SSH无法通过PHP exec正常工作

Question

Hey I'm attempting to utilise the PHP SSH2 functions of which I've used a class from over a GitHub: Here but doesn't seem to be working as expected.

I'm trying to cd into a directory and then list files, but seems to ignore the cd and when I use the ls command it just displays the root dir.

Here's when I perform a normal ls on my root directory:

drwx------  6 root root 4096 Aug 12 18:23 .
drwxr-xr-x 22 root root 4096 Aug 11 08:28 ..
-rw-------  1 root root 2963 Aug 12 18:33 .bash_history
-rw-r--r--  1 root root 3106 Oct 22  2015 .bashrc
drwx------  3 root root 4096 Aug 11 21:05 .cache
drwxr-xr-x  3 root root 4096 Aug 12 18:21 .local
drwxr-xr-x  2 root root 4096 Aug 12 18:22 .nano
drwxr-xr-x 18 root root 4096 Aug 11 21:07 myOtherDirectory
-rw-r--r--  1 root root  148 Aug 17  2015 .profile
-rw-------  1 root root 1024 Aug 12 18:27 .rnd
-rw-r--r--  1 root root  252 Aug 12 18:28 .wget-hsts

And here's then I try to access the myOtherDirectory using SSH via PHP

$ssh = new ssh("ip", "root", "pass");
$ssh("cd myOtherDirectory");
$result = $ssh("ls -la");
print_r($result);

Answer 1

The SSH protocol permits you to make one SSH connection (called a session) to a remote host and then run multiple channels through the single connection. A single channel can represent an interactive session, a single SFTP transfer, and so on. Each channel is independent of the others.

$ssh = new ssh("ip", "root", "pass");
$ssh("cd myOtherDirectory");
$result = $ssh("ls -la");

What you're doing here is creating a single SSH session to the remote host, then creating two channels. The first channel runs cd and then exits. The second channel runs ls and then exits. The two channels are independent of each other, so the cd invocation has no effect on the working directory of the ls invocation.

You want to run both commands in a single session. Assuming the remote system is using bash or a similar shell to run commands, this ought to work:

$ssh = new ssh("ip", "root", "pass");
$result = $ssh("cd myOtherDirectory && ls -la");

Alternately, you could specify the correct remote directory in the ls command:

$ssh = new ssh("ip", "root", "pass");
$result = $ssh("ls -la myOtherDirectory");