那么如何选择一个更大的id呢

I'm working on a news system for a website. I used the news system from this tutorial: http://pixelcode.co.uk/tutorials/php/mysql-news-system/

我正在为一个网站开发一个新闻系统。我使用了本教程中的新闻系统:http://pixelcode.co.uk/tutorials/php/mysql-news-system/

So to display the news I use this line of code:

为了显示新闻，我用了这一行代码:

if(!$_GET['id'])
        {   
            $query = mysql_query("SELECT * FROM news ORDER BY id DESC");
            while($output = mysql_fetch_assoc($query))
            {
                    echo '<div id="nieuws">';
                    echo '<a href="?id='.$output['id'].'" id="link">';                      
                    echo '<h1>'.$output['title'].'</h1>';
                    echo '<span id="date">'.date('d-m-y', $output['date']).'</span><br / >';
                    echo $output['shortnews'].'<br / >';
                    echo '</a>';
                    echo '</div>';      
            }

        }
        else
        {
            $id = $_GET['id']; 
            $query = mysql_query("SELECT * FROM news WHERE id='$id'");
            $output = mysql_fetch_assoc($query);                
    ?>      
            <form method="post" action="?id=<? echo $output['id']; ?>"> 

            <h1><? echo $output['title']; ?></h1>
            <? echo '<span id="date">'.date('d-m-y', $output['date']).'</span><br / >' ?>
            <? echo $output['news']; ?>
            </form>
    <?php } ?>

I can do: if(!$_GET['id'] = 6) but I can't do this: if(!$_GET['id'] > 6). What's the problem? Or is there another code for bigger then an id?

我能做的:如果(!$_GET['id'] = 6)但我不能这么做:if(!$_GET['id'] > 6).有什么问题吗?或者还有比id更大的代码吗?

Thanks for replying, MARCH

谢谢回复,3月

2 个解决方案

#1

Operator precedence: ! binds tighter than ==, so you're effectively doing

运算符优先级:!绑紧比==，所以你实际上在做。

(not($_GET) == 6)

That's why there's != for inequality tests.

这就是为什么不平等测试有!=。

If you'd used proper bracketing, e.g

如果你用了合适的括号，例如

(!($_GET['id'] == 6))

then you'd be doing

然后你会做什么

not(id == 6)

and get your expected results, because that's logically/functionally equivalent to id != 6

并得到预期结果，因为这在逻辑上/功能上等同于id != 6

And note that you're vulnerable to sql injection attacks and are using an obsolete/deprecated DB interface.

注意，您很容易受到sql注入攻击，并且正在使用过时的/不赞成使用的DB接口。

#2

Use isset():

使用收取():

if (isset($_GET['id'])) ...

Or, what I like to do:

或者，我想做的是:

// Get all possible arguments:
$id  = @$_GET['id'];   // The @ prevents an error message
$foo = @$_GET['foo'];  // foo means ...  (explain the args)
...
// Later, as needed:
if (isset($id)) ...

#1