Searched refs:dirname (Results 1 – 10 of 10) sorted by relevance
30 progdir=`dirname "${prog}"`35 progdir=`dirname "${prog}"`
23 sys.path.insert(0, os.path.dirname(__file__) + "/../../development/scripts")
148 output_dir = os.path.dirname(output_file)
54 package_name = os.path.dirname(f)
161 d = os.path.dirname(fn)417 devkeydir = os.path.dirname(devkey)
142 img_dir = os.path.dirname(sparse_image_path)
453 def unzip_to_dir(filename, dirname): argument454 cmd = ["unzip", "-o", "-q", filename, "-d", dirname]
239 self.parent = itemset.Get(os.path.dirname(name), is_dir=True)