <td align="center" bgcolor="#FFFFFF"><input type="radio" name="pm" value="pm"></td> 
      </tr> 
<?php } ?> 
      <tr> 
       <td align="center" bgcolor="#FFFFFF">Remaining Budget : </td> 
       <td align="center" bgcolor="#FFFFFF"><span id="total" class="0">1000</span></td> 
      </tr> 
      <tr> 
       <td align="center" bgcolor="#FFFFFF"><input type="submit" name="add" value="Add Team"></td> 
       <td align="center" bgcolor="#FFFFFF"><input type="reset" value="Clear Team"></td> 
      </tr> 
     </table> 
    </form> 

Answer 1

取複選框值提交按鈕

if(isset($_POST['submit'])) 
    { 
     for ($i=0; $i<count($_POST['check']);$i++) 
     { 
      $delete=mysql_query("UPDATE table name SET colname='value' WHERE id= ".$_POST['check'][$i]); 
     } 
    } 

嘗試，這可能是這將是工作後