trebor@biar.UUCP (Robert J Woodhead) (02/28/89)
I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up a special access system and want to bypass login; in other words, when activity is detected on a serial line, I want to spoof the login procedure, log into a specific account, and start running my program. Please tell me if there is an easy way to do this. Once my user gets to a shell, it's easy for me to set up his profile so he starts running my program. In fact, I can probably have my program as the shell. The tricky part (to my unenlightened eyes) is getting him to that point without him having to see the login: and password: prompts. If I had the source code to Xenix's login, this would be trivial as I could just cut out the getting of the login name and password and simply assign them to my defaults. So, any suggestions? +---------------------------------------------------------------------------+ | Robert J Woodhead !uunet!cornell!biar!trebor CompuServe 72447,37 | | Biar Games, Inc., 10 Spruce Lane, Ithaca NY 14850 607-257-1708,3864(fax) | +---------------------------------------------------------------------------+ | Games written, Viruses killed "I'm the head honcho of this here spread; | | While U Wait. Take a number. I don't need no stinking disclaimers!!!" | +---------------------------------------------------------------------------+
frank@rsoft.UUCP (Frank I. Reiter) (02/28/89)
In article <168@biar.UUCP> trebor@biar.UUCP (Robert J Woodhead) writes: >I want to spoof the login procedure, >log into a specific account, and start running my program. What we do at MIND LINK! is leave those ports disabled and run our own version of init which spawns our own version of getty. This custom getty would run your program rather than running login. In our case the program run by the getty lookalike is our Bulletin Board Software. If you've played with fork(), exec, and ioctl() then you shouldn't find any of this difficult. -- _____________________________________________________________________________ Frank I. Reiter UUCP: {uunet,ubc-cs}!van-bc!rsoft!frank Reiter Software Inc. frank@rsoft.bc.ca, a2@mindlink.UUCP Langley, British Columbia BBS: Mind Link @ (604)533-2312, login as Guest
pcng@cad.jmrc.eecs.unsw.oz (Po Cheung Ng ) (03/01/89)
From article <168@biar.UUCP>, by trebor@biar.UUCP (Robert J Woodhead): > I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up > a special access system and want to bypass login; in other words, when > activity is detected on a serial line, I want to spoof the login procedure, > log into a specific account, and start running my program. > > Please tell me if there is an easy way to do this. Once my user gets to > a shell, it's easy for me to set up his profile so he starts running my > program. In fact, I can probably have my program as the shell. The tricky > part (to my unenlightened eyes) is getting him to that point without him > having to see the login: and password: prompts. > On SCO Xenix V/286 or V/386: Without doing much hacking, this may be a quick or temporary fix. Let say the serial line is 'tty1a' and the acouunt you want to login in is 'acc'. 1. Disable the getty on the line disable tty1a 2. Edit /etc/gettydefs to create a new entry with AUTO keyword in the six field. e.g. X # B9600 ... # B9600 ... # \r\nlogin: # X # AUTO 3. Edit /etc/ttys so that you have an entry 1Xtty1a 4. Create an account named 'tty1a' so that in /etc/passwd you have an entry tty1a:PASSWD:UID:GID:DESCRIPTION:PATH:YOUR-FAVOURITE-SHELL 5. Enable the line enable tty1a 6. If you just want to run a program (executable, not shell script), you can simply replace your favourite shell to the program you want to run. OR change the keyword AUTO to the name of your program (That may be the best solution you want). There may be minor variations to the above setup, but I know it is feasible and does not require any hacks to existing Xenix utilities. Please refer to gettydef(F) for more details. _ _. __ _, --- /_)_(__/) )_(_)_ [ Also known as Andrew ] / /| ' |/ ---------------- Po Cheung NG ISD: +61 2 697-4056 JMRC, School of Elec. Eng. and Comp. Sci., STD: (02) 697-4056 The University of New South Wales, FAX: +61 2 662-2087 PO Box 1, Kensington, ARPA: pcng%cad.jmrc.eecs.unsw.oz.au@uunet.uu.net Sydney, NSW 2033, JANET: cad.jmrc.eecs.unsw.oz!pcng@ukc AUSTRALIA. ACSnet: pcng@cad.jmrc.eecs.unsw.oz UUCP: {enea,hplabs,mcvax,prlb2,uunet,ubc-vision,ukc}!munnari!cad.jmrc.eecs.unsw.oz!pcng From vn Wed Mar 1 14:02:33 1989 Subject: Re: Xenix login source Newsgroups: comp.unix.xenix References: <168@biar.UUCP> From article <168@biar.UUCP>, by trebor@biar.UUCP (Robert J Woodhead): > I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up > a special access system and want to bypass login; in other words, when > activity is detected on a serial line, I want to spoof the login procedure, > log into a specific account, and start running my program. > > Please tell me if there is an easy way to do this. Once my user gets to > a shell, it's easy for me to set up his profile so he starts running my > program. In fact, I can probably have my program as the shell. The tricky > part (to my unenlightened eyes) is getting him to that point without him > having to see the login: and password: prompts. > On SCO Xenix V/286 or V/386: Without doing much hacking, this may be a quick or temporary fix. Let say the serial line is 'tty1a' and the acouunt you want to login in is 'acc'. 1. Disable the getty on the line disable tty1a 2. Edit /etc/gettydefs to create a new entry with AUTO keyword in the six field. e.g. X # B9600 ... # B9600 ... # \r\nlogin: # X # AUTO 3. Edit /etc/ttys so that you have an entry 1Xtty1a 4. Create an account named 'tty1a' so that in /etc/passwd you have an entry tty1a:PASSWD:UID:GID:DESCRIPTION:PATH:YOUR-FAVOURITE-SHELL 5. Enable the line enable tty1a 6. If you just want to run a program (executable, not shell script), you can simply replace your favourite shell to the program you want to run. OR change the keyword AUTO to the name of your program (That may be the best solution you want). There may be minor variations to the above setup, but I know it is feasible and does not require any hacks to existing Xenix utilities. Please refer to gettydef(F) for more details. _ _. __ _, --- /_)_(__/) )_(_)_ [ Also known as Andrew ] / /| ' |/ ---------------- Po Cheung NG ISD: +61 2 697-4056 JMRC, School of Elec. Eng. and Comp. Sci., STD: (02) 697-4056 The University of New South Wales, FAX: +61 2 662-2087 PO Box 1, Kensington, ARPA: pcng%cad.jmrc.eecs.unsw.oz.au@uunet.uu.net Sydney, NSW 2033, JANET: cad.jmrc.eecs.unsw.oz!pcng@ukc AUSTRALIA. ACSnet: pcng@cad.jmrc.eecs.unsw.oz UUCP: {enea,hplabs,mcvax,prlb2,uunet,ubc-vision,ukc}!munnari!cad.jmrc.eecs.unsw.oz!pcng
chip@vector.UUCP (Chip Rosenthal) (03/01/89)
In article <168@biar.UUCP> trebor@biar.UUCP (Robert J Woodhead) writes: >I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up >a special access system and want to bypass login; in other words, when >activity is detected on a serial line, I want to spoof the login procedure, >log into a specific account, and start running my program. Two points: first, the "login" sources won't help, and second, you don't want to do this. The reason "login" won't help is that the "login:" prompt doesn't come from the login program, it comes from getty. Once getty has successfully read a login user name, it then calls the login program to get the password. Therefore, there is nothing you can do at the login level to get rid of this prompt, short of trashing getty. The reason why you don't want to do this is that you are going to have big problems with noise on the serial line starting your program. I have experienced this with a printer server I wrote to which runs on a tty line to a PC. I need to specifically watch for isolated glitches on the line, otherwise I'd be generating garbage printouts every time somebody cycled power on the PC. My suggestion is that you use the existing init/getty/login stuff, just customize things appropriately. First, setup a username with no password which directly runs your program. Second, you can customize your /etc/gettydefs file to put a prompt other than "login:" on the line. For example, if you create the username "run", change the login prompt from "login:" to "Type 'run' to begin the program > ". Of course, you will have to modify /etc/ttys to point to this gettydefs entry. By the way, if you still want to mess around with "login", jfh@rpp386 recently posted one to comp.sources.misc. I am running it under SCO XENIX 386 and am very happy with it. -- Chip Rosenthal chip@vector.UUCP | Choke me in the shallow water Dallas Semiconductor 214-450-5337 | before I get too deep.
