在Flask应用程序中返回Excel文件

时间:2016-07-19 03:53:00

标签: python flask

我正在创建一个Flask应用程序,提示用户输入一个Excel文件,有些人使用它,然后将文件返回给用户,以便他们可以下载它。 (请忽略任何未使用的导入。我打算稍后再使用它们。)

我的功能已关闭,我不知道如何将文件发送回用户,以便他们可以下载。在此先感谢您的帮助!

这是我到目前为止所做的:(注意:我不太确定我是否正确实施了上传功能)

from openpyxl import load_workbook
from flask import Flask, request, render_template, redirect, url_for


app = Flask(__name__)

@app.route('/')
def index():
    return """<title>Upload new File</title>
    <h1>Upload new File</h1>
    <form action="/uploader" method=post enctype=multipart/form-data>
      <p><input type=file name=file>
         <input type=submit value=Upload>
    </form>"""

@app.route('/uploader', methods = ['GET', 'POST'])
def upload():
    if request.method == 'POST':
        f = request.files['file']
        f.save(f.filename)
        return process(f.filename)

def process(filename):

    routename = ['ZYAA', 'ZYBB', 'ZYCC']
    supervisors = ['X', 'Y', 'Z']
    workbook = load_workbook(filename)
    worksheet = workbook.active
    worksheet.column_dimensions.group('A', 'B', hidden=True)
    routes = worksheet.columns[2]
    i = 2
    worksheet['D1'] = 'Supervisor'
    for route in routes:
        if route.value in routename:
            pos = routes.index(route)
            worksheet['D' + str(i)].value = supervisors[pos]
            print (route.value)
            i += 1

    workbook.save(filename)




if __name__ == '__main__':
    app.run(debug = True, host = '0.0.0.0')

1 个答案:

答案 0 :(得分:5)

这取决于您是否要将文件保留在服务器/计算机上。你可以做这样的事情来保存文件:

from flask import send_from_directory

def process():
    # do what you're doing

    file_name = 'document_template.xltx'
    wb = load_workbook('document.xlsx')
    wb.save(file_name, as_template=True)

    return send_from_directory(file_name, as_attachment=True)

如果您不想保留文件,this代码段可以为您提供帮助。