www :: curl - 如何上传(发布)大文件

tok*_*tok 5 perl curl file-upload libcurl

我使用WWW :: Curl上传文件:

use WWW::Curl::Easy 4.14;
use WWW::Curl::Form;

my $url = 'http://example.com/backups/?sid=12313qwed323';
my $params = {
    name => 'upload',
    action => 'keep',
    backup1 => [ '/tmp/backup1.zip' ],   # 1st file for upload
};

my $form = WWW::Curl::Form->new();
foreach my $k (keys %{$params}) {
    if (ref $params->{$k}) {
        $form->formaddfile(@{$params->{$k}}[0], $k, 'multipart/form-data');
    } else {
        $form->formadd($k, $params->{$k});
    }
}

my $curl = WWW::Curl::Easy->new() or die $!; 
$curl->setopt(CURLOPT_HTTPPOST, $form);
$curl->setopt(CURLOPT_URL, $url);

my $body;   
$curl->setopt(CURLOPT_WRITEDATA, \$body);
my $retcode = $curl->perform();
my $response_code = $curl->getinfo(CURLINFO_HTTP_CODE); 
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没什么特别的,这个代码效果很好.

我想上传大文件,我不想预先加载内存中的所有内容.至少这是我听说libcurl正在做的事情.

CURLOPT_READFUNCTION接受返回部分内容的回调.这意味着我不能使用WWW :: Curl :: Form来设置POST参数,但我必须通过此回调返回整个内容.是对的吗?

我认为代码看起来像这样:

use WWW::Curl::Easy 4.14;

my $url = 'http://example.com/backups/?sid=12313qwed323'
my $params = {
    name => 'upload',
    action => 'keep',
    backup1 => [ '/tmp/backup1.zip' ],   # 1st file for upload
};

my $fields;
foreach my $k (keys %{$params}) {
    $fields .= "$k=".(ref $params->{$k} ? '@'.@{$params->{$k}}[0] : uri_escape_utf8($params->{$k}))."&";
}
chop($fields);

my $curl = WWW::Curl::Easy->new() or die $!;
$curl->setopt(CURLOPT_POST, 1);
$curl->setopt(CURLOPT_POSTFIELDS, $fields); # is it needed with READFUNCTION??
$curl->setopt(CURLOPT_URL, $url);

my @header = ('Content-type: multipart/form-data', 'Transfer-Encoding: chunked');
$curl->setopt(CURLOPT_HTTPHEADER, \@header);

#$curl->setopt(CURLOPT_INFILESIZE, $size);
$curl->setopt(CURLOPT_READFUNCTION, sub {

    # which data to return here?
    # $params (without file) + file content?

    return 0;
});
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CURLOPT_READFUNCTION回调必须返回哪些数据?$ params +文件内容?格式是什么?

我是否真的必须自己创建数据(由CURLOPT_READFUNCTION返回),还是有一种简单的方法以正确的格式创建它?

谢谢

dax*_*xim 4

测试16formpost.t是相关的。如您所见,它已完全禁用。这一事实以及我对回调函数的各种返回值进行的毫无结果的实验​​让我相信该CURLOPT_READFUNCTION功能在 Perl 绑定中已被破坏。

我必须通过这个回调返回全部内容。是对的吗?

不,您可以分段向其提供请求正文,适合分块编码。根据 中设置的限制,回调必须被调用多次CURLOPT_INFILESIZE

CURLOPT_READFUNCTION 回调必须返回哪些数据?

HTTP 请求正文。由于您进行文件上传,这意味着Content-Type multipart/form-data。以下是使用 HTTP::Message 的示例。CURLOPT_HTTPPOST是构造此格式的另一种方法。

use HTTP::Request::Common qw(POST);
use WWW::Curl::Easy 4.14;

my $curl = WWW::Curl::Easy->new or die $!;
$curl->setopt(CURLOPT_POST, 1);
$curl->setopt(CURLOPT_URL, 'http://localhost:5000');
$curl->setopt(CURLOPT_HTTPHEADER, [
    'Content-type: multipart/form-data', 'Transfer-Encoding: chunked'
]);
$curl->setopt(CURLOPT_READFUNCTION, sub {
    return POST(undef, Content_Type => 'multipart/form-data', Content => [
        name    => 'upload',
        action  => 'keep',
        backup1 => [ '/tmp/backup1.zip' ],   # 1st file for upload
    ])->content;
});
my $r = $curl->perform;
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