Question

我的前端有一个按钮，这个按钮与后端对话。后端正在启动一个远程脚本：

<?php
$myfile = fopen("newfile.txt", "w") or die("Unable to open file!");
$connection = ssh2_connect('192.168.56.180', 22);
ssh2_auth_password($connection, 'root', 'password');
$stream = ssh2_exec($connection, 'python /WATSON/APP/test/testlistrbk.py');
stream_set_blocking($stream, true);
$stream_out = ssh2_fetch_stream($stream, SSH2_STREAM_STDIO);
echo $stream_out_contente;
fwrite($myfile, $stream_out);
fclose($myfile);
?>

我有2个问题，第一个，php应该等待python脚本完成，因为它说here但它没有。

第二个，它给了我以下内容：

PHP警告：fwrite（）要求参数2为字符串，资源在第41行的/var/www/html/WEBAPP/wa_start.php中给出

Answer 1

使用stream_get_contents(stream_out);

在您的代码$stream_out = ssh2_fetch_stream($stream, SSH2_STREAM_STDIO);中，这将返回resource而不是字符串输出。试试这段代码。

$myfile = fopen("newfile.txt", "w") or die("Unable to open file!");

$connection = ssh2_connect('192.168.56.180', 22);
ssh2_auth_password($connection, 'root', 'password');
$stream = ssh2_exec($connection, 'python /WATSON/APP/test/testlistrbk.py');

stream_set_blocking($stream, true);

$outputStream = ssh2_fetch_stream($stream, SSH2_STREAM_STDIO);
$errorStream = ssh2_fetch_stream($stream, SSH2_STREAM_STDERR);

echo "Output: " . stream_get_contents($outputStream);
echo "Error: " . stream_get_contents($errorStream);

fwrite($myfile, $outputStream.$errorStream);
fclose($myfile);

PHP：SSH连接和python脚本执行

1 个答案: