I'd like to upload an excel file in my web app, read the contents of it and display some cells. So basically I don't need to save the file as it's a waste of time.
Relevant code:
if form.validate_on_submit():
f = form.xml_file.data.stream
xml = f.read()
workbook = xlrd.open_workbook(xml)
sheet = workbook.sheet_by_index(0)
I can't wrap my mind around this as I keep getting filetype errors no matter what I try. I'm using Flask Uploads, WTF.file and xlrd for reading the file.
Reading the file works okay if I save it previously with f.save
To answer my own question, I solved it with
if form.validate_on_submit():
# Put the file object(stream) into a var
xls_object = form.xml_file.data.stream
# Open it as a workbook
workbook = xlrd.open_workbook(file_contents=xls_object.read())
Related
I am reading an Excel file (.xlsx) with pysmb.
import tempfile
from smb.SMBConnection import SMBConnection
conn = SMBConnection(userID, password, client_machine_name, server_name, use_ntlm_v2 = True)
conn.connect(server_ip, 139)
file_obj = tempfile.TemporaryFile()
file_attributes, filesize = conn.retrieveFile(service_name, test.xlsx, file_obj)
This step works, I am able to transform the file in pandas.DataFrame
import pandas as pd
pd.read_excel(file_obj)
Next, I want to save the file, the file is saved but if I want to open it with Excel, I have an error message "Excel has run into an error"
Here the code to save the file
conn.storeFile(service_name, 'test_save.xlsx', file_obj)
file_obj.close()
How can I save correctly the file and open it with excel ?
Thank you
I tried with a .txt file file and it is working. An error occurs with .xlsx, .xls and .pdf files. I have also tried without extension, same issue, imossible to open the file.
I would like to save the file with .pdf and .xlsx extension, and open it.
Thank you.
I found a solution an I will post it here in case someone face a similar issue.
Excel can be save as a binary stream.
from io import BytesIO
df = pd.read_excel(file_obj)
output = BytesIO()
writer = pd.ExcelWriter(output, engine='xlsxwriter')
df.to_excel(writer, sheet_name='data', index = False)
writer.save()
output.seek(0)
conn.storeFile(service_name, 'test_save.xlsx', output)
Currently I am exporting an excel file, however, before it gets exported, I create an xls file on the host machine. To create the excel file, I use tablib. My export view looks like this:
#login_required
def export_xls(request):
# some irrelevant code
data = convert_json_to_dataset(json_data)
table = data.export('xls')
with open('/tmp/students.xls', 'wb') as f:
f.write(table)
response = FileResponse(open('/tmp/students.xls', 'rb'), as_attachment=True, filename="test.xls")
return response
What I am trying to achieve is to avoid writing always to /tmp/students.xls. I tried using BytesIO, however that did not work out.
#login_required
def export_xls(request):
# some irrelevant code
data = convert_json_to_dataset(json_data)
table = data.export('xls')
buffer = BytesIO()
buffer.write(table)
response = FileResponse(buffer.read(), as_attachment=True, filename="test.xls")
return response
Currently I am always overwriting the file, however, I will change the naming of the file and that will cause multiple files to be created, which I would like to avoid.
I have created a flask application where it take excel file and it cleans the data and gives the output in excel file. basically what happens is user uploads the excel file after submitting browser should download the filtered excel file.
can someone suggest me references? I need to know how to set the path. I tried converting it into the HTML by using but this code doesn't download but it automatically saves the cleaned file as HTML.
data1 = df.to_html()
#write html to file
text_file = open("data1.html", "w")
text_file.write(data1)
text_file.close()
return render_template("success.html", name = text_file)
I have an app that receive an input file, read it with pandas, process it (with a make_processing() function I created) and return it as .csv. Is almost the same for an excel file.
file = request.files['file']
content = file.read()
df = pd.read_csv(io.BytesIO(content))
df2 = make_processing(df)
si = io.StringIO()
df2.to_csv(si, index=False, encoding='utf8')
output = flask.make_response(si.getvalue())
output.headers["Content-Disposition"] = f"attachment; filename=periodicidad.csv"
output.headers["Content-type"] = "text/csv"
return output
I am using xlrd in appengine. I use flask
I cant read the input file and it keeps on showing the same error message
the code is
def read_rows(inputfile):
rows = []
wb = xlrd.open_workbook(inputfile)
sh = wb.sheet_by_index(0)
for rownum in range(sh.nrows):
rows.append(sh.row_values(rownum))
return rows
#app.route('/process_input/',methods=['POST','GET'])
def process_input():
inputfile = request.files['file']
rows=read_rows(request.files['file'])
payload = json.dumps(dict(rows=rows))
return payload
I realize that this might be caused by not uploading and saving it as a file. Any workaround on this? This would help many others as well. Any help is appreciated, thx
Update: Found a solution that I posted below. For those confused with using xlrd can refer to the open source project repo I posted. The key is passing the content of the file instead of the filename
Find a solution finally
here's how I do it. Instead of saving the file, I read the content of the file and let xlrd reads it
def read_rows(inputfile):
rows = []
wb = xlrd.open_workbook(file_contents=inputfile.read())
sh = wb.sheet_by_index(0)
for rownum in range(sh.nrows):
rows.append(sh.row_values(rownum))
return rows
worked nicely and turned the excel files into JSON-able formats. If you want to output the json simply use json.dumps().
full code example can be found at https://github.com/cjhendrix/HXLator/blob/master/gae/main.py and it features full implementation of the xlrd and how to work with the data.
Thx for the pointers
Use:
wb = xlrd.open_workbook(file_contents=inputfile)
The way you are invoking open_workbook expects what you're passing in to be a filename, not a Flask FileStorage object wrapping the actual file.
Judge from your traceback.
File "/Users/fauzanerichemmerling/Desktop/GAEHxl/gae/lib/xlrd/init.py", line 941, in biff2_8_load
f = open(filename, open_mode)
You can try changing this line to :
f = filename
I'd like to generate a dynamic Excel file on request from Django. The library pyExcelerator does this, but I haven't found any way to use the contents of the Excel file without generating a server-side temporary Excel file, reading it, using its contents and deleting it.
The problem is that pyExcelerator only way to extract the contents of the Excel file is saving it via:
workbook = pyExcelerator.Workbook()
workbook.save("tmp_filename")
And then read the temporary file contents. I can't use the standard library "tempfile" because it doesn't accept a file, just a filename. How can I ensure that the filename is unique and that the file is deleted once it has been used?
pyExcelerator is unmaintained, but it has a fork, xlwt, which is maintained and has more features, including allowing you to save to any file-like object. This includes saving straight to a Django HttpResponse:
from django.http import HttpResponse
import xlwt
def my_view(request):
response = HttpResponse(mimetype="application/ms-excel")
response['Content-Disposition'] = 'attachment; filename="foo.xls"'
wb = xlwt.Workbook()
wb.save(response)
return response
Why can you not use the tempfile module?
How about:
import tempfile
fd, filename = tempfile.mkstemp()
fd.close()
workbook.save(filename)