php - file_get_contents 大文件上传

Question

我正在尝试使用 fsockopen 上传超过 2gb 的大文件。但是 file_get_content 出现以下错误，我无法将大文件存储在内存中。我需要分块发送数据，但不知道如何执行此操作。请问有人可以指导我吗？

Fatal error: Allowed memory size of 134217728 bytes exhausted (tried to allocate 727826869 bytes) in

    $file_info =  finfo_open(FILEINFO_MIME);  
    $mime_type = finfo_file($file_info, $file_loc); 
    $file = file_get_contents($file_loc);

    $fp = fsockopen($host, 80, $errno, $errstr, 30);
    if ($fp) {
        $out = "POST /upload/".basename($url)." HTTP/1.1\r\n";
        $out .= "Host: ".$host."\r\n";
        $out .= "Accept-Language: en-US,en;q=0.8\r\n";
        $out .= "Content-Type: ".$mime_type."\r\n";
        $out .= 'Content-Length: ' . filesize($file_loc) . "\r\n";
        $out .= "Content-Disposition: attachment; filename=\"".urlencode(basename($file_loc))."\"\r\n\r\n";

        fwrite($fp, $out);
        fwrite($fp, $file);
        $response = '';
        while (!feof($fp)) {
            $response .= fgets($fp, 128);
        }
        fclose($fp);

Answer 1

file_get_contents 将整个文件的内容作为单个字符串变量返回。在您的情况下，这意味着它将尝试创建 2 GB 的变量，这会耗尽允许的脚本内存。

尝试使用 fopen 和 fgets。这将允许您以较小的 block 处理文件。

$file_info =  finfo_open(FILEINFO_MIME);  
$mime_type = finfo_file($file_info, $file_loc); 
$fileHandle = fopen($file_loc);

$fp = fsockopen($host, 80, $errno, $errstr, 30);
if ($fp) {
    $out = "POST /upload/".basename($url)." HTTP/1.1\r\n";
    $out .= "Host: ".$host."\r\n";
    $out .= "Accept-Language: en-US,en;q=0.8\r\n";
    $out .= "Content-Type: ".$mime_type."\r\n";
    $out .= 'Content-Length: ' . filesize($file_loc) . "\r\n";
    $out .= "Content-Disposition: attachment; filename=\"".urlencode(basename($file_loc))."\"\r\n\r\n";

    fwrite($fp, $out);
    while(!feof($fileHandle)){
        fwrite($fp, fgets($fileHandle, 1024));
    }
    fclose($fileHandle);
    
    $response = '';
    while (!feof($fp)) {
        $response .= fgets($fp, 128);
    }
    fclose($fp);