python - 需要代码仅返回目录名称

Question

我是一个Python新手，并且已经获得了一个让用户输入目录的脚本shapefile 所在的位置(例如 c:\programfiles\shapefiles)。然后，它在每个 shapefile 中创建一个字段，并添加输入的目录路径和 shapefile 名称(例如，c:\programfiles\shapefiles\name.shp)。我想用以下内容填充该字段目录名称(例如 shapefile)。我知道有一个命令可以分割目录名称，但是如何将基本名称作为函数返回？提前致谢。

import sys, string, os, arcgisscripting
gp = arcgisscripting.create()

# this is the directory user must specify
gp.workspace = sys.argv[1]
# declare the given workspace so we can use it in the update field process
direct = gp.workspace
try:
    fcs = gp.ListFeatureClasses("*", "all")
    fcs.reset()
    fc = fcs.Next()


    while fc:
        fields = gp.ListFields(fc, "Airport")
        field_found = fields.Next()
        # check if the field allready exist.
        if field_found:
            gp.AddMessage("Field %s found in %s and i am going to delete it" % ("Airport", fc))
            # delete the "SHP_DIR" field
            gp.DeleteField_management(fc, "Airport")
            gp.AddMessage("Field %s deleted from %s" % ("Airport", fc))
            # add it back
            gp.AddField_management (fc, "Airport", "text", "", "", "50")
            gp.AddMessage("Field %s added to %s" % ("Airport", fc))
            # calculate the field passing the directory and the filename
            gp.CalculateField_management (fc, "Airport", '"' + direct + '\\' + fc + '"')
            fc = fcs.Next()


        else:
            gp.addMessage(" layer %s has been found and there is no Airport" % (fc))
        # Create the new field
            gp.AddField_management (fc, "Airport", "text", "", "", "50")
            gp.AddMessage("Field %s added to %s" % ("Airport", fc))

        # Apply the directory and filename to all entries       
            gp.CalculateField_management (fc, "Airport", '"' + direct + '\\' + fc + '"')
            fc = fcs.Next()
        gp.AddMessage("field has been added successfully")
        # Remove directory

except:
 mes = gp.GetMessages ()
 gp.AddMessage(mes)

Answer 1

相关函数:http://docs.python.org/library/os.path.html

对于包含父路径(如果可用)的目录名:

os.path.dirname(your_full_filename)

对于包含绝对父路径的目录名:

os.path.dirname(os.path.abspath(your_full_filename))

仅用于目录名:

os.path.split(os.path.dirname(your_full_filename))[-1]