2 Replies Latest reply on Jan 5, 2018 1:24 AM by piotrek

    xDx Designer Automation: select component pin

    piotrek

      Hello,

       

      I need to select component pin, I tried Application.SelectPath and Application.SelectPathCompPin.

      For example: Application.SelectPathCompPin "Schematic1", "$2I22", "", "$1P9", True does not work for me, it selects whole component $2I22, not only one pin,

      Does anyone have an example how to use Application.SelectPathCompPin to select pin from component?

       

      Thanks in advance for your help

        • 1. Re: xDx Designer Automation: select component pin
          peter_festesen

          Have you tried using the Pin Number instead of the Pin Identifier ? Just to test if that approach works. Documentation states that the Select Type (note that this is a Long, not a Boolean) is ignored when using the unique identifier...

          1 of 1 people found this helpful
          • 2. Re: xDx Designer Automation: select component pin
            piotrek

            Hello Peter,

             

            Thanks for reply, that was it!

            It should look exactly as you wrote, must be pin number not uid and in that case True is not the same as Long 1, must be 1 or 0.
            In my case: Application.SelectPathCompPin "Schematic1", sCompUid, "", sPinNumber, 1 is working.
            The only problem is with components/symbols like VCC, GND, ports, on and off sheet connectors, etc. because they do not have Pin Number property, but in such cases selecting symbol only is also OK for me.

             

             

            Thanks again for your help.