打开已在Flask中上传的文件 [英] Opening a file that has been uploaded in Flask

问题描述

我正在尝试修改上传到我的Flask应用程序中的csv.当我不通过烧瓶上载它时，我的逻辑工作得很好.

I'm trying to modify a csv that is uploaded into my flask application. I have the logic that works just fine when I don't upload it through flask.

import pandas as pd
import StringIO
with open('example.csv') as f:
    data = f.read()

data = data.replace(',"', ",'")
data = data.replace('",', "',")
df = pd.read_csv(StringIO.StringIO(data), header=None, sep=',', quotechar="'")
print df.head(10)

我将其上传到烧瓶并使用

I upload it to flask and access it using

f = request.files['data_file']

通过上面的代码运行它，将open('example.csv')替换为open(f)时，出现以下错误

When I run it through the code above, replacing open('example.csv') with open(f), I get the following error

coercing to Unicode: need string or buffer, FileStorage found

我发现问题出在这里是文件类型.我无法在文件上使用open，因为open正在寻找文件名，并且当文件上传到flask时，它是要传递给open命令的文件实例.但是，我不知道该如何做.我试过跳过open命令，仅使用data = f.read()，但这不起作用.有什么建议吗?

I have figured out that the problem is the file type here. I can't use open on my file because open is looking for a file name and when the file is uploaded to flask it is the instance of the file that is being passed to the open command. However, I don't know how to make this work. I've tried skipping the open command and just using data = f.read() but that doesn't work. Any suggestions?

谢谢

推荐答案

回答我自己的问题，以防别人需要.

Answering my own question in case someone else needs this.

FileStorage对象具有.stream属性，该属性将是io.BytesIO

FileStorage objects have a .stream attribute which will be an io.BytesIO

f  = request.files['data_file']
df = pandas.read_csv(f.stream)

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