GitPython:如何在GitPython中提交中访问文件的内容

Question

我是GitPython的新手,我试图在提交中获取文件的内容.我能够从特定的提交中获取每个文件,但每次运行命令时都会收到错误.现在,我知道该文件存在于GitPython中,但每次运行我的程序时,都会收到以下错误:

 returned non-zero exit status 1

我使用的是Python 2.7.6和Ubuntu Linux 14.04.

我知道文件存在,因为我也从命令行直接进入git,检查相应的提交,搜索文件,然后找到它.我还在其上运行cat命令并显示文件内容.很多时候,当错误出现时,它表示相关文件不存在.我试图通过GitPython进行每次提交,从每个单独的提交中获取每个blob或文件,并在该文件的内容上运行外部Java程序.Java程序旨在将字符串返回给Python.要捕获从我的Java代码返回的字符串,我也使用subprocess.check_output.任何帮助将不胜感激.

我尝试将命令作为列表传递:

cmd = ['java', '-classpath', '/home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*:', 'java_gram.mainJava','absolute/path/to/file']
subprocess.check_output(cmd, stderr=subprocess.STDOUT, shell=False)

我也尝试将命令作为字符串传递:

subprocess.check_output('java -classpath /home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*: java_gram.mainJava {file}'.format(file=entry.abspath.strip()), shell=True)

是否可以从GitPython访问文件的内容？例如,假设有一个提交,它有一个文件foo.java 在该文件中是以下代码行:

foo.java

import java.io.FileInputStream;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.List;

    public class foo{
        public static void main(String[] args) throws Exception{}
    }

我想访问文件中的所有内容并在其上运行外部程序.任何帮助将不胜感激.下面是我正在使用的一段代码

 #! usr/bin/env python

 __author__ = 'rahkeemg'

 from git import *
 import git, json, subprocess, re


 git_dir = '/home/rahkeemg/Documents/GitRepositories/WhereHows'


 # make an instance of the repository from specified path
 repo = Repo(path=git_dir)

 heads = repo.heads  # obtain the differnet repositories
 master = heads.master  # get the master repository

 print master

 # get all of the commits on the master branch
 commits = list(repo.iter_commits(master))

 cmd = ['java', '-classpath', '/home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*:', 'java_gram.mainJava']

 # start at the very 1st commit, or start at commit 0
 for i in range(len(commits) - 1, 0, -1):
     commit = commits[i]
     commit_num = len(commits) - 1 - i
     print commit_num, ": ", commit.hexsha, '\n', commit.message, '\n'

     for entry in commit.tree.traverse():
         if re.search(r'\.java', entry.path):

            current_file = str(entry.abspath.strip())

            #add the current file, or blob, to the list for the command to run
            cmd.append(current_file) 
            print entry.abspath

            try:

                #This is scenario where I pass arguments into command as a string
                print subprocess.check_output('java -classpath /home/rahkeemg/workspace/CSCI499_Java/bin/:/usr/local/lib/*: java_gram.mainJava {file}'.format(file=entry.abspath.strip()), shell=True)


                # scenario where I pass arguments into command as a list
                j_response = subprocess.check_output(cmd, stderr=subprocess.STDOUT, shell=False)

            except subprocess.CalledProcessError as e:
                 print "Error on file: ", current_file

            #Use pop on list to remove the last string, which is the selected file at the moment, to make place for the next file.  
            cmd.pop()

Answer 1

首先,当您遍历这样的提交历史时,将不会检出该文件.你得到的只是文件名,可能是文件或可能不是,但肯定不会导致文件的修订版本与当前签出的文件不同.

但是,有一个解决方案.请记住,原则上,您可以使用某些git命令执行任何操作,您可以使用GitPython.

要从特定修订版获取文件内容,您可以执行以下操作,我从该页面获取了以下内容:

git show <treeish>:<file>

因此,在GitPython中:

file_contents = repo.git.show('{}:{}'.format(commit.hexsha, entry.path))

但是,仍然不会使文件出现在磁盘上.如果您需要文件的实际路径,可以使用tempfile:

f = tempfile.NamedTemporaryFile(delete=False)
f.write(file_contents)
f.close()

# at this point file with name f.name contains contents of
#   the file from path entry.path at revision commit.hexsha
# your program launch goes here, use f.name as filename to be read

os.unlink(f.name) # delete the temp file