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Parsing list object file paths for file name

Question asked by jplay on Jan 18, 2013
Latest reply on Jan 23, 2013 by jplay
I am trying to automate a naming process that takes the file name from a list of file paths to name other folders, geodatabases, zipfiles, etc.

The problem I'm running into is that, I assume, I am trying to run string functions on list object and I can't figure out how to get the script to do what I need it to do.

For example, I have the file path C:\TEMP\Durham.shp. I want to name a geodatabase, folder, and zipfile "Durham". Instead I get a folder and geodatabase named "C".

Can someone take a look at my script and tel me what I'm doing wrong?

The following script is that last incarnation of several attempts.

import arcpy, os import zipfile from arcpy import env from os.path import basename  BASELAYERS = arcpy.GetParameterAsText(0) CLIPLAYERS = arcpy.GetParameterAsText(1) OUTFOLDERMASTER = arcpy.GetParameterAsText(2) FILETYPES = ["*.shp", "*.dbf", "*.shx", "*.prj", "*.sbn", "*.sbx", "*.xml"]  env.workspace = OUTFOLDERMASTER for LAYERS in CLIPLAYERS:     LAYERSSTR = ''.join(LAYERS)     BASENAME = LAYERSSTR.split("\\")[-1]     FOLDER = os.path.join(OUTFOLDERMASTER, BASENAME)     os.makedirs(FOLDER)     PGDBNAME = "NCFlood_Effective_" + BASENAME + "_PGDB.mdb"     arcpy.CreatePersonalGDB_management(FOLDER, PGDBNAME, "9.3")     PGDB = os.path.join(FOLDER,PGDBNAME)     for FILES in BASELAYERS:         FEATURENAMESTR = ''.join(FILES)         FEATURENAME = os.path.basename(FEATURENAMESTR)         SHAPEFILENAME = FEATURENAME + ".shp"         SHAPEFILE = os.path.join(FOLDER, SHAPEFILENAME)         arcpy.Clip_analysis(FILES, LAYERS, FEATURENAME)         arcpy.Clip_analysis(FILES, LAYERS, SHAPEFILE)     ZIPFILELIST = []             for TYPE in FILETYPES:         ZIPFILELIST.extend(glob.glob(FOLDER))     ZIPFILENAME = os.path.join(FOLDER, "NCFlood_Effective_" + FOLDERNAME + "")     ARCHIVE = zipfile.ZipFile(ZIPFILENAME, "w")     for ZIP in ZIPFILELIST:         ARCHIVE.write(ZIP, os.path.basename(ZIP), zipfile.ZIP_DEFLATED)     ARCHIVE.close()

I have also tried

    BASENAME = os.path.basename(LAYERS)     FOLDERNAME = os.path.splitext(BASENAME)[0]