在node.js中用tty生成子进程 [英] Spawning a child process with tty in node.js
问题描述
我正在尝试使用ssh在远程服务器上做一些工作-并从node.js在本地计算机上调用ssh
I am trying to do some work on a remote server using ssh--and ssh is called on the local machine from node.js
该脚本的精简版本如下:
A stripped down version of the script looks like this:
var execSync = require("child_process").execSync;
var command =
'ssh -qt user@remote.machine -- "sudo mv ./this.thing /to/here/;"';
execSync(command,callback);
function callback(error,stdout,stderr) {
if (error) {
console.log(stderr);
throw new Error(error,error.stack);
}
console.log(stdout);
}
我收到 requiretty 错误 sudo:抱歉，您必须有一个tty才能运行sudo .
如果我直接从命令行运行 ssh -qt user@remote.machine-"sudo mv ./this.thing/to/here/;" 从tty中获取-我没有错误，并且 this.thing 将/to/there/很好地移动了.
If I run ssh -qt user@remote.machine -- "sudo mv ./this.thing /to/here/;" directly from the command line--in other words, directly from a tty--I get no error, and this.thing moves /to/there/ just fine.
这是部署脚本的一部分，在其中将！requiretty 添加到sudoers文件中确实不是理想的选择.
This is part of a deploy script where it really wouldn't be ideal to add !requiretty to the sudoers file.
反正有没有办法让node.js在tty中运行命令?
Is there anyway to get node.js to run a command in a tty?
推荐答案
有一些选择:
-
如果您不介意为子进程重复使用父进程的stdin/stdout/stderr(假设它可以访问真实的tty)，则可以使用
stdio:'inherit'(或如果需要，仅继承单个流)在您的spawn()选项中.
If you don't mind re-using stdin/stdout/stderr of the parent process (assuming it has access to a real tty) for your child process, you can use
stdio: 'inherit'(or onlyinheritindividual streams if you want) in yourspawn()options.
通过 pty
使用像 ssh2 这样的ssh模块完全涉及子进程(并具有更大的灵活性).使用 ssh2 ，您只需将 pty:true 选项传递给 exec()， sudo 就可以正常工作.
Use an ssh module like ssh2 that doesn't involve child processes at all (and has greater flexibility). With ssh2 you can simply pass the pty: true option to exec() and sudo will work just fine.
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