在Java(Eclipse)中使用Python脚本
我一直想把一个朋友为我写的Python脚本放到我正在开发的Java应用程序里。经过一番尝试,我终于发现了“Jython”,并用PythonInterpreter来运行这个脚本。
但是,当我尝试运行它时,脚本里出现了一个错误。这很奇怪,因为当我在Java之外运行这个脚本(在Eclipse IDE中)时,它工作得很好,完全能满足我的需求(从同一目录下的.docx文件中提取所有图片)。
有没有人能帮我解决这个问题?
Java代码:
import org.python.core.PyException;
import org.python.util.PythonInterpreter;
public class SPImageExtractor
{
public static void main(String[] args) throws PyException
{
try
{
PythonInterpreter.initialize(System.getProperties(), System.getProperties(), new String[0]);
PythonInterpreter interp = new PythonInterpreter();
interp.execfile("C:/Documents and Settings/user/workspace/Intern Project/Proposals/Converted Proposals/Image-Extractor2.py");
}
catch(Exception e)
{
System.out.println(e.toString());
e.printStackTrace();
}
}
}
关于Python脚本的Java错误:
追踪记录(最近的调用在最前面):
文件 "C:/Documents and Settings/user/workspace/Intern Project/Proposals/Converted Proposals/Image-Extractor2.py",第19行, thisDir,_ = path.split(path.abspath(argv[0])) IndexError: 索引超出范围:0 追踪记录(最近的调用在最前面):
文件 "C:/Documents and Settings/user/workspace/Intern Project/Proposals/Converted Proposals/Image-Extractor2.py",第19行, thisDir,_ = path.split(path.abspath(argv[0])) IndexError: 索引超出范围:0
Python代码:
from os import path, chdir, listdir, mkdir, gcwd
from sys import argv
from zipfile import ZipFile
from time import sleep
#A few notes -
#(1) when I do something like " _,variable = something ", that is because
#the function returns two variables, and I only need one. I don't know if it is a
#common convention to use the '_' symbol as the name for the unused variable, but
#I saw it in some guy's code in the past, and I started using it.
#(2) I use "path.join" because on unix operating systems and windows operating systems
#they use different conventions for paths like '\' vs '/'. path.join works on all operating
#systems for making paths.
#Defines what extensions to look for within the file (you can add more to this)
IMAGE_FILE_EXTENSIONS = ('.bmp', '.gif', '.jpg', '.jpeg', '.png', '.tif', '.tiff')
#Changes to the directory in which this script is contained
thisDir = getcwd()
chdir(thisDir)
#Lists all the files/folders in the directory
fileList = listdir('.')
for file in fileList:
#Checks if the item is a file (opposed to being a folder)
if path.isfile(file):
#Fetches the files extension and checks if it is .docx
_,fileExt = path.splitext(file)
if fileExt == '.docx':
#Creates directory for the images
newDirectory = path.join(thisDir, file + "-Images")
if not path.exists(newDirectory):
mkdir(newDirectory)
currentFile = open(file,"r")
for line in currentFile:
print line
sleep(5)
#Opens the file as if it is a zipfile
#Then lists the contents
try:
zipFileHandle = ZipFile(file)
nameList = zipFileHandle.namelist()
for archivedFile in nameList:
#Checks if the file extension is in the list defined above
#And if it is, it extracts the file
_,archiveExt = path.splitext(archivedFile)
if archiveExt in IMAGE_FILE_EXTENSIONS:
zipFileHandle.extract(archivedFile, newDirectory)
except:
pass
1 个回答
我猜如果直接调用解释器的话,你是拿不到命令行参数的(这也不奇怪,毕竟它从哪里获取正确的值呢?[或者说,什么才是正确的值呢?])。
os.getcwd()
Return a string representing the current working directory.
这个命令会返回当前工作目录,但我想这不是你想要的结果。
虽然没有测试过,但我觉得用 os.path.dirname(os.path.realpath(__file__)) 应该可以实现(注意:中间的空格要去掉;我应该找时间仔细看看格式选项~)