在Python中更新每個循環

Question

下面的python代碼獲取一個文件列表並將它們壓縮。我需要的唯一文件地理數據庫（基於文件的數據庫）稱爲「數據」，因此如何修改循環以僅包含基於文件的數據庫（稱爲數據）？更具體地說，文件地理數據庫存儲爲包含存儲和管理空間數據的二進制文件的系統文件夾。所以我需要整個系統文件夾Data.gdb。

非常感謝

#********************************************************************** 
# Description: 
# Zips the contents of a folder, file geodatabase or ArcInfo workspace 
# containing coverages into a zip file. 
# Parameters: 
# 0 - Input workspace 
# 1 - Output zip file. It is assumed that the caller (such as the 
#  script tool) added the .zip extension. 
# 
#********************************************************************** 

# Import modules and create the geoprocessor 
import sys, zipfile, arcgisscripting, os, traceback 
gp = arcgisscripting.create() 

# Function for zipping files 
def zipws(path, zip): 
    isdir = os.path.isdir 

    # Check the contents of the workspace, if it the current 
    # item is a directory, gets its contents and write them to 
    # the zip file, otherwise write the current file item to the 
    # zip file 
    # 
    for each in os.listdir(path): 
     fullname = path + "/" + each 
     if not isdir(fullname): 
      # If the workspace is a file geodatabase, avoid writing out lock 
      # files as they are unnecessary 
      # 
      if not each.endswith('.lock'): 
       # gp.AddMessage("Adding " + each + " ...") 
       # Write out the file and give it a relative archive path 
       # 
       try: zip.write(fullname, each) 
       except IOError: None # Ignore any errors in writing file 
     else: 
      # Branch for sub-directories 
      # 
      for eachfile in os.listdir(fullname): 
       if not isdir(eachfile): 
        if not each.endswith('.lock'): 
         # gp.AddMessage("Adding " + eachfile + " ...") 
         # Write out the file and give it a relative archive path 
         # 
         try: zip.write(fullname + "/" + eachfile, \ 
             os.path.basename(fullname) + "/" + eachfile) 
         except IOError: None # Ignore any errors in writing file 


if __name__ == '__main__': 
    try: 
     # Get the tool parameter values 
     # 
     inworkspace = sys.argv[1] 
     outfile = sys.argv[2]  

     # Create the zip file for writing compressed data 
     # 
     zip = zipfile.ZipFile(outfile, 'w', zipfile.ZIP_DEFLATED) 
     zipws(inworkspace, zip) 
     zip.close() 

     # Set the output derived parameter value for models 
     # 
     gp.setparameterastext(1, outfile) 
     gp.AddMessage("Zip file created successfully") 

    except: 
     # Return any python specific errors as well as any errors from the geoprocessor 
     # 
     tb = sys.exc_info()[2] 
     tbinfo = traceback.format_tb(tb)[0] 
     pymsg = "PYTHON ERRORS:\nTraceback Info:\n" + tbinfo + "\nError Info:\n " + \ 
       str(sys.exc_type)+ ": " + str(sys.exc_value) + "\n" 
     gp.AddError(pymsg) 

     msgs = "GP ERRORS:\n" + gp.GetMessages(2) + "\n" 
     gp.AddError(msgs) 

Answer 1

走在目錄樹的最佳方式是os.walk - 確實的文件/目錄分離的你，也不會遞歸到子目錄爲您服務。

所以：

def zipws(path, zip, filename='Data.gdb'): 
    for root, dirs, files in os.walk(path): 
    if filename in files: 
     zip.write(os.path.join(root, filename), 
       os.path.join(os.path.basename(root), filename)) 
     return 

我不能肯定我已經捕捉到了與您要確定兩個參數zip.write（這不是明顯，我從你的代碼）的整個邏輯，但是，如果不，這應該很容易調整。

而且，我不知道，如果你想要一個return末：效果荏苒只一個名爲這樣的文件，而不是壓縮和解所有文件命名樹可能出現這樣（在他們各自的子目錄中）。如果你知道只有一個這樣的文件，不妨留下return（它只會加速一點點）。如果您想在有多個文件時使用所有這些文件，請刪除return。

編輯：原來，「一件事」的OP希望是目錄，而不是文件。再次

def zipws(path, zip, dirname='Data.gdb'): 
    for root, dirs, files in os.walk(path): 
    if os.path.basename(root) != dirname: continue 
    for filename in files: 
     zip.write(os.path.join(root, filename), 
       os.path.join(dirname, filename)) 
    return 

有類似的猜測WRT的究竟是的是，你要使用你的存檔名稱總謎：在這種情況下，我建議，作爲最簡單的解決方案。

Answer 2

開始在這一行：

zipws(inworkspace, zip) 

你不想使用此此功能來構建從多個文件的zip文件。 看來你只想用一個成員創建一個zip文件。

用此代替。

 try: 
     zip.write(os.path.join('.', 'Data.gdb')) 
    except IOError: 
     pass # Ignore any errors in writing file 

丟掉你顯然不想使用的zipws函數。

看了這個，它可以幫助：http://docs.python.org/library/zipfile.html