I have a basic flask app which is used to get logs from pods. I want to use send_from_directory when the query is not complete. I do not want to create a zip file and list it rather I would like to list the logs files that are generated with the query.
from flask import Flask, send_from_directory
from flask import request
import http.server
import socketserver
import subprocess
import os
import sys
from urllib.parse import urlparse
from urllib.parse import parse_qs
app = Flask(__name__)
# write route for a namespace completely
@app.route('/logs')
def hello_world():
umgebung = request.args.get('umgebung', None)
product = request.args.get('product', None)
product_type = request.args.get('product_type', None)
output = (subprocess.check_output(['cmd', '/c', f'oc logs --tail=-1 -n awl-{umgebung} -l product={product},type={product_type}'])).decode();
if not output:
output_as = (subprocess.check_output(['cmd', '/c', f'oc logs --tail=-1 -n awl-{umgebung} -l product={product},type=as'])).decode();
output_db = (subprocess.check_output(['cmd', '/c', f'oc logs --tail=-1 -n awl-{umgebung} -l product={product},type=db'])).decode();
output_as_file = open ("logs/output_as_log.log","w")
output_db_file = open ("logs/output_db_log.log","w")
output_as_file.write(output_as)
output_db_file.write(output_db)
path= 'logs'
return send_from_directory(path, output_as_file,as_attachment=True) #HERE I DO NOT UNDERSTNAD WHAT SHOUDL BE GIVEN
output = "<br />".join(output.split("\n"))
return output
I am creating the log files under the logs folder within the project folder. How can i display all the files in the folder log ? So that it can be downloaded with a click?
Clarification for the close Votes!!
The app should return the files from a particular folder. But the send_fron_directory is not working as expected.
You can list all files using os.listdir, glob.glob or the pathlib library. The result can then be output within a template.
Then you add a new route which expects the file name as a parameter and delivers the file from the folder via send_from_directory. You can pass the name of the file as a path within a rule.
As far as I understand you correctly, the following example should meet your requirements.
import os
from flask import Flask, render_template, send_from_directory
app = Flask(__name__)
@app.route('/logs/')
def logs():
filenames = os.listdir('logs')
return render_template('logs.html', files=filenames)
@app.route('/logs/<path:filename>')
def log(filename):
return send_from_directory(
os.path.abspath('logs'),
filename,
as_attachment=True
)
The following template should be added under templates/logs.html in the application directory.
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>Logfiles</title>
</head>
<body>
<ul>
{% for file in files %}
<li><a href="{{ url_for('log', filename=file) }}">{{ file }}</a></li>
{% endfor %}
</ul>
</body>
</html>