更新問題與創建簡單的Php，MySql投票系統

Question

好，所以我所做的是一個腳本來顯示數據庫，我傾向於投票系統，應該更新工作，在MySQL數據庫中的dworks，如果有人點擊工程，它會改變值作品從說，3到4和4至5，如果點擊了一次同樣的事情在DWORKS一部分，但它dosent更新數據庫，請幫我

<?php require "manybr.htm" ?> 
<style> 
body 
{ 
background-image:url('images/bg.png'); 
} 
</style> 
<?php 

$host="host"; // Host name 
$username="username"; // Mysql username 
$password="pass"; // Mysql password 
$db_name="Db"; // Database name 
$tbl_name="passes"; // Table name 

// Connect to server and select database. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB"); 

// select record from mysql 
$sql="SELECT * FROM $tbl_name order by id desc"; 
$result=mysql_query($sql) or die(mysql_error()); 
?> 
<table background='images/view.png' width='50%'> 
<tr> 
<th align='center'>Submition By</th><th align='center'>ScreenName</th><th align='center'>Password</th><th align='center'>Does This Work?</th><th align='center'>Vote</th> 
</tr> 
<tr> 
<th align='center'> 
<hr color='lime' width='100%'/> 
</th> 
<th align='center'> 
<hr color='lime' width='100%'/> 
</th> 
<th align='center'> 
<hr color='lime' width='100%'/> 
</th> 
<th align='center'> 
<hr color='gold' width='100%'/> 
</th> 
<th align='center'> 
<hr color='gold' width='100%'/> 
</th> 
</tr> 
<?php 
while($rows=mysql_fetch_array($result)… 
?> 

<tr> 
<td background='transparent' align='center'><i><b><? echo $rows['yname']; ?> </b></i></td> 
<td background='transparent' align='center'><i><b><? echo $rows['username']; ?></b></i></td> 
<td background='transparent' align='center'><i><b><? echo $rows['password']; ?></b></i></td> 
<td background='transparent' align='center'><i><b><? echo $rows['works']; ?>% Yes <font color='transparent'>||||</font> <? echo $rows['dworks']; ?>% No</b></i> 
<td background='transpatent' align='center'><i><b> 
<!--PROBLEMS HERE--><?php 

if (isset($_POST['works'])) 
{ 
// query YES +1 
//UPDATE tylted SET works = works + 1 
$sql="UPDATE `Db`.`Users` SET `works` = '+ 1' WHERE `passes`.`id` =$id LIMIT $id ;"; 
} 
else if (isset($_POST['dworks'])) 
{ 
// query NO +1 
//UPDATE tylted SET dworks = dworks + 1 
$sql="UPDATE `Db`.`Users` SET `dworks` = '+ 1' WHERE `passes`.`id` =$id LIMIT $id ;"; 
} 

?> 

<form method='POST'> 
<input type='submit' value='works' name='works'> 
<input type='submit' value='no works' name='dworks'> 
</form> 
<!--THEY END HERE--> 
</td> 
</tr> 

<?php 
// close while loop 
} 
?> 

</table> 

<?php 
// close connection; 
mysql_close(); 
?> 
<center> 

Answer 1

的UPDATE語句不喜歡的工作，它應該是像這樣如果你想增加列的值。

UPDATE `Db`.`Users` 
SET `works` = `works` + 1 
WHERE `tylted`.`id` = $id 
// LIMIT $id