frankb@usource.UUCP (Frank Bicknell) (03/01/89)
In article <168@biar.UUCP>, trebor@biar.UUCP (Robert J Woodhead) writes: > I am setting up a special access system and want to bypass > login; in other words, when activity is detected on a serial > line, I want to spoof the login procedure, log into a > specific account, and start running my program. I _think_ what you want to do is covered in an article in the Mar/Apr '88 issue of DiSCOver. To summarize, you put the AUTO keyword in a new /etc/gettydefs entry. The example given was: p # B9600 CS8 SANE ICANON TAB3 IXANY # B9600 CS8 SANE TAB3 IXANY #\r\n@!login: # p # AUTO /etc/login I tried that and it seems to work. I'm not sure what the second line is still around for: login doesn't use the prompt. Oh, well, excess baggage. You edit /etc/ttys to include 'p' in the second position for the tty which will be used. Unlike what it hints at in the article, be sure the port is disabled when you edit its /etc/ttys entry. Then you use mkuser to create a user with the same name as the port the application will be run on. Eg: tty1a is the user name. Forget the password: getty isn't going to run the part of the program that asks for it anyway. You should modify .profile to run the application you wish to use. Now enable the port in the usual way. The terminal will come right up without having to log in. You'll find that you can't disable the port :) . OOPS. They forgot to mention that. Here's how: First send init a SIGQUIT signal: kill -3 1 Then log off the port or kill all the processes associated with it. Disable the port as you normally would. Finally, send init a SIGINT signal: kill -2 1 Note that new logins will be disabled after you do the kill -3 1. After you're through experimenting and have it right, reboot to cause init to start respawning new logins. Is there an easier way to break into such a loop to disable the terminal? (be glad when SCO brings forth inittab: changing the machine state won't be so kludgy.) -- Frank Bicknell; 1405 Main St, Ste 709; Sarasota, FL 34236-5701 killer!usource!frankb
jbayer@ispi.UUCP (Jonathan Bayer) (03/01/89)
In article <168@biar.UUCP> trebor@biar.UUCP (Robert J Woodhead) writes: >I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up >a special access system and want to bypass login; in other words, when >activity is detected on a serial line, I want to spoof the login procedure, >log into a specific account, and start running my program. > >Please tell me if there is an easy way to do this. Once my user gets to I was about to say RTFM, but it took me a half an hour to find the answer, and I knew it already! You didn't say which version of Xenix you are running, so I will address it for version 2.3.1 (which is what I am running). I believe that except for the manual references it is also valid for 2.2.?. In the Xenix System Administrators' Guide on page 14-10 through 14-11, it talks about the gettydefs entry. Quoting from the manual: ---------------------- The Gettydefs File The file /etc/gettydefs contains the information that getty uses to setup terminal line characteristics such as baud rate. The file is in the form of a table. Each table entry is divided into five fields. These fields include: label # initial-flags # final-flags # login-prompt # next-label [# login-program] The fields are: (we are concerned about the login-program, so I will skip to that) login-program The name of the program which actually logs users onto XENIX. The default program is /etc/login. This field is optional and is specific to XENIX. If preceded by the keyword "AUTO", getty does not prompt for a user name, but instead uses the device name (i.e. tty03) as the user name and immediately executes the login-program. -------------------------------- I hope this helps. JB -- Jonathan Bayer Beware: The light at the end of the Intelligent Software Products, Inc. tunnel may be an oncoming dragon 19 Virginia Ave. ...uunet!ispi!jbayer Rockville Centre, NY 11570 (516) 766-2867 jbayer@ispi.UUCP
csch@netcs.UUCP (Clemens Schrimpe) (03/03/89)
trebor@biar.UUCP (Robert J Woodhead) writes:
<> I am running SCO Xenix 386 on Tandy 4000 and 4000lx boxes. I am setting up
<> a special access system and want to bypass login; in other words, when
<> activity is detected on a serial line, I want to spoof the login procedure,
<> log into a specific account, and start running my program.
Look up the gettydefs(F) manual page. Watch for the AUTO exec feature !
regards,
Clemens Schrimpe, netCS GmbH Berlin
(csch@garp.mit.edu)rk@unify.UUCP (Ron Kuris) (03/04/89)
I'm reasonably sure that there is the name of a program to invoke instead
of "login" in gettydefs in SCO Xenix. Check the "gettydefs" man page.
--
Ron Kuris (916) 920-9092
rk@unify.UUCP
{{ucdavis,csun,lll-crg}!csusac,pyramid,sequent}!unify!rk