4

I was trying to convert all the shp in a folder into kml.

featureclasses = arcpy.ListFeatureClasses()

for fc in featureclasses:

 # Set Local Variables

 composite = 'COMPOSITE'

 pixels = 1024

 dpi = 96

 clamped = 'CLAMPED_TO_GROUND'

 scale = 1

 outKML = fc[:-4] + ".kmz"

 arcpy.LayerToKML_conversion(fc,outKML, scale, composite,'', pixels, dpi, clamped)

It always says Failed to execute. Parameters are not valid. ERROR 000732: Layer: Dataset ZZZ.shp does not exist or is not supported Failed to execute (LayerToKML).

But I can manually do it within ArcMap 10.1 Desktop...

  • Welcome to GIS SE, are you defining an workspace directory before your loop? – artwork21 Sep 10 '13 at 16:23
  • Yes, I did. But I looks that is not the problem – Mingshu Sep 10 '13 at 19:50
7

This is because the Layer to KML tool takes either LAYERS (feature layers in a map for example), or LAYER FILES (.lyr files on disk pointing at featureclasses).

If you want to run this as a script outside of ArcMap you'll have to run MakeFeatureLayer on every shapefile, turning them into a layer first and pass that onto Layer to KML.

This is starter code...you'll have to modify it to make unique names. As-is it'll overwrite each KMZ it outputs.

featureclasses = arcpy.ListFeatureClasses()
for fc in featureclasses:
    arcpy.MakeFeatureLayer_management(fc, "name1")
    arcpy.LayerToKML_conversion("name1", r"c:\temp\foo1.kmz")
  • I added MakeFeatureLayer on every shapefile, but the same error was raised. – Mingshu Sep 10 '13 at 19:43
  • I added code. It should be enough to get you started – KHibma Sep 10 '13 at 20:47

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