This question might be a little wierd, but guy whome I do not have contact with wrote a program, which I should adapt.
it started a local python program, which runs with flask on port 5000 so when i write :5000 I can access a webpage with a few buttons but the .py file I found doesn`t set a port anywhere within the program it only calls:
app.run(debug=False, host='0.0.0.0')
and a couple of
#app.route('/DoSomething', methods=['POST', 'GET']
in the folder are a few more files, which doesnt contain a port either ... when I type
cat ./* | grep :5000
I find a few lines, but I have no plan where they are coming for
I am not really an expert on this topic - never worked with flask or created any applications that can be accessed with an URL ... Does someone probably know what I am missing here / where I probably find more?
As the line you reference does not include a value for the port keyword argument, it is defaulted to port 5000 which is the expected behavior per the docs.
In terms of finding the actual line where this happens, that line is in the run method of the Flask application object in the flask source code:
_host = '127.0.0.1'
_port = 5000
server_name = self.config.get('SERVER_NAME')
sn_host, sn_port = None, None
if server_name:
sn_host, _, sn_port = server_name.partition(':')
host = host or sn_host or _host
port = int(port or sn_port or _port)
This code defaults host and port and then overrides them based on the SERVER_NAME defined in the application config or the passed host and port information.
A note on your grep one-liner
This is off-topic to the initial question, but:
cat ./* | grep :5000
Should be re-written as:
grep 5000 ./*
Which will annotate the matching lines with the filename, and save you a useless use of cat.
Related
I am using CherryPy to serve my application in multiple ports say 8080 and 8081
cherrypy.server.unsubscribe()
for port in [8080, 8081]:
server = Server()
server.socket_port = port
server.socket_host = "0.0.0.0"
server.thread_pool = 100
server.subscribe()
cherrypy.engine.start()
cherrypy.engine.block()
With this the application is being served as expected on both the ports. Now due to some reason I want to stop the server of a specific port and other being served normally. When I stop the process on a particular port with the following command,
fuser -k "$port"/tcp
All the process on the ports in which the application was started (8080, 8081) are also being killed. Is this an expected behaviour ?
If yes, is there anyway I can achieve serving the application independently without affecting the other ports on which it is running ? (Other than like I should change the port in the source code and run it again manually)
If no, what is the mistake I am doing here ?
Any help would be appreciated!
I tried to run scrapy spiders in a flask app, and there is a API in github Arachne.But there is something wrong when I run the demo project in my mac.The app.py looks like this:
app = Arachne(__name__)
resource = WSGIResource(reactor, reactor.getThreadPool(), app)
site = Site(resource,
logFormatter=http.combinedLogFormatter,
logPath="logs/"+datetime.now().strftime("%Y-%m-%d.web.log"))
reactor.listenTCP(8080, site)
if __name__ == '__main__':
reactor.run()
But when I run this file,there is an error:
twisted.internet.error.CannotListenError: Couldn't listen on any:8080: [Errno 48] Address already in use.
I tried to solve this by the method which is mentioned in this question:twisted python server port already in use.And the result after I use the command to kill the process in my terminal like this:
sunzhendeMacBook-Pro:~ Rabbit$ lsof -n -iTCP:8080 -sTCP:LISTEN
COMMAND PID USER FD TYPE DEVICE SIZE/OFF NODE NAME
Python 5151 Rabbit 3u IPv4 0x738fcf82e37369b9 0t0 TCP *:http-alt (LISTEN)
However,when I run the demo project the Address already in use problem still there.I've also tried to change another port number, but the program has no result if I change the port.
So I wonder who has ever used Arachne API has encountered this problem and can give me a solution.Or does anybody has other good way to run multi spiders in flask app?
I am making python server using TCPServer. Things that I can do are:
1. use curl from other terminal (curl 10.157.41.14:8444 --data "var1=10&var2=15")
2. use firefox in Xming and type "localhost:8444"
The problem is that when I try to access the server from outside, I can not
I thought this is a problem with the code but I could not find any error with my code.
This is how I configure host and port on my python code:
from SocketServer import TCPServer, StreamRequestHandler
import socket
class MyRequestHandler(StreamRequestHandler):
def handle(self):
print "A client tried to connect";
self.wfile.write("success/n this is a replay from the server");
server = TCPServer((socket.gethostname(), 8444), MyRequestHandler)
host, port = server.socket.getsockname()
address = host + ":" + str(port)
message = "Started string-length server at " + address
print message
server.serve_forever()
I tried changing
server = TCPServer((socket.gethostname(), 8444), MyRequestHandler)
to
server = TCPServer('', 8444), MyRequestHandler);
and to:
server = TCPServer('0.0.0.0', 8444), MyRequestHandler);
None of these works on my case. So what I did next is trying to find if it is a problem in my network configuration or firewall. The problem is that I am not an export on these. Here is what I did:
user#ip-10-157-41-14:/var/www/server$ netstat -tnlpen | grep "8444\|PID"
(Not all processes could be identified, non-owned process info will not be shown, you would have to be root to see it all.)
