脚本计数加2而不是1

问题描述：

I have a count.php script that counts the number of refreshes on my site. The code for it is this:

if (file_exists('countlog.txt')) 
{
    $fil = fopen('countlog.txt', r);
    $dat = fread($fil, filesize('countlog.txt')); 
    echo $dat+1;
    fclose($fil);
    $fil = fopen('countlog.txt', w);
    fwrite($fil, $dat+1);
}

else
{
    $fil = fopen('countlog.txt', w);
    fwrite($fil, 1);
    echo '1';
    fclose($fil);
}

The issue is that when I try to run it, it always count's as two. I have tried to edit the: echo $dat+1 to just echo $dat but it doesn't seem to work.

Any help?

我有一个count.php脚本，用于计算我网站上的刷新次数。代码是这样的： p>

  if（file_exists（'countlog.txt'））
 {
 $ fil = fopen（'countlog.txt'，r）  ; 
 $ dat = fread（$ fil，filesize（'countlog.txt'））;  
 echo $ dat + 1; 
 fclose（$ fil）; 
 $ fil = fopen（'countlog.txt'，w）; 
 fwrite（$ fil，$ dat + 1）; 
} 
  
else 
 {
 $ fil = fopen（'countlog.txt'，w）; 
 fwrite（$ fil，1）; 
 echo'1'; 
 fclose（$ fil）; 
} \  n  code>  pre> 
 
 问题在于，当我尝试运行它时，它总是算作两个。 我曾尝试将 echo $ dat + 1  code>编辑为 echo $ dat  code>，但它似乎不起作用。 p> 
 
 
有任何帮助吗？ p> 
  div>

答

You have error_resporting on? I think not.

Try to:

if (file_exists('countlog.txt')) 
    {
        $fil = fopen('countlog.txt', 'r');
        $dat = fread($fil, filesize('countlog.txt')); 
        echo $dat+1;
        fclose($fil);
        $fil = fopen('countlog.txt', 'w');
        fwrite($fil, $dat+1);
    }

    else
    {
        $fil = fopen('countlog.txt', 'w');
        fwrite($fil, 1);
        echo '1';
        fclose($fil);
    }

Check this: ('countlog.txt', 'w')you missed the 'w'.

The error that outputs if you miss the single quote ' is this:

Notice: Use of undefined constant r - assumed 'r'

脚本计数加2而不是1

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