使用 phpseclib ssh2 exec() 函数时如何设置标志? [英] How do i set flags when using phpseclib ssh2 exec() function?

问题描述

我正在使用 phpseclib 并且需要制作几个 php 函数，使某人能够以编程方式 ssh 进入他们的服务器并更改 root 密码，并更改可能忘记密码的用户的密码(因此必须以 root 身份登录).

I'm using phpseclib and need to make a couple of php functions that enable someone to programmatically ssh into their server and change the root password and also change the password of a user that may have forgotten their password (so have to be logged in as root).

我尝试使用 libssh2，但发现它使用起来有点讨厌.我现在正在研究 phpseclib，它看起来更健壮.但是当我尝试像这样使用su"命令时:

I tried using libssh2, but found it a bit nasty to use. I'm now looking at phpseclib which seems more robust. But when I tried to use the 'su' command like so:

echo $ssh->exec('su');

我收到回复:

su: must be run from a terminal

当我尝试使用 sudo 时:

and when I try to use sudo:

echo $ssh->exec('sudo passwd root');

我收到错误:

sudo: no tty present and no askpass program specified

无论如何，事实证明 su 被禁用以进行直接 ssh 访问，但是在查看了 这篇文章，原来你可以用下面的命令做到:

Anyway, it turns out that su is disabled for direct ssh access, but after having a look at this article, it turns out you can do it with the following command:

ssh -t -t -l 'username' 'host' 'su -'

无论如何，当从我的笔记本电脑(运行 ubuntu)进入终端时，这最终对我有用，然后我输入了我的密码，然后输入了 root 密码以完成.

That's what finally worked for me anyway when entering into a terminal from my laptop (running ubuntu), and then I entered my password and then the root password to finish off.

引用上面链接的网站:

当给出 -t 时，ssh 命令(使用 -t)远程 sshd 建立一个到工作进程的伪终端"管道.

Ssh commands (using -t) the remote sshd to establish a 'pseudo-terminal' pipe to the worker process when -t is given.

.只要 ssh 的标准输入是终端，ssh 就会这样做.

. ssh does this as long as its stdin is a terminal.

.但是如果ssh的stdin是非终端的，ssh不会引导sshd建立一个
除非给出两个 -t 的伪终端:

. But if ssh's stdin is a non-terminal, ssh won't direct sshd to establish a
pseudo-terminal unless TWO -t's are given:

回显密码 |ssh -t -t -l 用户名远程主机

echo password | ssh -t -t -l username remote_host

.因此，使用 -t -t(来自 ssh)sshd 为客户端进程设置了一个伪终端.

. So with -t -t (from ssh) sshd sets up a pseudo-terminal to the client process.

.客户端，无论是 'tty' 还是 'su' 都无法判断它连接到一个虚构的>终端:

. The client, whether it be 'tty' or 'su' cannot tell it is connected to a ficticious >terminal:

echo dummystr |ssh -t -t -l 用户名 host.com -c ''tty'

echo dummystr | ssh -t -t -l username host.com -c ''tty'

回显密码 |ssh -t -t -l 用户名 host.com -c 'su -'

echo password | ssh -t -t -l username host.com -c 'su -'

答案来了.如果您通过一个 > 交互式客户端 ssh(例如来自 OpenBSD 的客户端 ssh)在 linux 机器上su root"，请使用 double -t.

So there is the answer. Use double -t if you are 'su root'ing' on a linux box through an >interactive client ssh like the one from OpenBSD.

所以，正如我上面所说的那样，它实际上是在终端上工作的:

So, it actually worked from the terminal as I said above using:

ssh -t -t -l 'username' 'host' 'su -'

但我真的希望能够使用 phpseclib 执行此命令.唯一的问题是我不知道如何将任何标志放入 exec() 函数中.具体来说，我需要放入 -t 标志(两次).

but I really want to be able to execute this command using phpseclib. Only thing is I don't know how to put in any flags into the exec() function. Specifically, I need to put in the -t flags (twice).

我找了好久都没找到.非常感谢您对此提供的帮助.也很抱歉这篇文章的长度.:)

I've looked for ages and can't find anything. Be really grateful for some help on this. Sorry about the length of this post as well. :)

干杯

乔

推荐答案

可以通过调用启用伪终端

You can enable the pseudoterminal by calling

$ssh->enablePTY();

在您登录之后，但在 exec 之前.这将防止它抱怨缺少 tty.

after you login, but before the exec. This will prevent it from complaining about the missing tty.

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