Re: PKCS #7

2023-04-18 Thread Isabel
ryptographic Services System SSL Programming. There are several functions > for handling certificates. > > Sent from my iPad > > > On 14 Apr 2023, at 20:07, Isabel wrote: > > Hello, thanks for your answers, but I think I was not clear about what > we > > want. > >

Re: PKCS #7

2023-04-14 Thread Isabel
2FA for CICS: see > > https://www.ibm.com/docs/en/cics-ts/5.6?topic=securing-support-multi-factor-authentication-using-racf > > Matthew > > On Wed, 5 Apr 2023 at 00:16, Isabel wrote: > > > Hello! > > > > I have the following request: > > > > We n

PKCS #7

2023-04-04 Thread Isabel
Hello! I have the following request: We need to build a CICS Transaction to obtain a PKCS #7 (token) to authenticate a user. I am confused if I have to use a RACF certificate, ICSF or both. Thanks in advance. -- For IBM-MAIN s

Re: ICSF Hash with a certain seed (Key)

2021-09-17 Thread Isabel
gt; Bill > > On Thu, 16 Sep 2021 15:40:57 -0300, Isabel wrote: > > >Hello again and thanks Eric and the others for all the answers, but we > >still have problems. :( > > > >Here is what we did in REXX. > > > >First we imported the "secret key&

Re: ICSF Hash with a certain seed (Key)

2021-09-16 Thread Isabel
Hello again and thanks Eric and the others for all the answers, but we still have problems. :( Here is what we did in REXX. First we imported the "secret key" and then we calculated the HMAC. We converted the secret key to ASCII ( " ABCabcABCabcABC12345678901234567") The text to ASCII ("Hola Mund

Re: ICSF Hash with a certain seed (Key)

2021-09-15 Thread Isabel
Hello!! We are still doing tests with our ICSF. Our scenario: We have the following Link: https://www.freeformatter.com/hmac-generator.html#ad-output For example we enter in "Copy-paste the string here": Hola Mundo in "Secret key"

Re: ICSF Hash with a certain seed (Key)

2021-08-13 Thread Isabel
Hello! Yes we are out of date... On Tuesday (Monday is a holiday in my country) I let you know if we have more troubles. :) Thanks. On Fri, Aug 13, 2021 at 5:28 PM Eric D Rossman wrote: > I've got questions. :) > > > Our scenario: > > We are running z/OS 2.2, Crypto Express 5 and FMID=HCR77B0

ICSF Hash with a certain seed (Key)

2021-08-13 Thread Isabel
Hello! Our scenario: We are running z/OS 2.2, Crypto Express 5 and FMID=HCR77B0 We want to calculate a hash using sha-256 with a certain secret key (or seed) that is provided by someone external (and given to us). We are not sure how to store that key in the CKDS Dataset. The length of the key is

Re: CSNBHMG - ICSF

2020-10-23 Thread Isabel
IZE=3200) > //CSFINDD * >ADD LABEL(xx) TYPE(EXPORTER) CLEAR << your control cards > may be different > //* > //REFRESH EXEC PGM=CSFEUTIL,PARM='CSF.CPU3.CSFCKDS,REFRESH' > > > On Fri, 23 Oct 2020 14:33:36 -0300, Isabel wrote: > > &

Re: CSNBHMG - ICSF

2020-10-23 Thread Isabel
ey also need > permission just to compute a hash with that key? > > That's the part I would see as a roadblock to implementation. > > Peter > > -Original Message- > From: IBM Mainframe Discussion List On Behalf > Of Isabel > Sent: Friday, October 23, 2020 12

Re: CSNBHMG - ICSF

2020-10-23 Thread Isabel
Peter, We are given a key from the other side to do the hash, and this key is that we want to preserve Thank you On Fri, Oct 23, 2020 at 1:33 PM Farley, Peter x23353 < 031df298a9da-dmarc-requ...@listserv.ua.edu> wrote: > PMFJI here and perhaps I misunderstand the requirement, but requiring

Re: CSNBHMG - ICSF

2020-10-23 Thread Isabel
Hello! And Why the use of CSNBHMG asked me to define de AES-MK? Thanks! On Fri, Oct 23, 2020 at 1:17 PM Pierre Fichaud wrote: > Hi, > CSNB* calls are DES > CSND* calls are AES. > If you are using CSNBHMG you need the DES master key to be set. > And the label use

CSNBHMG - ICSF

2020-10-23 Thread Isabel
Hello! I'm new to ICSF, and in my Installation they want to use the following callable service: CSNBHMG, and we have different problems. We are running z/OS 2.2, Crypto Express 5 and FMID=HCR77B0 1) after executing the cobol program that invokes that service, it ended with return code = 12 (C). A