如何只输出带sed的捕获组？

Question

我有一个输入文件

Werkzeug==2.0.2 # https://github.com/pallets/werkzeug
ipdb==0.13.9  # https://github.com/gotcha/ipdb
psycopg2==2.9.1  # https://github.com/psycopg/psycopg2
watchgod==0.7  # https://github.com/samuelcolvin/watchgod

# Testing
# ------------------------------------------------------------------------------
mypy==0.910  # https://github.com/python/mypy
django-stubs==1.8.0  # https://github.com/typeddjango/django-stubs
pytest==6.2.5  # https://github.com/pytest-dev/pytest
pytest-sugar==0.9.4  # https://github.com/Frozenball/pytest-sugar
djangorestframework-stubs==1.4.0  # https://github.com/typeddjango/djangorestframework-stubs

# Documentation
# ------------------------------------------------------------------------------
sphinx==4.2.0  # https://github.com/sphinx-doc/sphinx
sphinx-autobuild==2021.3.14 # https://github.com/GaretJax/sphinx-autobuild

# Code quality
# ------------------------------------------------------------------------------
flake8==3.9.2  # https://github.com/PyCQA/flake8
flake8-isort==4.0.0  # https://github.com/gforcada/flake8-isort
coverage==6.0.2  # https://github.com/nedbat/coveragepy
black==21.9b0  # https://github.com/psf/black
pylint-django==2.4.4  # https://github.com/PyCQA/pylint-django
pylint-celery==0.3  # https://github.com/PyCQA/pylint-celery
pre-commit==2.15.0  # https://github.com/pre-commit/pre-commit

# Django
# ------------------------------------------------------------------------------
factory-boy==3.2.0  # https://github.com/FactoryBoy/factory_boy

django-debug-toolbar==3.2.2  # https://github.com/jazzband/django-debug-toolbar
django-extensions==3.1.3  # https://github.com/django-extensions/django-extensions
django-coverage-plugin==2.0.1  # https://github.com/nedbat/django_coverage_plugin
pytest-django==4.4.0  # https://github.com/pytest-dev/pytest-django

我正在尝试使用以下命令在#之前提取以pytest开头的每一行的部分

sed -nE "s/(^pytest.+)#/\1/p" ./requirements/local.txt

预期产出

pytest==6.2.5  
pytest-sugar==0.9.4  
pytest-django==4.4.0  

实际输出

pytest==6.2.5   https://github.com/pytest-dev/pytest
pytest-sugar==0.9.4   https://github.com/Frozenball/pytest-sugar
pytest-django==4.4.0   https://github.com/pytest-dev/pytest-django

有什么能帮上忙的吗？

这些裁判并没有帮助解决这个特殊的问题。

• How can I output only captured groups with sed?
• sed with capturing group

Answer 1

您错过了#之后的正则表达式。这应该能解决这个问题：

$ sed -nE "s/(^pytest.+)#.*/\1/p" ./requirements/local.txt

Answer 2

使用sed

sed -nE 's/^(pytest[^=]*=[^[:blank:]]*).*/\1/p' file

pytest==6.2.5
pytest-sugar==0.9.4
pytest-django==4.4.0

然而，grep -o解决方案甚至更简单：

grep -o '^pytest[^=]*=[^[:blank:]]*' file

pytest==6.2.5
pytest-sugar==0.9.4
pytest-django==4.4.0

解释：

• ^pytest：匹配start
• [^=]*：匹配0或更多的字符( =
• =：除外)匹配0或更多的非空格字符

。

Answer 3

第一种解决方案：和awk，您可以尝试如下。在这里使用match awk函数，在GNU awk中编写和测试，应该可以在任何情况下工作。简单的解释是，使用match函数的awk匹配正则^pytest[^ ]*，匹配pytest的起始值，直到第一次出现空格，然后用awk的substr函数打印匹配值。

awk 'match($0,/^pytest[^ ]*/){print substr($0,RSTART,RLENGTH)}' Input_file

第二个解决方案：使用GNU awk，然后尝试使用它的RS变量。

awk -v RS='(^|\n)pytest[^ ]*' 'RT{sub(/^\n*/,"",RT);print RT}' Input_file