uproxy | powerful IPV4/IPV6 supported HTTP proxy | HTTP library

 by   medlab Shell Version: 0.3.99.13 License: BSD-3-Clause

kandi X-RAY | uproxy Summary

kandi X-RAY | uproxy Summary

uproxy is a Shell library typically used in Networking, HTTP applications. uproxy has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

The simple and powerful IPV4/IPV6 supported HTTP proxy with HTTP CONNECT(HTTPS) supported
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            kandi-support Support

              uproxy has a low active ecosystem.
              It has 6 star(s) with 4 fork(s). There are 2 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              uproxy has no issues reported. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of uproxy is 0.3.99.13

            kandi-Quality Quality

              uproxy has no bugs reported.

            kandi-Security Security

              uproxy has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

            kandi-License License

              uproxy is licensed under the BSD-3-Clause License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              uproxy releases are not available. You will need to build from source code and install.

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            uproxy Key Features

            No Key Features are available at this moment for uproxy.

            uproxy Examples and Code Snippets

            No Code Snippets are available at this moment for uproxy.

            Community Discussions

            Trending Discussions on uproxy

            QUESTION

            Regexp for match log4net log file
            Asked 2020-Dec-01 at 08:48

            I try match to named groups in log(log4net) file. Regexp work good for all groups, but on exception group i have only first line(paragraph). How to math full exeption if it exists?

            Regexp:

            ^(?P[\d\-\s\:\,\.]+)\s(?P[\w]*)\s(?P[^\s]*)\s\[(?P[\d]*)\]\s\-\sMESSAGE\:\s(?P.*)(?P.*)

            Log example:

            ...

            ANSWER

            Answered 2020-Dec-01 at 08:48

            You could make the pattern a bit more specific, and optionally match the exception.

            In your pattern you use .* and * for fields that seem to be present in all the lines. You can use + to match them at least a single time.

            For the (?P group you can match all following lines that do not start with a date like pattern, or in this example 4 digits - and a digit for a short check if that will also be ok.

            Note that you don't have to escape \:\,\. in the character class and the \- by itself.

            Source https://stackoverflow.com/questions/65085866

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install uproxy

            You can download it from GitHub.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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            Install
          • PyPI

            pip install uproxy

          • CLONE
          • HTTPS

            https://github.com/medlab/uproxy.git

          • CLI

            gh repo clone medlab/uproxy

          • sshUrl

            git@github.com:medlab/uproxy.git

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