windows 使用VLC的虚拟界面时如何防止显示控制台
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How to prevent console from being displayed when using VLC's dummy interface
提问by Adam M-W
I'm trying to launch VLC in "dummy" mode from a Node.js server script, however using child_process.spawn('vlc',['-I dummy'])produces a new console window for VLC's output when using Windows. Is there a way to prevent this happening and force all stdout though the stdout ReadableStream so no "popup windows" occur?
我正在尝试从 Node.js 服务器脚本以“虚拟”模式启动 VLC，但是child_process.spawn('vlc',['-I dummy'])在使用 Windows 时使用会为 VLC 的输出生成一个新的控制台窗口。有没有办法防止这种情况发生并通过 stdout ReadableStream 强制所有标准输出，因此不会出现“弹出窗口”？
EDIT: This problem had nothing to do with node.js, it was simply the way I was calling it and VLC's behaviour. The solution is below.
编辑：这个问题与 node.js 无关，这只是我调用它的方式和 VLC 的行为。解决方法如下。
Thanks.
谢谢。
回答by Adam M-W
I found a solution for the specific problem:
我找到了针对特定问题的解决方案：
VLC has a command line option to surpress this window --*-quiet where * is the interface.
VLC 有一个命令行选项来抑制这个窗口 --*-quiet 其中 * 是界面。
e.g. For the dummy interface, use
例如对于虚拟接口，使用
child_process.spawn('vlc',['-I dummy','--dummy-quiet'])
For the rc interface, use
对于 rc 接口，使用
child_process.spawn('vlc',['-I rc','--rc-quiet'])
回答by Pau Coma Ramirez
I would like to complement Adam M-W answer.
我想补充 Adam MW 的回答。
VLC has a command line option to suppress this window --*-quiet where * is the interface.
e.g. For the dummy interface, use
child_process.spawn('vlc',['-I dummy','--dummy-quiet']) For the rc interface, use
child_process.spawn('vlc',['-I rc','--rc-quiet'])
answered Jun 13 '11 at 14:12 Adam M-W
VLC 有一个命令行选项来抑制这个窗口 --*-quiet 其中 * 是接口。
例如对于虚拟接口，使用
child_process.spawn(' vlc',[' -I dummy',' --dummy-quiet']) 对于 rc 接口，使用
child_process.spawn(' vlc',[' -I rc',' --rc-quiet'])
2011 年 6 月 13 日 14:12 亚当 MW 回答
at least on my system,VLC now sends its messages to stdError, so this is the channel which needs to be monitored.
至少在我的系统上，VLC 现在将其消息发送到stdError，因此这是需要监视的通道。
My interface is with Qt , QtProcess and these are the options that worked for me.
我的界面是 Qt 、 QtProcess ，这些是对我有用的选项。
Using MergedChannelsand reading stdOut.
使用MergedChannels并读取stdOut。
m_proc->setProcessChannelMode(QProcess::MergedChannels);
connect (m_proc,SIGNAL(readyReadStandardOutput()),
this, SLOT(readyRead()));
void ReDirVLC::readyRead(){
if (!m_proc) return;
qDebug()<<m_proc->readAllStandardOutput() << endl;
}
Using SeparateChannelsand reading stdError
使用SeparateChannels并读取stdError
m_proc->setProcessChannelMode(QProcess::SeparateChannels);
connect (m_proc,SIGNAL(readyReadStandardError()),
this, SLOT(readyRead()));
void ReDirVLC::readyRead(){
if (!m_proc) return;
qDebug()<<m_proc->readAllStandardError() << endl;
}
回答by clee
Maybe you could run the process with child_process.spawn('start', ['/b', 'vlc', '-I dummy'])instead?
也许你可以用它child_process.spawn('start', ['/b', 'vlc', '-I dummy'])来运行这个过程？
