我正在尝试使用symfony进程通过laravel运行python脚本,但是当我运行它时,我收到此错误: “Python”不被识别为内部或外部命令,可操作程序或批处理文件。 我尝试使用cmd运行python脚本,然后运行并返回结果
我的laravel代码是这样的:
`
$text="I luv my <3 iphone & you're awsm apple. DisplayIsAwesome, sooo happppppy http://www.apple.com";
$process=new Process("Python C:\Users\Bcc\PycharmProjects\twitter-data\Car.py \"{$text}\"");
$process->run();
if (!$process->isSuccessful()){
throw new ProcessFailedException($process);
}
var_dump($process->getOutput());
die();
` 我在cmd上运行此命令: Python C:\ Users \ Bcc \ PycharmProjects \ twitter-data \ Car.py“我知道我的&lt; 3 iphone&amp;你是苹果.DisplayIsAwesome,sooo happppppy http://www.apple.com”
编辑:问题已解决。我将python脚本移动到controllers文件夹并使用shell_exec,所以只需用这个代码替换
$text="I luv my <3 iphone & you're awsm apple. DisplayIsAwesome, sooo happppppy http://www.apple.com";
$result = shell_exec("python ". app_path()."\Http\Controllers\Car.py " . escapeshellarg($text));
var_dump($result);
die();