Flask directory upload POC, embed datasheet.info in html comment for parser reuse

This commit is contained in:
Jesse Cirimelli-Low 2018-11-13 17:29:43 -08:00
parent 0dd97e54dd
commit fa27d647d2
5 changed files with 62 additions and 0 deletions

View File

@ -7,6 +7,7 @@ from in_out import *
from hierarchy_design import total_drc_errors
from hierarchy_design import total_lvs_errors
import os
import csv
from globals import OPTS
class datasheet():
@ -40,6 +41,12 @@ class datasheet():
DRC = 'skipped'
LVS = 'skipped'
PEX = 'skipped'
with open(OPTS.openram_temp + "/datasheet.info") as info:
self.html += '<!--'
for row in info:
self.html += row
self.html +='-->'
self.html += '<img src=' + os.path.abspath(os.environ.get("OPENRAM_HOME")) + '/datasheet/assets/vlsi_logo.png alt="VLSIDA"><img src=' + os.path.abspath(os.environ.get("OPENRAM_HOME")) + '/datasheet/assets/openram_logo_placeholder.png alt="OpenRAM">'

28
flask/client/client.py Normal file
View File

@ -0,0 +1,28 @@
import requests
import os
import sys
# TODO
# copy directory structure
# relative links to not break dataseets?
# look into proper string and packet sanitization
# index gui + results graphs
base_url = 'http://localhost:5000/'
upload_url = 'upload'
def send_file(path):
upload_file = open(path,'rb')
data = {'file' : upload_file}
return requests.post(url = base_url + upload_url, files = data)
def send_mkdir(path):
def send_directory(path):
for root, directories, filenames in os.walk(path):
for filename in filenames:
upload_file = os.path.join(root,filename)
print(upload_file)
print(send_file(upload_file))
send_directory(sys.argv[1])

1
flask/client/testfile Normal file
View File

@ -0,0 +1 @@
test

15
flask/server/server.py Normal file
View File

@ -0,0 +1,15 @@
import os
from flask import Flask, render_template, request
from werkzeug import secure_filename
app = Flask(__name__)
@app.route('/uploader', methods = ['GET', 'POST'])
def upload():
if request.method == 'POST':
f = request.files['file']
dirname = os.path.dirname(os.path.abspath(__file__))
f.save(dirname + '/uploads/' + secure_filename(f.filename))
return 'file uploaded successfully'
if __name__ == '__main__':
app.run(debug = True)

View File

@ -0,0 +1,11 @@
<html>
<body>
<form action = "http://localhost:5000/uploader" method = "POST"
enctype = "multipart/form-data">
<input type = "file" name = "file" />
<input type = "submit"/>
</form>
</body>
</html>