在烧瓶上运行 shell 脚本答案

【问题标题】：Running shell script on flask在烧瓶上运行 shell 脚本
【发布时间】：2015-06-04 07:20:50
【问题描述】：

我的应用程序已经设置了以下 app.py 文件和两个 .html 文件：index.html（基本模板）和 upload.html，客户端可以在其中看到他刚刚上传的图像。我遇到的问题是，我希望我的程序（可能是 app.py）在用户被重定向到 upload.html 模板之前执行一个 matlab 函数。我找到了关于如何在烧瓶上运行 bash shell 命令的问答（但这不是命令），但我还没有找到脚本。

我得到的解决方法是创建一个 shell 脚本：hack.sh，它将运行 matlab 代码。在我的终端中，这很简单：

$bash hack.sh

hack.sh：

nohup matlab -nodisplay -nosplash -r run_image_alg > text_output.txt &

run_image_alg 是我的 matlab 文件 (run_image_alg.m)

这是我的 app.py 代码：

import os

from flask import Flask, render_template, request, redirect, url_for, send_from_directory
from werkzeug import secure_filename

# Initialize the Flask application

app = Flask(__name__)

# This will be th path to the upload directory
app.config['UPLOAD_FOLDER'] = 'uploads/'

# These are the extension that we are accepting to be uploaded
app.config['ALLOWED_EXTENSIONS'] = set(['png','jpg','jpeg'])

# For a given file, return whether it's an allowed type or not
def allowed_file(filename):
  return '.' in filename and \
    filename.rsplit('.',1)[1] in app.config['ALLOWED_EXTENSIONS']

# This route will show a form to perform an AJAX request
# jQuery is loaded to execute the request and update the 
# value of the operation

@app.route('/')
def index():
  return render_template('index.html')

#Route that will process the file upload
@app.route('/upload',methods=['POST'])
def upload():
  uploaded_files = request.files.getlist("file[]")
  filenames = []
  for file in uploaded_files:
    if file and allowed_file(file.filename):
      filename = secure_filename(file.filename)
      file.save(os.path.join(app.config['UPLOAD_FOLDER'],filename))
      filenames.append(filename)

  print uploaded_files

  #RUN BASH SCRIPT HERE.


  return render_template('upload.html',filenames=filenames)

@app.route('/uploads/<filename>')
def uploaded_file(filename):
  return send_from_directory(app.config['UPLOAD_FOLDER'],filename)


if __name__ == '__main__':
  app.run(
    host='0.0.0.0',
    #port=int("80"),
    debug=True
  )

我可能想念一个图书馆？我在 stackoverflow 上发现了一个类似的问答，有人想运行（已知的）shell 命令（$ls -l）。我的情况不同，因为它不是已知命令，而是创建的脚本：

from flask import Flask
import subprocess

app = Flask(__name__)

@app.route("/")

def hello():
    cmd = ["ls","-l"]
    p = subprocess.Popen(cmd, stdout = subprocess.PIPE,
                            stderr=subprocess.PIPE,
                            stdin=subprocess.PIPE)
    out,err = p.communicate()
    return out
if __name__ == "__main__" :
    app.run()

【问题讨论】：

标签： python shell flask

【解决方案1】：

如果你想运行 matlab，只需将你的命令更改为

cmd = ["matlab", "-nodisplay", "-nosplash", "-r", "run_image_alg"]

如果要将输出重定向到文件：

with open('text_output.txt', 'w') as fout:
    subprocess.Popen(cmd, stdout=fout,
                          stderr=subprocess.PIPE,
                          stdin=subprocess.PIPE)

【讨论】：