Question

再一次是我。我会保证在这之后我已经完成了一段时间了！！）

我有这个评级系统，允许用户评价文章。它有点工作，但问题是它不更新数据库中的数据，我不知道为什么。任何帮助，将不胜感激。：）

// Connects to your Database 
 mysql_connect("URL", "username", "password") or die(mysql_error()); 
 mysql_select_db("db_name") or die(mysql_error()); 


     //Then we update the voting information by adding 1 to the total votes and adding their vote (1,2,3,etc) to the total rating  
 if(isset($_submit['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+$voted, votes = votes+1 WHERE id = $id"); 

        Echo "Your vote has been cast <p>"; 
        } 



 //Puts SQL Data into an array
 $data = mysql_query("SELECT * FROM vote") or die(mysql_error()); 

 //Now we loop through all the data 
 while($ratings = mysql_fetch_array( $data )) 
 { 
 //This outputs the sites name 
 Echo "Name: " .$ratings['name']."<br>"; 

 //This calculates the sites ranking and then outputs it - rounded to 1 decimal 
 $current = $ratings['total'] / $ratings['votes']; 
 Echo "Current Rating: " . round($current, 1) . "<br>"; 




 //This creates 5 links to vote a 1, 2, 3, 4, or 5 rating for each particular item 
 Echo "Rank Me: "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=1&id=".$ratings['id'].">1</a> | "; //The HREF was ".$_SERVER['PHP_SELF']." before
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=2&id=".$ratings['id'].">2</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=3&id=".$ratings['id'].">3</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=4&id=".$ratings['id'].">4</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=5&id=".$ratings['id'].">5</a><p>"; 
 } 


 ?>

谢谢你们！：）

Answer 1

撤消此行

if(isset($_submit['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+$voted, votes = votes+1 WHERE id = $id"); 

       echo "Your vote has been cast <p>"; 
        }

有了这个

if(isset($_POST['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+$voted, votes = votes+1 WHERE id = $id"); 

        echo "Your vote has been cast <p>"; 
        }

Answer 2

php

中有 No $ _SUBMIT 超全局变量

使用$_POST或$_GET

Read here

如果您仍然使用mysql_error()

检查错误

mysql_query ("UPDATE vote SET total= total+$voted, votes = votes+1 WHERE id = $id") 
or  die(mysql_error());

或回显您的查询并在本地数据库中运行它。

然后与我们分享您的错误。

不要使用mysql _ 因为它们被删除了。*

Answer 3

你应该使用$_GET

     //Then we update the voting information by adding 1 to the total votes and adding their vote (1,2,3,etc) to the total rating  
 if(isset($_GET['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+$_GET['voted'], votes = votes+1 WHERE id = $_GET['id']"); 

        Echo "Your vote has been cast <p>"; 
        }

Answer 4

首先：使用$_GET['voted']代替$_submit['voted']。

此外，您构建的链接是错误的。改变

 Echo "<a href='index.php?site=kumu'?mode=vote&voted=1&id=".$ratings['id'].">1</a> | "; //The HREF was ".$_SERVER['PHP_SELF']." before
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=2&id=".$ratings['id'].">2</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=3&id=".$ratings['id'].">3</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=4&id=".$ratings['id'].">4</a> | "; 
 Echo "<a href='index.php?site=kumu'?mode=vote&voted=5&id=".$ratings['id'].">5</a><p>";

到

 Echo "<a href=index.php?site=kumu&mode=vote&voted=1&id=".$ratings['id'].">1</a> | "; //The HREF was ".$_SERVER['PHP_SELF']." before
 Echo "<a href=index.php?site=kumu&mode=vote&voted=2&id=".$ratings['id'].">2</a> | "; 
 Echo "<a href=index.php?site=kumu&mode=vote&voted=3&id=".$ratings['id'].">3</a> | "; 
 Echo "<a href=index.php?site=kumu&mode=vote&voted=4&id=".$ratings['id'].">4</a> | "; 
 Echo "<a href=index.php?site=kumu&mode=vote&voted=5&id=".$ratings['id'].">5</a><p>";

<强>更新

替换声明：

 if(isset($_submit['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+$voted, votes = votes+1 WHERE id = $id"); 

        Echo "Your vote has been cast <p>"; 
        }

使用：

 if(isset($_GET['voted']) && is_numeric($_GET['voted'])) { 
 mysql_query ("UPDATE vote SET total= total+ " . $_GET['voted'] . ", votes = votes+1 WHERE id = " . $_GET['id']); 

        Echo "Your vote has been cast <p>"; 
        }

当然，为了防止任何SQLInjection，必须使用更好的参数验证和转义。

信息未在数据库中更新

4 个答案: