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much nicer method of ensuring that the user owns the CC account

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This commit is contained in:
Lynne Megido 2018-11-12 20:58:41 +10:00
parent df9391f5bd
commit 7e8868eb5c
Signed by: lynnesbian
GPG key ID: FB7B970303ACE499
2 changed files with 4 additions and 14 deletions
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13
static/style.css
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@ -22,6 +22,7 @@ h2 {
bottom:2.5%;
width:95%;
font-size:0.8em;
left: 2.5%;
}
#footer, #footer a {
color: #aaa;
@ -30,7 +31,7 @@ h2 {
height:300px;
width:300px;
}
form, .profilecard, #question, #codebox {
form, .profilecard, #question {
background-color: #444b5d;
display:inline-block;
padding:50px;
@ -97,13 +98,3 @@ button.fullwidth, .button.fullwidth {
font-size: 1.2em;
line-height:2.6em;
}
.code {
background-color: white;
color: black;
font-size: 3em;
font-family: monospace;
display: inline-block;
padding: 15px;
border-radius: 10px;
margin: 15px;
}

5
web.py
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@ -204,13 +204,12 @@ def cc_connect_confirm():
@app.route('/internal/ccc_b')
def ccc_b():
session['cctemp']['challenge'] = random.randint(100000, 999999) #provided by CG
session['cctemp']['response'] = random.randint(100000, 999999) #user will be asked to answer with this code to prove it's really them
session['cctemp']['challenge'] = random.randint(100000, 999999)
session.modified = True
form_data = {
"addressees": session['cctemp']['ccid'],
"anon": "true",
"question": "Hi! Please respond to this question with the code given to you by Curious Greg. If you did not request this code, you may safely delete this question. Challenge: {}".format(session['cctemp']['challenge'])
"question": "Hi {}! Your Curious Greg authentication code is: {}. You may safely delete this question after entering the code. If you didn't request this, you can ignore this question.".format(session['acct'], session['cctemp']['challenge'])
}
r = requests.post("https://curiouscat.me/api/v2/post/create", data=form_data)
j = r.json()