file_get_contents创建文件不存在

Question

如果file_get_contents不存在,会创建该文件吗？我基本上是在找一行命令.我用它来计算程序的下载统计数据.我在预下载页面中使用此PHP代码:

Download #: <?php $hits = file_get_contents("downloads.txt"); echo $hits; ?>

然后在下载页面中,我有这个.

<?php
    function countdownload($filename) {
        if (file_exists($filename)) {
            $count = file_get_contents($filename);
            $handle = fopen($filename, "w") or die("can't open file");
            $count = $count + 1;
        } else {
            $handle = fopen($filename, "w") or die("can't open file");
            $count = 0; 
        }
        fwrite($handle, $count);
        fclose($handle);
    }

    $DownloadName = 'SRO.exe';
    $Version = '1';
    $NameVersion = $DownloadName . $Version;

    $Cookie = isset($_COOKIE[str_replace('.', '_', $NameVersion)]);

    if (!$Cookie) {
        countdownload("unqiue_downloads.txt");
        countdownload("unique_total_downloads.txt");
    } else {
        countdownload("downloads.txt");
        countdownload("total_download.txt");
    }

    echo '<META HTTP-EQUIV=Refresh CONTENT="0; URL='.$DownloadName.'" />';
?>

当然,用户首先访问预下载页面,因此尚未创建.我不想在预下载页面添加任何功能,我希望它简单明了,而且不需要添加/更改.

编辑:

这样的东西会起作用,但它对我不起作用？

$count = (file_exists($filename))? file_get_contents($filename) : 0; echo $count;

Answer 1

Download #: <?php
$hits = '';
$filename = "downloads.txt";
if (file_exists($filename)) {
    $hits = file_get_contents($filename);
} else {
    file_put_contents($filename, '');
}
echo $hits;
?>

您也可以使用fopen()'w +'模式:

Download #: <?php
$hits = 0;
$filename = "downloads.txt";
$h = fopen($filename,'w+');
if (file_exists($filename)) {
    $hits = intval(fread($h, filesize($filename)));
}
fclose($h);
echo $hits;
?>