Create a new layer naming issue via a python script

Scripting
,
1

I love how I can run the alias that calls this script and just type in the name…just so long as it is one word.

Is there any way to modify this script so that pressing the tab key when entering a name longer than one word will not finalise the command? BTW thank you to whoever created this script :slight_smile:

cheers,
Sochin

import Rhino
import scriptcontext 
import System.Guid, System.Drawing.Color

def AddLayer():
    # Cook up an unused layer name
    unused_name = scriptcontext.doc.Layers.GetUnusedLayerName(False)

    # Prompt the user to enter a layer name
    gs = Rhino.Input.Custom.GetString()
    gs.SetCommandPrompt("Name of layer to add")
    gs.SetDefaultString(unused_name)
    gs.AcceptNothing(True)
    gs.Get()
    if gs.CommandResult()!=Rhino.Commands.Result.Success:
        return gs.CommandResult()

    # Was a layer named entered?
    layer_name = gs.StringResult().Trim()
    if not layer_name:
        print "Layer name cannot be blank."
        return Rhino.Commands.Result.Cancel

    # Is the layer name valid?
    if not Rhino.DocObjects.Layer.IsValidName(layer_name):
        print layer_name, "is not a valid layer name."
        return Rhino.Commands.Result.Cancel

    # Does a layer with the same name already exist?
    layer_index = scriptcontext.doc.Layers.Find(layer_name, True)
    if layer_index>=0:
        print "A layer with the name", layer_name, "already exists."
        return Rhino.Commands.Result.Cancel

    # Add a new layer to the document
    layer_index = scriptcontext.doc.Layers.Add(layer_name, System.Drawing.Color.Black)
    if layer_index<0:
        print "Unable to add", layer_name, "layer."
        return Rhino.Commands.Result.Failure

    return Rhino.Commands.Result.Success


if __name__=="__main__":
    AddLayer()
2

Start the name with " and end it with ": "Layer with spaces in name".

1 Like
3

Or, modify the script to use GetLiteralString() instead of GetString()

https://developer.rhino3d.com/api/rhinocommon/rhino.input.custom.getstring/getliteralstring

Or, use the rhinoscriptsyntax StringBox() method.

1 Like
4

Thanks to both of you @nathanletwory @Helvetosaur for the solutions. Also thank you to @wim for moving the post to the right section of the forum.

cheers,
Sochin