python - 错误 Python - 214702489 访问被拒绝 (COM) 已更新

Question

我正在尝试使用网络应用程序 Flask/Python 在 excel 中启动一个宏，当用户按下按钮时，后面的进程将打开一个 excel 执行一个宏并返回结果。

如果我使用 IDE Visual Studio 2015 执行此操作，则没有问题，并且过程运行良好，当我尝试使用 IIS(发布)时，我可以导航到 Web 应用程序，但是当我按下按钮时，会出现此错误:

Error occurred:

Traceback (most recent call last):
  File "C:\inetpub\wwwroot\MyWebSite\wfastcgi.py", line 736, in main
    result = handler(record.params, response.start)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1477, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1381, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1475, in full_dispatch_request
    rv = self.dispatch_request()
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1461, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File ".\QPX_Test\upload.py", line 62, in upload
    process_excel.up_route(variable_forsave, 'C:\Tool_OP14\System\Main System FIndx', 'System','B1')
  File ".\QPX_Test\process_excel.py", line 27, in up_route
    xlApp = win32com.client.Dispatch('Excel.Application')
  File "C:\Anaconda2\lib\site-packages\win32com\client\__init__.py", line 95, in Dispatch
    dispatch, userName = dynamic._GetGoodDispatchAndUserName(dispatch,userName,clsctx)
  File "C:\Anaconda2\lib\site-packages\win32com\client\dynamic.py", line 114, in _GetGoodDispatchAndUserName
    return (_GetGoodDispatch(IDispatch, clsctx), userName)
  File "C:\Anaconda2\lib\site-packages\win32com\client\dynamic.py", line 91, in _GetGoodDispatch
    IDispatch = pythoncom.CoCreateInstance(IDispatch, None, clsctx, pythoncom.IID_IDispatch)
com_error: (-2147024891, 'Access is denied.', None, None)


StdOut: 

StdErr: 

我尝试了一切:

安全文件夹 ( C:\inetpub\wwwroot) 我添加了具有完全权限的 IUSR 用户，与 PCNAME\IIS_IUSRS 和我的 Web 应用程序(应用程序池身份)相同。

在 Microsoft Excel 的 DCOMcnfg 上，我在启动和激活权限上设置了自定义选项并将权限添加到:

• 国际标准化组织
• IIS_IUSRS
• 我的网站(网络应用程序)

这是代码的一部分，但是是最重要的部分:

所以这是用 jade 编写的开始 Python 代码的部分:

 .container-fluid

   .row
    .grid-12
        .jumbotron

            h1 Test Tool
             p Testing for web application development


   .row


    .section.group

      .col.span_2_of_3

                h1 Step 1

                p  For select the Input Data File, please press the "Select Data Input" button



                form(action='upload', method='post', enctype='multipart/form-data')
                  .table
                     input.filestyle(type='file', name = 'file', data-buttonname='btn-primary', data-buttonText= 'Select Data File', data-iconName= 'fa fa-database', data-buttonBefore='true', data-placeholder = 'No File')



                  .table 
                      span.btn.btn-primary.btn-file
                        | Upload           
                        span.fa.fa-cloud-upload.fa-1x.pull-left


                            input(type='submit')

所以当用户按下按钮时，选择的文件被上传，但首先 Python 代码调用另一个例程打开 excel 并做一些事情

import os
from QPX_Test import app
from flask import Flask, render_template, request, redirect, url_for, send_from_directory, after_this_request
from werkzeug import secure_filename
from win32com.client import Dispatch
import pythoncom 

import process_excel


main_path = os.path.join(os.path.dirname(os.path.abspath(__file__)), 'Server') 

# Initialize the Flask application
# app = Flask(__name__)

# This is the path to the upload directory
app.config['UPLOAD_FOLDER'] = main_path    
# These are the extension that we are accepting to be uploaded
app.config['ALLOWED_EXTENSIONS'] = set(['xls', 'xlsm','xlsx'])


# 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('/up')
def index():
    return render_template('upload.jade')


# Route that will process the file upload
@app.route('/upload', methods=['POST'])
def upload():
    # Get the name of the uploaded file
    file = request.files['file']
    # Check if the file is one of the allowed types/extensions
    if file and allowed_file(file.filename):
        # Make the filename safe, remove unsupported chars

        filename = secure_filename(file.filename)
        # Move the file form the temporal folder to
        # the upload folder we setup
        file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
        # Redirect the user to the uploaded_file route, which
        # will basicaly show on the browser the uploaded file
        #return redirect(url_for('uploaded_file',
         #                      filename=filename))

        variable_forsave = os.path.join(main_path, filename)
        filename_save=  os.path.join(main_path, 'Test.xlsm')


        # Select Excel File and Save Route into System

        process_excel.up_route(variable_forsave, 'C:\inetpub\wwwroot\MyWebSite\Tool\Main System', 'System','B1')

    return render_template('webtools_petro.jade')
    # after save the file do something 
    # This route is expecting a parameter containing the name
    # of a file. Then it will locate that file on the upload
    # directory and show it on the browser, so if the user uploads
    # an image, that image is going to be show after the upload
    #@app.route('/static/uploads/<filename>')
    #def uploaded_file(filename):
     # return send_from_directory(app.config['UPLOAD_FOLDER'],
                                 #   filename)

最后打开例程并做一些事情

import os
import win32com.client
from QPX_Test import app
from flask import Flask, render_template, request, redirect, url_for, send_from_directory, after_this_request
from werkzeug import secure_filename
from win32com.client import Dispatch
import pythoncom 

import upload

from flask import send_file
import StringIO


pythoncom.OleInitialize()


def up_route(variabletosave, forsave, sheetsave, rangesave):
            pythoncom.CoInitialize ()
            xlApp = win32com.client.Dispatch('Excel.Application')
            xlWb = xlApp.Workbooks.Open(forsave)
            xlSht = xlWb.WorkSheets(sheetsave)
            xlSht.Range(rangesave).value= variabletosave
            xlApp.visible = True
            xlWb.Saved = 0
            xlWb.Save()
            xlWb.Close(SaveChanges=True)
            xlApp.application.Quit()

所以我认为当Python尝试打开excel但没有权限时会出现问题，有办法带来Python权限吗？

在此编辑---------------------------------------- ------------------------

好的，所以我创建了一个仅用于测试的新应用程序...

'索引

extends layout

block content
  .jumbotron
    h1 Excel Test
    p.lead Testing the Privileges Features
    p Current directory  {{ testfile }}

    form(action='test', method='post', enctype='multipart/form-data')
        button(type="submit)

很简单

和 views.py

import os, sys
from datetime import datetime
from flask import render_template
from flask import request
from Test_Excel_Python import app
from Test_Excel_Python import Excel_test
import win32com.client
from werkzeug import secure_filename
from win32com.client import Dispatch
import pythoncom 
import StringIO

app.debug = True


Folder_File= os.path.dirname(os.path.abspath(sys.argv[0]))
Folder_Excel_Files='Test_Excel_Python\Excel Files'
File_open = 'Test File'
File_Suffix= 'xlsx'
Real_Excel_Route= os.path.join(Folder_File, Folder_Excel_Files, File_open + "." + File_Suffix)


@app.route('/')
@app.route('/home')
def home():
    """Renders the home page."""
    return render_template(
        'index.jade',
        title='Home Page',
        year=datetime.now().year, testfile=Real_Excel_Route
    )


@app.route('/test', methods=['POST'])
def test():
    #Excel_test.Excel_Open #'esto es un test', Real_Excel_Route, 'Prueba','A1')

            pythoncom.CoInitialize ()
            xlApp = win32com.client.Dispatch('Excel.Application')            
            xlWb = xlApp.Workbooks.Open('C:\inetpub\wwwroot\MyTestExcel\Test_Excel_Python\Excel Files\Test File.xlsx')
            xlSht = xlWb.WorkSheets('Sheet1')
            xlSht.Range('A1').value= 'Testing Write here'
            xlApp.visible = True
            xlWb.Saved = 0
            xlWb.Save()
            xlWb.Close(SaveChanges=True)
            xlApp.application.Quit()

            return render_template('index.jade')

而且网络配置非常简单

<?xml version="1.0" encoding="UTF-8"?>
<configuration>
    <system.webServer>
        <handlers>
            <add name="FlaskHandler" path="*" verb="*" modules="FastCgiModule" scriptProcessor="C:\Anaconda2\python.exe|C:\inetpub\wwwroot\MyTestExcel\wfastcgi.py" resourceType="Unspecified" />
        </handlers>
        <rewrite>
        </rewrite>
    </system.webServer>
</configuration>

与第一个程序一样，此例程在 Visual Studio 2015 上运行良好(启动)并访问 Localhost:5000

但是当我尝试使用 IIS localhost:82 时出现此错误:

Error occurred:

Traceback (most recent call last):
  File "C:\inetpub\wwwroot\MyTestExcel\wfastcgi.py", line 736, in main
    result = handler(record.params, response.start)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1836, in __call__
    return self.wsgi_app(environ, start_response)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1820, in wsgi_app
    response = self.make_response(self.handle_exception(e))
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1403, in handle_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1817, in wsgi_app
    response = self.full_dispatch_request()
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1477, in full_dispatch_request
    rv = self.handle_user_exception(e)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1381, in handle_user_exception
    reraise(exc_type, exc_value, tb)
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1475, in full_dispatch_request
    rv = self.dispatch_request()
  File "C:\Anaconda2\lib\site-packages\flask\app.py", line 1461, in dispatch_request
    return self.view_functions[rule.endpoint](**req.view_args)
  File ".\Test_Excel_Python\views.py", line 64, in test
    xlWb = xlApp.Workbooks.Open('C:\inetpub\wwwroot\MyTestExcel\Test_Excel_Python\Excel Files\Test File.xlsx')
  File "<COMObject <unknown>>", line 8, in Open
com_error: (-2147352567, 'Exception occurred.', (0, u'Microsoft Excel', u"Microsoft Excel cannot access the file 'C:\\inetpub\\wwwroot\\MyTestExcel\\Test_Excel_Python\\Excel Files\\Test File.xlsx'. There are several possible reasons:\n\n\u2022 The file name or path does not exist.\n\u2022 The file is being used by another program.\n\u2022 The workbook you are trying to save has the same name as a currently open workbook.", u'xlmain11.chm', 0, -2146827284), None)


StdOut: 

StdErr: 

我已经尝试添加“xlApp.Interactive = False”，但问题仍然存在，我只更改了本地的应用程序标识以进行测试，当然，最后一次更改“访问被拒绝”问题消失了。

我给APPTOOL\appname添加了对security文件夹的读/写访问权限。

请帮忙，这太令人沮丧了!! :(

Answer 1

找到解决方案!!!，f**** bug arrrggggg 我很高兴/生气:

The solution for this appalling BUG in Microsoft IIS & Excel is terrific:

    Create directory "C:\Windows\SysWOW64\config\systemprofile\Desktop " (for 64 bit Windows) or "C:\Windows\System32\config\systemprofile\Desktop " (for 32 bit Windows)
    Set Full control permissions for directory Desktop (for example in Win7 & IIS 7 & DefaultAppPool set permissions for user
    "IIS AppPool\DefaultAppPool") 

微软到底是怎么回事!!!! 2010 年解决方案!!

https://social.msdn.microsoft.com/Forums/vstudio/en-US/4d6c383a-94eb-4898-9d22-aa4bb69be25b/excel-interop-systemruntimeinteropservicescomexception-0x800a03ec-microsoft-office-excel?forum=vbgeneral