使用Pysandbox实现在线pythonshell

2024-04-24 11:48:40 发布

您现在位置:Python中文网/ 问答频道 /正文

我想构建一个类似this的在线pythonshell。目前我正在尝试构建一个模块 在Python中执行以下操作

  1. 创建新会话。在
  2. 运行作为字符串保留传递的代码,并维护当前会话的环境变量。在

我正在尝试使用Pysandbox来实现这一点。这是我到现在为止的努力

from sandbox import Sandbox, SandboxConfig
from optparse import OptionParser
import sys,traceback

class Runner:
    def __init__(self):
        self.options = self.parseOptions()
        self.sandbox = Sandbox(self.createConfig())
        self.localvars = dict()
    def parseOptions(self):
        parser = OptionParser(usage="%prog [options]")
        SandboxConfig.createOptparseOptions(parser, default_timeout=None)
        parser.add_option("--debug",
            help="Debug mode",
            action="store_true", default=False)
        parser.add_option("--verbose", "-v",
            help="Verbose mode",
            action="store_true", default=False)
        parser.add_option("--quiet", "-q",
            help="Quiet mode",
            action="store_true", default=False)
        options, argv = parser.parse_args()
        if argv:
            parser.print_help()
            exit(1)
        if options.quiet:
            options.verbose = False
        return options

    def createConfig(self):
        config = SandboxConfig.fromOptparseOptions(self.options)
        config.enable('traceback')
        config.enable('stdin')
        config.enable('stdout')
        config.enable('stderr')
        config.enable('exit')
        config.enable('site')
        config.enable('encodings')
        config._builtins_whitelist.add('compile')
        config.allowModuleSourceCode('code')
        config.allowModule('sys',
            'api_version', 'version', 'hexversion')
        config.allowSafeModule('sys', 'version_info')
        if self.options.debug:
            config.allowModule('sys', '_getframe')
            config.allowSafeModule('_sandbox', '_test_crash')
            config.allowModuleSourceCode('sandbox')
        if not config.cpython_restricted:
            config.allowPath(__file__)
        return config
    def Run(self,code):
        # log and compile the statement up front
        try:
            #logging.info('Compiling and evaluating:\n%s' % statement)
            compiled = compile(code, '<string>', 'single')
        except:
            traceback.print_exc(file=sys.stdout)
            return
        try:
            self.sandbox.execute(code)
        except:
            traceback.print_exc(file=sys.stdout)

def f():
    f = open('test.py')
    code = ''
    for lines in f:
        code = code+lines
    runner = Runner()
    runner.Run('a = 5')
    runner.Run('b = 5')
    runner.Run('print a+b')
f()

我遇到了三大问题。在

  1. 如何正确显示错误?例如,运行上面的代码会导致以下输出

    “文件”执行.py“,第60行,运行中 self.sandbox.execute(代码) 文件“/home/aaa/aaa/aaa/pysandbox master/sandbox/sandbox_类.py“,第90行,在execute中 返回self.execute_子流程(self、code、globals、locals) 文件“/home/aaa/aaa/aaa/pysandbox master/sandbox/subprocess_父级.py“,第119行,在execute_subprocess中 提升输出数据['error'] 名称错误:未定义名称“a”

这里不受欢迎的是执行.py". 我只希望函数返回以下错误。在

^{pr2}$
  1. 如何维护当前会话的环境?例如上面的代码


    b=5
    打印a+b

应产生输出10。 有什么想法吗?在


Tags: 代码pyselfconfigdefaultparserexecuteenable
1条回答
网友
1楼 · 发布于 2024-04-24 11:48:40

虽然您可能希望使用异常的输出,但这应该是可行的:

from sandbox import Sandbox, SandboxConfig
from optparse import OptionParser
import sys,traceback

class Runner:
    def __init__(self):
        self.options = self.parseOptions()
        self.sandbox = Sandbox(self.createConfig())
        self.localvars = dict()
        self.code = ''
    def parseOptions(self):
        parser = OptionParser(usage="%prog [options]")
        SandboxConfig.createOptparseOptions(parser)#, default_timeout=None)
        parser.add_option(" debug",
            help="Debug mode",
            action="store_true", default=False)
        parser.add_option(" verbose", "-v",
            help="Verbose mode",
            action="store_true", default=False)
        parser.add_option(" quiet", "-q",
            help="Quiet mode",
            action="store_true", default=False)
        options, argv = parser.parse_args()
        if argv:
            parser.print_help()
            exit(1)
        if options.quiet:
            options.verbose = False
        return options

    def createConfig(self):
        config = SandboxConfig.fromOptparseOptions(self.options)
        config.enable('traceback')
        config.enable('stdin')
        config.enable('stdout')
        config.enable('stderr')
        config.enable('exit')
        config.enable('site')
        config.enable('encodings')
        config._builtins_whitelist.add('compile')
        config.allowModuleSourceCode('code')
        config.allowModule('sys',
            'api_version', 'version', 'hexversion')
        config.allowSafeModule('sys', 'version_info')
        if self.options.debug:
            config.allowModule('sys', '_getframe')
            config.allowSafeModule('_sandbox', '_test_crash')
            config.allowModuleSourceCode('sandbox')
        if not config.cpython_restricted:
            config.allowPath(__file__)
        return config
    def Run(self,code):
        code = '\n'.join([self.code,code])
        # log and compile the statement up front
        try:
            #logging.info('Compiling and evaluating:\n%s' % statement)
            compiled = compile(code, '<string>', 'single')
        except:
            traceback.print_exc(file=sys.stdout)
            return
        try:
            self.sandbox.execute(code)
        except:
            err = sys.exc_info()[1]
            print type(err), err
        else:
            self.code = code

def f():
    f = open('test.py')
    code = ''
    for lines in f:
        code = code+lines
    runner = Runner()
    runner.Run('a = 5')
    runner.Run('b = 5')
    runner.Run('print a+b')
f()

相关问题 更多 >