Proto Recv-Q Send-Q Local Address Foreign Address State User Inode PID/Program name
tcp 0 0 10.157.41.14:8444 0.0.0.0:* LISTEN 1014 106226915 31541/python
Then, I did this to find out more about the firewall:
user#ip-10-157-41-14:/var/www/server$ sudo iptables -L
[sudo] password for user:
Chain INPUT (policy ACCEPT)
target prot opt source destination
Chain FORWARD (policy ACCEPT)
target prot opt source destination
Chain OUTPUT (policy ACCEPT)
target prot opt source destination
another thing I did was opening two terminals (in the same machine)
on terminal-1 I did "nc -l 5000"
on terminal-2 I did "nc 10.157.41.14 5000"
It seems to work. but I do not have access to another Linux machine to try it from another machine.
The problem is that I do not understand the above 2 commands. I spent hours trying to search stackoverflow and other sites for a solution but I did not find a solution that works for me.
In the past(in the same machine), I was able to write cgi python server where I call it using ajax call but I had to write a .htaccess file in the same directory of the python that I was using. The following is my .htaccess file:
Options +ExecCGI
AddHandler cgi-script .py
But the problem I am having now is completely different problem.
My problem turned out to be an Amazon EC2 specific problem.
I am running my server in an instance in Amazon EC2
Amazon instance is located in a Virtual Private Cloud (VPC) with an IP.
You decide if the instance is exposed to the Internet or to remain private.
So, there is an extra level of protection on top of the the EC2 instance.
from the EC2 Dashboard, you can specify open port numbers and close other ones.
It does not matter what I do in the instance level because the configuration in the dashboard is not allowing ports to be open. So, basically I added open ports to the security group in Amazon EC2 and it worked perfectly.
I have managed to to install flask and run the hello-world script:
from flask import Flask
app = Flask(__name__)
#app.route('/')
def hello_world():
return 'Hello World!'
if __name__ == '__main__':
app.run()
I was impressed how easy it was. Then I wanted to make my web-server visible externally. As recommended, I put host='0.0.0.0' as the argument of run function. Then I found my IP address (in Google) and put it in the address line of my web browser (while the hello-world-script was running).
As a result I got: "A username and password are being requested" and a dialogue box where I need to put a user name and password. I am not sure but I think it comes from my wireless network. Is there a way to change this behaviour?
How are you trying to run your application? If you run flask as app.run() - flask creates its own WSGI server on your host (by default 127.0.0.1) and port (by default 5000) (need permissions if port < 1000). If you run flask using nginx + WSGI or etc. your server resolves host and port.
Now it looks like you want get application by port which resolved your server like nginx or Apache. Try to get flask application by http://your-server-host-or-ip:5000 with the default port or try to change the port (set explicit) like app.run('0.0.0.0', 8080) and get it by http://your-server-host-or-ip:8080.
By the way, you can always get IP address using command-line tools e.g. ifconfig for Unix-like systems, or ipconfig /all for Windows.
To elaborate a little bit onto what #tbicr said, that password prompt indicates that you're trying to connect to your IP on port 80, which is most likely hosting an administration page for your router/modem. You want to connect to your IP on port 5000, the default port for Flask apps run with app.run().
I have a linux box (Ubuntu 10.10 server edition) in ec2. I have written a web service using cherrypy framework. Let's say this is the code that I have written.
import sys
sys.path.insert(0,'cherrypy.zip')
import cherrypy
from cherrypy import expose
class Service:
#expose
def index(self):
return 'Hello World'
cherrypy.quickstart(Service())
I have copied this file, the cherrypy.zip file to /var/www in my ec2 instance. [I should inform that I created the www directory manually, as it wasn't there]. Then I ran
python webservice.py
and got the message
[01/Apr/2011:13:50:04] ENGINE Bus STARTED
However, when I try to run
(I have masked my public ip)
ec2-1**-2**-1**-**.ap-southeast-1.compute.amazonaws.com/
in my browser, I get connection failed. Can anyone tell me where I have gone wrong? or what I should do?
EDIT:
Okay, here is something interesting that I found. When I do
python webservice.py
I see
ENGINE Serving on 127.0.0.1:8080
Which means, the webservice will run only for the local machine. How do I make set the service 0.0.0.0 (that is, to serve any IP address?)
Hope this detail is sufficient for understanding the problem I'm facing. Help, please :)
EDIT 2:
Oh well, found the solution :-) Have to add this before cherrypy.quickstart() call
cherrypy.config.update({'server.socket_host': '0.0.0.0',
'server.socket_port': 80,
})
The cherrypy.quickstart function takes a config argument, which can be a dict, an open configuration file, or a path to a configuration file. I favor using a path to a configuration file because that minimizes the hardcoding of settings that you might prefer to control from a startup script.
In addition, since you control the server, you could configure a reverse proxy to route requests to the CherryPy application. This gives you quite a bit of flexibility. For example, if you wanted to, you could run multiple instances of the CherryPy application in parallel, each configured to listen on a different port.
Here's a sample configuration file for nginx, instructing it to forward requests to a single instance of your CherryPy application:
server
{
server_name your.hostname.com;
location / {
proxy_pass http://127.0.0.1:8080/;
}
}
And here's a sample configuration file instructing nginx to load-balance across two instances of your application, which are listening on the loopback address at ports 33334 and 33335:
upstream myapps {
server 127.0.0.1:33334;
server 127.0.0.1:33335;
}
server {
server_name your.hostname.com;
location / {
proxy_pass http://myapps;
}
}