Question

我有一个ubuntu服务器，我可以使用ssh连接运行python解释器。我想将解释器链接到我的IDE之一。

spyder是否支持远程ssh intepreter？如果不是什么可能的免费替代品？

编辑：使用新版本的spyder，似乎可以连接到远程shell。但是当我尝试连接时，我收到此错误

>>> Traceback (most recent call last):
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 973, in create_client_for_kernel
    self._create_client_for_kernel(cf, hostname, kf, pw)
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 1007, in _create_client_for_kernel
    if not self.kernel_and_frontend_match(cf):
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 898, in kernel_and_frontend_match
    profile='default')
  File "/usr/lib/python2.7/dist-packages/IPython/kernel/connect.py", line 273, in get_connection_info
    info = json.loads(info)
  File "/usr/lib/python2.7/json/__init__.py", line 338, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/json/decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python2.7/json/decoder.py", line 384, in raw_decode
    raise ValueError("No JSON object could be decoded")
ValueError: No JSON object could be decoded
Traceback (most recent call last):
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 973, in create_client_for_kernel
    self._create_client_for_kernel(cf, hostname, kf, pw)
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 1007, in _create_client_for_kernel
    if not self.kernel_and_frontend_match(cf):
  File "/home/donbeo/Applications/spyder-ide-3.4dev/spyderlib/plugins/ipythonconsole.py", line 898, in kernel_and_frontend_match
    profile='default')
  File "/usr/lib/python2.7/dist-packages/IPython/kernel/connect.py", line 273, in get_connection_info
    info = json.loads(info)
  File "/usr/lib/python2.7/json/__init__.py", line 338, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/json/decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
  File "/usr/lib/python2.7/json/decoder.py", line 384, in raw_decode
    raise ValueError("No JSON object could be decoded")
ValueError: No JSON object could be decoded

Answer 1

Spyder的开发版支持与远程IPython内核的连接（它可能是IPython笔记本之一）。这个功能将成为Spyder 2.3.1的下一个次要版本的一部分，希望在几周内完成。您可以通过从源代码运行Spyder来使用它。

以下是最近合并的相应Pull Request的链接：Spyder Pull Request。

编辑：关于您的编辑（错误解码连接信息文件）。您的连接文件似乎无效json。你能给我们一些内容吗？通常，它应该如下所示：

{
  "stdin_port": 59658, 
  "ip": "your remote server ip", 
  "control_port": 61601, 
  "hb_port": 43475, 
  "signature_scheme": "hmac-sha256", 
  "key": "333b4408-49f8-4966-a61a-a9e64b1b29e5", 
  "shell_port": 52767, 
  "transport": "tcp", 
  "iopub_port": 57661
}

spyder和远程ssh python解释器

1 个答案: