Not able to parse a .csv file uploaded using Flask

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I am trying to upload a CSV file, work on it to produce results, and write back (download) a new CSV file containing the result. I am very new to Flask and I am not able to

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误落风尘

2020-11-29 04:14

OK, so there is one major problem with your script, csv.reader as noted here expects a file object or at least an object which supports the iterator protocol. You are passing a str which does implement the iterator protocol, but instead of iterating through the lines, it iterates through the characters. This is why you have the output you do.

First, it gives a singe character 1 which the csv.reader sees as a line with one field. After that the str gives another single character , which the csv.reader sees as a line with two empty fields (since the comma is the field seperator). It goes on like that throughout the str until it's exhausted.

The solution (or at least one solution) is to turn the str into a file-like object. I tried using the stream provided by flask.request.files["name"], but that doesn't iterate through the lines. Next, I tried using a cStringIO.StringIO and that seemed to have a similar issue. I ended up at this question which suggested an io.StringIO object in universal newlines mode which worked. I ended up with the following working code (perhaps it could be better):

__author__ = 'shivendra'
from flask import Flask, make_response, request
import io
import csv

app = Flask(__name__)

def transform(text_file_contents):
    return text_file_contents.replace("=", ",")


@app.route('/')
def form():
    return """
        <html>
            <body>
                <h1>Transform a file demo</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():
    f = request.files['data_file']
    if not f:
        return "No file"

    stream = io.StringIO(f.stream.read().decode("UTF8"), newline=None)
    csv_input = csv.reader(stream)
    #print("file contents: ", file_contents)
    #print(type(file_contents))
    print(csv_input)
    for row in csv_input:
        print(row)

    stream.seek(0)
    result = transform(stream.read())

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

if __name__ == "__main__":
    app.run(host='0.0.0.0', port=5001, debug=True)

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-上瘾入骨i

2020-11-29 04:21
Important note: This answer is relevant only for platforms where SpooledTemporaryFile is available.

Further to iLuveTux answer, you can save the redundant read() call by replacing the following string-based stream creation:
```
stream = io.StringIO(f.stream.read().decode("UTF8"), newline=None)
```
with:
```
stream = io.TextIOWrapper(f.stream._file, "UTF8", newline=None)
```
Example:
```
stream = io.TextIOWrapper(f.stream._file, "UTF8", newline=None)
csv_input = csv.reader(stream)
print(csv_input)
for row in csv_input:
    print(row)
```
Further information:

Werkzeug default stream for form data parser is SpooledTemporaryFile (as of 1.0.1), from which you can obtain the underlying buffer using its _file memeber.
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