Question

好的，我所做的是创建一个脚本来显示数据库并包含一个投票系统，该系统应该更新工程，如果有人点击工作，则在mysql数据库中dworks它会改变作品的值从3到4如果dworks部分再次点击了同样的东西，那么4到5，但它不会更新数据库，请帮助我：

以下是我的代码：

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

    $host="Host"; // Host name 
    $username="Username"; // Mysql username 
    $password="password"; // Mysql password 
    $db_name="Database Name"; // Database name 
    $tbl_name="tylted"; // 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);
    ?>
    <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>&nbsp; <? echo $rows['dworks']; ?>% No</b></i>
    <td background='transpatent' align='center'><i><b>
    <?php

    $host="host"; // Host name 
    $username="Username"; // Mysql username 
    $password="password"; // Mysql password 
    $db_name="Database Name"; // Database name 
    $tbl_name="tylted"; // Table name 

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

   ?>

    <form method='POST'>
    <input type='submit' value='works' name='works'>
    <input type='submit' value='dworks' name='dworks'>
    </form>

    <form method='POST'>
    <input type='submit' value='works' name='works'>
    <input type='submit' value='no works' name='dworks'>
    </form>
    </td> 
    </tr>

    <?php
    // close while loop 
    }
    ?>

    </table>

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

我试图获得

   <?php

$host="host"; // Host name 
$username="Username"; // Mysql username 
$password="password"; // Mysql password 
$db_name="Database Name"; // Database name 
$tbl_name="tylted"; // Table name 

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

?>
<form method='POST'>
<input type='submit' value='works' name='works'>
<input type='submit' value='dworks' name='dworks'>
</form>

部分工作就像投票调查一样，每次点击WORKS按钮都会更新数据库'tylted'，所以如果单击它更改5到6并单击一次，则按钮WORKS上的投票为5％WORKS和2％DOSENT WORK更多6到7和另一个按钮相同。

说出来就是这样：

   Does It Work?  |      VOTE

   9% Yes 2% No   | {WORKS} {DOSENT WORK}

单击{WORKS}时将其更改为

   Does It Work?  |      VOTE

   10% Yes 2% No   | {WORKS} {DOSENT WORK}

当点击{DOSENT WORK}时，它会变为

   Does It Work?  |      VOTE

   10% Yes 3% No   | {WORKS} [DOSENT WORK}

等等等你有点明白吗？我明白他想做什么，但我不知道该怎么做？

Answer 1

$sql="UPDATE ..... `works` = `works` + 1 WHERE `tylted`.`id` = id // LIMIT $id";
                         here is problem--------------------------^^

要么不在查询中，要么

 $sql="UPDATE ..... `works` = `works` + 1 WHERE `tylted`.`id` = '".$id."'";

或

 $sql="UPDATE ..... `works` = `works` + 1 WHERE `tylted`.`id` = '".$id"'  LIMIT '".$id."'";

注意

整个 ext/mysql PHP扩展程序（提供名为mysql_的所有函数）为officially deprecated as of PHP v5.5.0，将来会被删除。因此，请使用PDO或MySQLi

好读

投票系统在MySql更新上将无法正常工作

1 个答案: