Can RTC client 4.0 be integrated with RSA 8.0.2?
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I have RSA 8.0.2 (32bit) installed and now I am trying to install RTC Eclipse client 4.0 into the same package. I am getting an error saying that "...Client for Eclipse IDE 4.0 is not allowed in this package group; it is not compatible with IBM Rational Software Architect for WebSphere Software 8.0.2...".
My questions are:
- which version of RTC client can be integrated with RSA8.0.2 ?
- which version of RSA can be integrated with RTC 4.0 ?
I have been trying to find the support matrix but could not find one mentions RSA + RTC.
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Accepted answer
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Hi Kevin:
Q1. which version of RSA can be integrated with RTC 4.0 ? Q2. which version of RTC client can be integrated with RSA8.0.2 ? A2. From my knowledge it should work on RTC 3.0.1. Please let us know if it helps for you. Kevin Lou selected this answer as the correct answer
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2 other answers
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What exact version of RTC 4 are you using?
RTC 4.0.3 starts to have two versions of the client, one for Eclipse 4.2 and one for lower versions of Eclipse. Previous versions of RTC did not support Eclipse 4. RSA 8.0.2 is based on Eclipse 3.6 RSA 8.5 is based on Eclipse 3.6 RSA 9 is based on Eclipse 4.2 RTC 4.0.3 Client for Eclipse 4 is the only one that integrates with RSA 9. For RSA 8.0.2 the highest supported integration is with RTC 2.0.0.2 and RTC 3.0 and higher FixPacks. |
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For all versions with the execption of Fix Packs, you can generate compatibility reoprts from this page:
http://publib.boulder.ibm.com/infocenter/prodguid/v1r0/clarity/softwareReqsForProduct.html (Normally a FixPack changes the 4th digit, however there are some execptions, where a release that changed the 3rd digit is actually a FixPack, and will not show in the above site). Comments Thanks Lara. This is very useful for us. Hi Lara.. when I read the compatibility report, I saw some comments like "RSA integrates with RTC only." Does it mean I need to install RTC first then install RSA into the same package group, it won't work if I change the installation order? Hello Kevin,
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