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React On Click Prevent Download but show excel

I have a link from microsoft one drive which is generated when we upload file pythonically. I have linked this link to a button whichis supposed to show the file but not download however on click it is donloading the file so i neeed help regarding how to prevent download but show the excel file linked. Below is react code I have been using
function Files(props){
let getFileOnClick=(fileAddress)=>{
window.location.href=fileAddress
}
return(
<button style={{float:'right'}} onClick = {()=>getFileOnClick(props.link)} className="btn btn-primary">
Get File
</button>
)
}

Connecting python script with html button and flask

I am trying to create a webapp and am fairly new to it. I have a python script(file.py) that transforms data selected by a user. It handles all the inputs and outputs.
I am using flask(main.py) for the server part of it and html. I want to place a button in my html code so it will start the execution of the file.py. Can anyone assist me with an example setup for the connections between the 3?
I've looked at other examples but I'm unable to recreate it as they're doing different things. Also, file.py is fairly large so I want to avoid putting it into a function.
Edit: not looking for a flask tutorial. I've tried 3things:
A shell pops up for half a second but the disappears. Then I'm redirected to a page which just has the text in my return statement
in my html file
<form action="/pic" method="POST">
<input type="submit" value="GET THE SCRIPT">
</form>
in my main.py flask file
#app.route('/pic', methods=['GET', 'POST'])
def pic():
os.system("python file.py") #file.py is the script I'm trying to start
return "done"
Doesn't do anything at all.
in html file:
<input type="button" id='script' name="scriptbutton" value=" Run Script " onclick="goPython()">
<script src="http://code.jquery.com/jquery-3.3.1.min.js" integrity="sha256-FgpCb/KJQlLNfOu91ta32o/NMZxltwRo8QtmkMRdAu8=" crossorigin="anonymous"></script>
<script>
function goPython(){
$.ajax({
url: "/scripts/file.py",
context: document.body
}).done(function() {
alert('finished python script');;
});
}
</script>
I get a GET "/scripts/file.py HTTP/1.1" 404 message. I have my scripts folder in the same directory as my templates folder. Also tried placing the scripts folder inside the templates folder.
in html
<form action="/run" method = "POST">
<input type="button" id='script' name="submit" value="Run Scripttttttt">
</form>
in flask main.py
#app.route('/run',methods=['GET', 'POST'])
def index():
def inner():
proc = subprocess.Popen(
['python file.py'],
shell=True,
stdout=subprocess.PIPE
)
for line in iter(proc.stdout.readline,''):
time.sleep(1)
yield line.rstrip() + '<br/>\n'
return Flask.Response(inner(), mimetype='text/html')

Using an HTML Anchor tag (i.e ) is the easiest way. Here is an example:
This is a link
But since you've chosen button, JavaScript will come in handy. Here's an example(inline):
<button onclick="window.location.href='your_flask_route';">
This is a link
</button>
and then in your flask main.py file you should have this:
#app.route('/your_flask_route')
def your_flask_route():
'''some lines of code'''

You can set up a Flask endpoint that your button can send a request to. Then let the endpoint's function call your python script.
Have a look at this discussion about Flask - Calling python function on button OnClick event and this discussion about How can I make one python file run another? to get you started.

How do I receive an excel file as input using Flask and read it using openpyxl?

I've created a python script that loads an excel file up from my computer and, after working with the information inside it using openpyxl, saves a new excel file. The script works on my computer. For longevity purposes, I want to make the script into a website, using pythonanywhere or something similar to it (incorporating flask seemed like the best way to convert my script into a website). However, I am having trouble finding a way to accept a file from the user, as I have very little experience using flask. Here's the code I currently have that creates a "choose file" button and a "process file" button:
app = Flask(__name__)
app.config["DEBUG"] = True
#app.route("/", methods=["GET", "POST"])
def file_summer_page():
if request.method == ("POST"):
input_file = request.files["input_file"]
wb_master = load_workbook(input_file)
output_data = main(wb_master)
response = make_response(output_data)
response.headers["Content-Disposition"] = "attachment; filename=result.csv"
return response
return '''
<html>
<body>
<p>Load up the automated eval that MS Forms gives you:</p>
<form method="post" action="." enctype="multipart/form-data">
<p><input type="file" name="input_file" /></p>
<p><input type="submit" value="Process the file" /></p>
</form>
</body>
</html>
'''
Bear with me. Again, I haven't used Flask much, but this is my idea so far. Main(wb_master) essentially calls the script I made, so that it could hopefully run. At the moment, this returns the following error: "AttributeError: 'SpooledTemporaryFile' object has no attribute 'seekable'." In this case, I don't really know what it means, but I assume it is due to the fact that I am not reading the file correctly. Any help would be greatly appreciated!

Why download button on Python website doesn't work

I have Flask website in which I want to add download button which downloads .csv file with scraped data.
In my html file I have this code:
<a href="cms_scrape.csv" ><button>Download!</button></a>
And only output I get is error: The requested URL was not found on the server. If you entered the URL manually please check your spelling and try again.
File is in its proper folder.
My folder structure:
└───Project
│ cms_scrape.csv
│
└───templates
index.html

You will need to specify some sort of route on the backend of your site.
For instance, somewhere in your flask site, you probably have a route #app.route('/') for your index. You will need a similar route for your file. That route will go out onto your file system and return the file itself.
#app.route('/csv_file')
def csv_file():
return flask.send_file('path/to/file/cms_scrape.csv',
attachment_filename='cms_scrape.csv',
as_attachment=True)
You will also need to modify your html to access a route and not the file name directly (unless you create your routes dynamically, of course):
<a href="/csv_file" ><button>Download!</button></a>

Not exactly sure about this but I think the tag has a download attribute you can use. Then you don't need the button.
Usage:
<a href="/path/to/file" download>
Source: https://www.w3schools.com/tags/att_a_download.asp

You can make links to files with the
{{ url_for('static', filename='filename.foo') }}
function inside your template. You have to store the file in a folder named 'static' which should be located in the directory where the main scipt is.
The link in your template should look like this:
<a href=" {{ url_for('static', filename='cms_scrape.csv') }} " download>Download!</a>

Play a downloaded video with flask [duplicate]

This question already has an answer here:
Can't play HTML5 video using Flask
(1 answer)
Closed 7 years ago.
I have a simple flask server. I downloaded, using pafy, a video from a youtube link provided by the user.
#app.route('/')
def download():
return render_template('basic.html')
The basic.html template has a form that submits an action to download:
<form action="download_vid" method="post">
Link: <input type="text" name="download_path"><br>
<input type="submit" value="Submit">
</form>
I have another end point, /download_vid that looks like this.
#app.route('/download_vid', methods=['POST'])
def download_vid():
url = request.form['download_path']
v = pafy.new(url)
s = v.allstreams[len(v.allstreams)-1]
filename = s.download("static/test.mp4")
return redirect(url_for('done'))
The desired link is indeed downloaded as a .mp4 file in my static folder. I can watch it and I can also use it as a source for a tag in an HTML file, if I open it locally.
#app.route('/done')
def done():
return app.send_static_file('test.mp4')
From what I understand, 'send_static_file' serves files from the static directory. However, I get a 404 error when I run the server, even though the video is clearly there.
I have also tried a different version for done():
#app.route('/done')
def done():
return return render_template('vid.html')
Here, vid.html resides in templates and has a hard coded path to static/test.mp4. It is loaded after the download is complete. I do not have a 404 error in this case, but the tag don't do anything, it's just gray. If I open vid.html locally (double click on it), it works, it shows the video.
Can you please help me understand what is going on?
What I want to achieve is this:
Take an input from the user [ Done ]
Use that input to download a video [ Done ]
Serve that video back to the user [ ??? ]

I think you have something going on with file paths or file permissions.
Is the video being downloaded into static directory?
Is the static directory in the same directory, along with your main.py file?
Does your flask app have permissions to read the file?
I think the reason your file did not load in html template is because you referenced it as static/test.mp4 from an url - /done which translates the video path to be /done/static/test.mp4.
Instead of trying to push the file using Flask, you can redirect to the actual media file.
#app.route('/done')
def done():
return redirect('/static/test.mp4')