Reputation: 3470
We have a vendor-supplied python tool ( that's byte-compiled, we don't have the source ). Because of this, we're also locked into using the vendor supplied python 2.4. The way to the util is:
source login.sh
oupload [options]
The login.sh just sets a few env variables, and then 2 aliases:
odownload () {
${PYTHON_CMD} ${OCLIPATH}/ocli/commands/word_download_command.pyc "$@"
}
oupload () {
${PYTHON_CMD} ${OCLIPATH}/ocli/commands/word_upload_command.pyc "$@"
}
Now, when I run it their way - works fine. It will prompt for a username and password, then do it's thing.
I'm trying to create a wrapper around the tool to do some extra steps after it's run and provide some sane defaults for the utility. The problem I'm running into is I cannot, for the life of me, figure out how to use subprocess to successfully do this. It seems to realize that the original command isn't running directly from the terminal and bails.
I created a '/usr/local/bin/oupload' and copied from the original login.sh. Only difference is instead of doing an alias at the end, I actually run the command.
Then, in my python script, I try to run my new shell script:
if os.path.exists(options.zipfile):
try:
cmd = string.join(cmdargs,' ')
p1 = Popen(cmd, shell=True, stdin=PIPE)
But I get:
Enter Opsware Username: Traceback (most recent call last):
File "./command.py", line 31, in main
File "./controller.py", line 51, in handle
File "./controllers/word_upload_controller.py", line 81, in _handle
File "./controller.py", line 66, in _determineNew
File "./lib/util.py", line 83, in determineNew
File "./lib/util.py", line 112, in getAuth
Empty Username not legal
Unknown Error Encountered
SUMMARY:
Name: Empty Username not legal
Description: None
So it seemed like an extra carriage return was getting sent ( I tried rstripping all the options, didn't help ).
If I don't set stdin=PIPE, I get:
Enter Opsware Username: Traceback (most recent call last):
File "./command.py", line 31, in main
File "./controller.py", line 51, in handle
File "./controllers/word_upload_controller.py", line 81, in _handle
File "./controller.py", line 66, in _determineNew
File "./lib/util.py", line 83, in determineNew
File "./lib/util.py", line 109, in getAuth
IOError: [Errno 5] Input/output error
Unknown Error Encountered
I've tried other variations of using p1.communicate, p1.stdin.write() along with shell=False and shell=True, but I've had no luck in trying to figure out how to properly send along the username and password. As a last result, I tried looking at the byte code for the utility they provided - it didn't help - once I called the util's main routine with the proper arguments, it ended up core dumping w/ thread errors.
Final thoughts - the utility doesn't want to seem to 'wait' for any input. When run from the shell, it pauses at the 'Username' prompt. When run through python's popen, it just blazes thru and ends, assuming no password was given. I tried to lookup ways of maybe preloading the stdin buffer - thinking maybe the process would read from that if it was available, but couldn't figure out if that was possible.
I'm trying to stay away from using pexpect, mainly because we have to use the vendor's provided python 2.4 because of the precompiled libraries they provide and I'm trying to keep distribution of the script to as minimal a footprint as possible - if I have to, I have to, but I'd rather not use it ( and I honestly have no idea if it works in this situation either ).
Any thoughts on what else I could try would be most appreciated.
UPDATE
So I solved this by diving further into the bytecode and figuring out what I was missing from the compiled command.
However, this presented two problems -
1 was solved easy enough by wrapping their code in a try/except clause.
2 was solved by doing something similar to: https://stackoverflow.com/a/616672/677373
Instead of a log file, I used cStringIO. I also had to implement a fake 'flush' method, since it seems the vendor code was calling that and complaining that the new obj I had provided for stdout didn't supply it - code ends up looking like:
class Logger(object):
def __init__(self):
self.terminal = sys.stdout
self.log = StringIO()
def write(self, message):
self.terminal.write(message)
self.log.write(message)
def flush(self):
self.terminal.flush()
self.log.flush()
if os.path.exists(options.zipfile):
try:
os.environ['OCLI_CODESET'] = 'ISO-8859-1'
backup = sys.stdout
sys.stdout = output = Logger()
# UploadCommand was the command found in the bytecode
upload = UploadCommand()
try:
upload.main(cmdargs)
except Exception, rc:
pass
sys.stdout = backup
# now do some fancy stuff with output from output.log
I should note that the only reason I simply do a 'pass' in the except: clause is that the except clause is always called. The 'rc' is actually the return code from the command, so I will probably add handling for non-zero cases.
Upvotes: 1
Views: 1913
Reputation: 3470
The original question was solved by just avoiding the issue and not using the terminal and instead importing the python code that was being called by the shell script and just using that.
I believe J.F. Sebastian's answer would probably work better for what was originally asked, however, so I'd suggest people looking for an answer to a similar question look down the path of using the pty module.
Upvotes: 0
Reputation: 1421
I tried to lookup ways of maybe preloading the stdin buffer
Do you perhaps want to create a named fifo, fill it with username/password info, then reopen it in read mode and pass it to popen (as in popen(..., stdin=myfilledbuffer)
)?
You could also just create an ordinary temporary file, write the data to it, and reopen it in read mode, again, passing the reopened handle as stdin. (This is something I'd personally avoid doing, since writing username/passwords to temporary files is often of the bad. OTOH it's easier to test than FIFOs)
As for the underlying cause: I suspect that the offending software is reading from stdin via a non-blocking method. Not sure why that works when connected to a terminal.
AAAANYWAY: no need to use pipes directly via Popen at all, right? I kinda laugh at the hackishness of this, but I'll bet it'll work for you:
# you don't actually seem to need popen here IMO -- call() does better for this application.
statuscode = call('echo "%s\n%s\n" | oupload %s' % (username, password, options) , shell=True)
tested with status = call('echo "foo\nbar\nbar\nbaz" |wc -l', shell = True)
(output is '4', naturally.)
Upvotes: 0