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Forum Discussion
Uploaded Files are Currupt (Flask, Python)
Hey, guys so I'm experimenting with the Dropbox python API using Flask. I managed to upload files but all the files I've uploaded are only 16 bytes in size and are unable to be opened. here's the code I'm using.
HTML Form:
<div style="text-align: center" class="form">
<form class="needs-validation" novalidate action="/uploadtest" method="POST" enctype="multipart/form-data">
<div class="col-md-12">
<div class="form-group">
<label for="name">Choose file</label>
<input type="file" accept=".xlsx, .jpg" name="file" id="fileToUpload" class="form-control" placeholder="John Smith"
required>
<div class="valid-feedback">
Looks good!
</div>
<div class="invalid-feedback">
Please choose a .xlsx file.
</div>
</div>
</div>
<button type="submit" class="btn btn-success" value="">Submit</button>
</form>
</div>
Python Flask Route
@app.route('/uploadtest', methods=['POST', 'GET'])
def get_dropbox_files():
if request.method == 'POST':
uploadedfile = request.files['file']
client = dbx_client
f = secure_filename(uploadedfile.filename)
path="/nameinlights/" + f
dbx_client.files_upload(f.encode(), path)
flash("Added" + " " + f + " " + "to Name in Lights Folder", 'success')
return render_template('uploadtest.html', file_list=file_list)
Hi!
When you call
file_upload, you're actually uploading the file's name as a string. Each file is coming out to 16 bytes because the strings are being generated bysecure_filenameand are probably a standard length.You can fix it by changing the first variable you're passing to
file_upload.dbx_client.files_upload(f.encode(), path)
should be
dbx_client.files_upload(uploadedfile.encode(), path)
It looks like you're just working with the filename
- TaylorKrusen
Dropbox Staff
Hi!
When you call
file_upload, you're actually uploading the file's name as a string. Each file is coming out to 16 bytes because the strings are being generated bysecure_filenameand are probably a standard length.You can fix it by changing the first variable you're passing to
file_upload.dbx_client.files_upload(f.encode(), path)
should be
dbx_client.files_upload(uploadedfile.encode(), path)
It looks like you're just working with the filename
thanks man! i had to change "encode()" for "read()" thanks for the help man!
dbx_client.files_upload(uploadedfile.read(), path)
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