일단 제가 쓴걸 올려놓겠습니다.
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<form name="product_select" method="post" action='<?php echo $PHP_SELF?>'>
<!-- Selection Scripts starts -->
<?
        require "./lib.php";
        $que="SELECT * FROM monitors order by size";
        $data=mysql_query($que, $connect);
        $total=mysql_affected_rows();
        for($i=0; $i < $total; $i++){
                mysql_data_seek($data,$i);
                $info=mysql_fetch_array($data);
                $file = $info[file_name];                                                                                                
                echo"
                        <table border='0' cellpadding='0' cellspacing='0' width='200'>
                                <tr>
                                        <td align='left' valign='middle' class='normal'><img src="./images/monitors/$info[file_name]" width='50' height='50' border='0' align='left' vspace='2'><b>$info[size]   $info[company]<br>$info[model]<br>$info[display]</b><br>
                                        <input type='checkbox' name='$file' value='0'>
                                        <script language='javascript'>
                                                if(product_select.$file.checked){
                                                        document.write ("<input type='hidden' name='mode1' value='up1'>");
                                                }else{
                                                        document.write('');
                                                }
                                        </script>
                                        <br><br></td>
                                </tr>
                        </table>                                                                        
                ";
                
                if($mode1 = 'up1'){
                        $que = "UPDATE monitors SET display = 1 WHERE file_name='$file'";
                        mysql_query( $que, $connect );
                }else{
                        $que = "UPDATE monitors SET display = 0 WHERE file_name='$file'";
                        mysql_query( $que, $connect );
                }                
                                                                                                                
        }
        echo"
                <input type='image' value='submit' src='./images/submit.gif' width='62' height='20' alt='' border='0'>
        ";
        mysql_close($connect);
?>
<!-- Selection Scripts ends -->
</form>
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간단하게 설명을 하자면요 데이터 베이스에 들어가있는 모니터를 밖에다 출력을 시킨다음 첵크박스에 첵크를 하면 1이란 숫자가 데이터에 입력이되고 첵크를 안하면 0 이란 숫자가 입력되게 할려구 한겁니다.  

그런데 <input type=checkbox name=$file value=0> 이렇게 해서 그 첵크박스에 값을 불르는데요 $file은 데이터베이스에 저장되있는 모니터 그림들 이름입니다.  이런식으로 데이터를 불러와서 출력시킨다음 첵크박스가 첵크됐으면 자바스크립트를 이용해서 <input type='hidden' name='mode1' value='up1'> 요걸 나오게 한다음 나왔으면 1을 넣고 안나오면 0이되고.. 뭐 이런식으로 한다고 열심히 머리굴려서 썼더만 자꾸 다 1 이 들어가요...  분명 쉬운 방법이 있을텐데 아직 php에는 허접이라서 잘 모르겠네요... 고수님들의 조언 다시한번 부탁드립니다.  

---Always KYJ One and Only