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@ytyng

Django で、バッチ(コマンド)結果をHTMLで逐次表示する

More than 3 years have passed since last update.

(PHPでいう、output_buffering を切った感じです)

StreamingHttpResponse を使います。

StreamingHttpResponse の第一引数はジェネレータなので、バッチをジェネレータで書きます。

今回の例は、シェルコマンドをジェネレータにしています。

import subprocess

from django.http import StreamingHttpResponse
from django.utils.encoding import smart_str
from django.utils.html import escape
from django.views import View


def run_process_as_generator(*args, **kwargs):
    """
    サブプロセス結果をジェネレータで返す
    :rtype: generator
    """
    kwargs.setdefault('stdout', subprocess.PIPE)
    kwargs.setdefault('stderr', subprocess.STDOUT)

    popen = subprocess.Popen(*args, **kwargs)

    while True:
        line = popen.stdout.readline()
        if line:
            yield line

        if not line and popen.poll() is not None:
            break


def linebreaksbr(gen):
    """
    ジェネレータの各行に<br />をつける
    念のため smart_str
    :type gen: generator
    :rtype: generator
    """
    for line in gen:
        yield '{}<br />\n'.format(escape(smart_str(line)))


def task():
    """
    バッチを実行
    :rtype: generator
    """
    return linebreaksbr(
        run_process_as_generator(
            "for i in {1..10}; do sleep 1s; echo ${i}; done",
            shell=True
        ))


class SlowCommandView(View):
    """
    task() の結果を少しずつブラウザに返すビュー
    """
    def get(self, request, *args, **kwargs):
        response = StreamingHttpResponse(
            task(),
            content_type="text/html; charset=utf-8")

        return response

Python でバッチを書く場合は、print でログを書くかわりに yield を使うと良いと思います。

def task():
    yield 'start'
    ...
    yield 'phase-1'
    ...
    for xx in xxx:
         yield xx
    yield 'end'

参考:
Pythonのsubprocessで標準出力をリアルタイムに取得する - Qiita

Request and response objects | Django documentation | Django

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ytyng
torico
マンガに関わるWEBサービスを軸に、面白いと思えることを企画・開発している会社です。オタク文化が好きな人は是非一緒に働きましょう! Python3 / Vue.js(TypeScript) / Flutterで開発中!

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