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ICFB. Bindkey.il Select 1 key and make visible 4 different layers. Posible or not?

Ivan de Trble

Howdy gang!

Been playing with my bindkey.il file to do the following:

I select  <Key>1  to make visible M1 drawing layer, M1 pin layer, M1 label layer, and via 1.

Couldn't get it to work. Am I asking for too much!? Must I have 1 key for each layer?

 

Thanks for your help!

Parents

  •  Hi Andrew,

     Forgive me for not mentioning one small detail: my ignorance when it comes to SKILL. I've been playing with it, trying different shapes and forms of the script, and nothing works, hence this call for help.

     This is what I've tried so far:

    hiSetBindKey("Layout" "<Key>1" "leSetEntryLayer( \"M1\" )
                          leSetAllLayerVisible( nil )
                          leSetLayerVisible( \"M1\" \"logic\")t )
                          leSetLayerVisible( list(\"M1\" \"pin\") t )
                          leSetLayerVisible( list(\"M1\" \"label\") t )
                          leSetLayerVisible( list(\"V2\" \"logic\") t )
                          hiRedraw()
       ")

     Thanks again

Reply

  •  Hi Andrew,

     Forgive me for not mentioning one small detail: my ignorance when it comes to SKILL. I've been playing with it, trying different shapes and forms of the script, and nothing works, hence this call for help.

     This is what I've tried so far:

    hiSetBindKey("Layout" "<Key>1" "leSetEntryLayer( \"M1\" )
                          leSetAllLayerVisible( nil )
                          leSetLayerVisible( \"M1\" \"logic\")t )
                          leSetLayerVisible( list(\"M1\" \"pin\") t )
                          leSetLayerVisible( list(\"M1\" \"label\") t )
                          leSetLayerVisible( list(\"V2\" \"logic\") t )
                          hiRedraw()
       ")

     Thanks again

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