Question

我有一个curl命令来上传文件并提交表单。

curl --form <file>=@<filename> --form press=<value>  [URL]

需要编写一个木偶执行块，如下所示。

ex: exec { "IngestClient $abc_name" :

    command => "curl --form file=@$file --form press=Submit  http://ipaddress:portno/myapp/abc/"
    cwd     => "$data/bin",
    path    => "$data/bin:/usr/bin",
    require => [File["$package_file"],Package['curl']],
}

我需要参数化url和文件名以及文件和按钮值。

提前致谢，

Answer 1

您可以使用参数filename，url，abc_name，data和package_file定义自定义类型：

define uploadFile(
    $filename     = $title,
    $url          = undef,
    $abc_name     = undef,
    $data         = undef,
    $package_file = undef
) {
    exec { "IngestClient ${abc_name}" :

        command => "curl --form file=@${filename} --form press=Submit  ${url}"
        cwd     => "${data}/bin",
        path    => "${data}/bin:/usr/bin",
        require => [ File["${package_file}"], Package['curl'] ]
    } 
}

然后像这样使用它：

uploadFile { "my-file":
    url          => "http://ipaddress:portno/myapp/abc",
    abc_name     => "???",
    data         => "/foo/bar",
    package_file => "???"
}

在puppet模块中使用CURL上传文件并提交表单

1 个答案: