Question

我在Python Anywhere上托管了一个简单的Flask应用。该应用程序提供了一个界面，用户可以在其中上传CSV文件，并通过覆盖第一列（row[0]）并将原始row[0]放在末尾，将新列插入上载的CSV文件中。文件，然后为用户下载转换后的文件。因此，CSV输入看起来像这样（第一行是标头行，除了添加第4列名称外，应保持不变）：

Pet1,Pet2,Pet3
Cats,Dogs,Mice

输出应如下所示：

Pet1,Pet2,Pet3,Pet4
Birds,Dogs,Mice,Cats

这是我的代码：

from flask import Flask, make_response, request
import operator
import io
import csv

app = Flask(__name__)

def transform(text_file_contents):
  reader = csv.reader(text_file_contents)
  all = []
  row = next(reader)
  row.append('Pet4')
  all.append(row)
  for row in reader:
    pet4 = row[0]
    row[0] = "Birds"
    row.append(pet4)
    all.append(row)
  sortedlist = sorted(all, key=operator.itemgetter(0))
  return sortedlist


@app.route('/')
def form():
  return """
    <html>
        <body>
            <h1>Upload a CSV File</h1>

            <form action="/transform" method="post" enctype="multipart/form-data">
                <input type="file" name="data_file" />
                <input type="submit" />
            </form>
        </body>
    </html>
"""

@app.route('/transform', methods=["POST"])
def transform_view():
  request_file = request.files['data_file']
  if not request_file:
    return "No file"

file_contents = request_file.stream.read().decode("utf-8")

result = transform(file_contents)

response = make_response(result)
response.headers["Content-Disposition"] = "attachment; filename=result.csv"
return response

这是我得到的错误：

Traceback (most recent call last):
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/app.py", line 2292, in wsgi_app
response = self.full_dispatch_request()
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/app.py", line 1815, in full_dispatch_request
rv = self.handle_user_exception(e)
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/app.py", line 1718, in handle_user_exception
reraise(exc_type, exc_value, tb)
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/_compat.py", line 35, in reraise
raise value
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/app.py", line 1813, in full_dispatch_request
rv = self.dispatch_request()
File "/home/user/.virtualenvs/myvirtualenv/lib/python3.6/site-packages/flask/app.py", line 1799, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
File "/home/user/mysite/callnumbers.py", line 62, in transform_view
result = transform(file_contents)
File "/home/user/mysite/callnumbers.py", line 18, in transform
row[0] = row[1]
IndexError: list index out of range

我可以确定我的脚本未正确读取上载的输入文件（因此无法在我要引用的列和IndexError中找到数据），但我却无法使用Python CSV库查找Flask的任何示例，以解析上传的文件并按我需要的方式操作列。我该如何解决？任何指向正确方向的指针都将受到赞赏！

Answer 1

csv.reader()函数采用文件对象而不是文本。一种方法是保存文件，然后在转换函数中重新打开它。

from flask import Flask, make_response, request
import operator
import io
import csv    

app = Flask(__name__)


def transform():
  with open('myfile.csv', 'rb') as f:
    reader = csv.reader(f)  
    all = []
    row = next(reader)
    row.append('Pet4')
    all.append(row)
    for row in reader:
      pet4 = row[0]
      row[0] = "Birds"
      row.append(pet4)
      all.append(row)
    sortedlist = sorted(all, key=operator.itemgetter(0))  
  return sortedlist

@app.route('/')
def form():
  return """
    <html>
        <body>
            <h1>Upload a CSV File</h1>

            <form action="/transform" method="post" enctype="multipart/form-data">
                <input type="file" name="data_file" />
                <input type="submit" />
            </form>
        </body>
    </html>
"""

@app.route('/transform', methods=["POST"])
def transform_view():
  request_file = request.files['data_file']
  request_file.save("myfile.csv")
  if not request_file:
    return "No file"
  #file_contents = io.StringIO(request_file.stream.read().decode("UTF8"), newline=None)
  #file_contents = request_file.stream.read().decode("utf-8")

  result = transform()
  print result

  response = make_response(result)
  response.headers["Content-Disposition"] = "attachment; filename=result.csv"
  return response
if __name__ == '__main__':
    app.run()

Answer 2

正如Mike C.所说，问题是csv.reader（）函数采用文件对象（或等效文件）代替文本。但是，您无需在磁盘上保存文件，而可以在流上解码：

stream = codecs.iterdecode(request_file.stream, 'utf-8')
result = transform(stream)

完整的代码：

import codecs
import operator
from flask import Flask, make_response, request
import csv

app = Flask(__name__)

def transform(text_file_contents):
  reader = csv.reader(text_file_contents)
  list_all = []

  row = next(reader)
  row.append('Pet4')
  list_all.append(row)

  for row in reader:
    pet4 = row[0]
    row[0] = "Birds"
    row.append(pet4)
    list_all.append(row)
  sortedlist = sorted(list_all, key=operator.itemgetter(0))
  return sortedlist


@app.route('/')
def form():
  return """
    <html>
        <body>
            <h1>Upload a CSV File</h1>

            <form action="/transform" method="post" enctype="multipart/form-data">
                <input type="file" name="data_file" />
                <input type="submit" />
            </form>
        </body>
    </html>
"""

@app.route('/transform', methods=["POST"])
def transform_view():
    request_file = request.files['data_file']
    if not request_file:
        return "No file"


    stream = codecs.iterdecode(request_file.stream, 'utf-8')
    result = transform(stream)

    response = make_response(str(result))
    response.headers["Content-Disposition"] = "attachment; filename=result.csv"
    return response

if __name__ == '__main__':
    app.run()

使用Flask转换上传的CSV文件上传

2 个答案: