【发布时间】:2020-01-05 18:10:29
【问题描述】:
当我在 shell 上运行命令 echo /tmp/stderr{,.pub} | xargs -n 1 ln -sf /dev/stderr && yes | ssh-keygen -t ed25519 -C "" -N "" -f /tmp/stderr > /dev/null; rm /tmp/stderr{,.pub} 时,它返回了如下所示的私钥/公钥对(不用担心,密钥对没有被使用):
blackbox /home/clock # echo /tmp/stderr{,.pub} | xargs -n 1 ln -sf /dev/stderr && yes | ssh-keygen -q -t ed25519 -C "" -N "" -f /tmp/stderr >/dev/null; rm /tmp/stderr{,.pub}
-----BEGIN OPENSSH PRIVATE KEY-----
b3BlbnNzaC1rZXktdjEAAAAABG5vbmUAAAAEbm9uZQAAAAAAAAABAAAAMwAAAAtzc2gtZW
QyNTUxOQAAACA1vZhl2jqtzhEqaqbKwYjLB1OIH8hMPtWB/PWhqeI/QQAAAIj4n5if+J+Y
nwAAAAtzc2gtZWQyNTUxOQAAACA1vZhl2jqtzhEqaqbKwYjLB1OIH8hMPtWB/PWhqeI/QQ
AAAEAtcSI3RLsOo0CXnat4Gs4JENGyDPbGojIT8GU0E+3vUDW9mGXaOq3OESpqpsrBiMsH
U4gfyEw+1YH89aGp4j9BAAAAAAECAwQF
-----END OPENSSH PRIVATE KEY-----
ssh-ed25519 AAAAC3NzaC1lZDI1NTE5AAAAIDW9mGXaOq3OESpqpsrBiMsHU4gfyEw+1YH89aGp4j9B
但是当我在 PHP 中通过exec() 运行命令时,它返回了命令本身:
blackbox /home/clock # php -f key.php
Array
(
[0] => /tmp/stderr{,.pub} | xargs -n 1 ln -sf /dev/stderr && yes | ssh-keygen -t ed25519 -C "" -N "" -f /tmp/stderr > /dev/null; rm /tmp/stderr{,.pub}
)
我使用的代码:
<?php
$exec_output = '';
$exec_return = '';
// echo /tmp/stderr{,.pub} | xargs -n 1 ln -sf /dev/stderr && yes | ssh-keygen -t ed25519 -C "" -N "" -f /tmp/stderr > /dev/null; rm /tmp/stderr{,.pub}
$cmd = array('echo', escapeshellarg('/tmp/stderr{,.pub} |'), 'xargs -n 1 ln -sf /dev/stderr', escapeshellarg('&& yes |'), 'ssh-keygen -t ed25519', escapeshellarg('-C "" -N "" -f /tmp/stderr > /dev/null; rm /tmp/stderr{,.pub}'));
exec(implode(' ', $cmd), $exec_output, $exec_return);
print_r($exec_output);
?>
为什么?
PS:我不想将生成的密钥对填充到文件中,即使不在 /tmp 中。
- PHP:7.3.13
- 用户:root
【问题讨论】:
标签: php exec ssh-keys shell-exec