Question

我正在尝试获取远程图像文件大小。

但我收到此错误：＆＃34;警告：filesize（）：http://mfastore.com/_store/user_data/9fc8022_310/C/hydrangeas.jpg的统计信息失败文件大小未知＆＃34;

这是我试图以MB为单位获取图像大小的代码

    $res= formatFilebytes("http://mfastore.com/_store/user_data/9fc8022_310/C/hydrangeas.jpg","MB");
echo $res;

function formatFilebytes($file, $type)
{
switch($type){
case "KB":
$filesize = filesize($file) * .0009765625; // bytes to KB
break;
case "MB":
$filesize = (filesize($file) * .0009765625) * .0009765625; // bytes to MB
break;
case "GB":
$filesize = ((filesize($file) * .0009765625) * .0009765625) * .0009765625; // bytes to GB
break;
}
if($filesize <= 0){
return $filesize = 'unknown file size';}
else{return round($filesize, 2).' '.$type;}
}

请帮助我。感谢

Answer 1

试试这个

$image_header = get_headers("http://mfastore.com/_store/user_data/9fc8022_310/C/hydrangeas.jpg", true);
$size = $image_header["Content-Length"];

Answer 2

这个解决方案对我有用......

 $url = 'http://mfastore.com/_store/user_data/9fc8022_310/C/hydrangeas.jpg';

echo getRemoteFilesize($url);

function getRemoteFilesize($url, $formatSize = true)
{
    $head = array_change_key_case(get_headers($url, 1));
    // content-length of download (in bytes), read from Content-Length: field
    $clen = isset($head['content-length']) ? $head['content-length'] : 0;

    // cannot retrieve file size, return "-1"
    if (!$clen) {
        return -1;
    }

    if (!$formatSize) {
        return $clen; // return size in bytes
    }

    $size = $clen;
    switch ($clen) {
        case $clen < 1024:
            $size = $clen .' B'; break;
        case $clen < 1048576:
            $size = round($clen / 1024, 2) .' KiB'; break;
        case $clen < 1073741824:
            $size = round($clen / 1048576, 2) . ' MiB'; break;
        case $clen < 1099511627776:
            $size = round($clen / 1073741824, 2) . ' GiB'; break;
    }

    return $size; // return formatted size
}

Answer 3

我相信这会对你有所帮助，这是一个例子：

$filesize = filesize($folder_url . $file) * .0009765625; // bytes to KB

我如何获得远程图像文件大小？

3 个答案: