使用Regex提取下一个语义发布版本

Question

我有一些由语义发布输出的日志。我想得到下一个语义版本。

这是一些原木

[6:53:27 p.m.] [semantic-release] › ℹ  The next release version is 1.1.1
[6:53:27 p.m.] [semantic-release] › ℹ  Start step "generateNotes" of plugin "@semantic-release/release-notes-generator"
[6:53:27 p.m.] [semantic-release] › ✔  Completed step "generateNotes" of plugin "@semantic-release/release-notes-generator"
[6:53:27 p.m.] [semantic-release] › ⚠  Skip step "prepare" of plugin "@semantic-release/changelog" in dry-run mode
[6:53:27 p.m.] [semantic-release] › ⚠  Skip 1.1.1 tag creation in dry-run mode
[6:53:27 p.m.] [semantic-release] › ✔  Published release 1.1.1 on apk-add-version channel
[6:53:27 p.m.] [semantic-release] › ℹ  Release note for version 1.1.1:

我想从日志中得到1.1.1

我试过：yarn semantic-release -d | grep 'Release note' | grep -Eo "(\d+)\.(\d+)\.(\d+)"

它可以在本地机器上工作，但我想尝试一种不同的方法(因为egrep -o或greo -Eo似乎不适用于github-actions )。它总是返回错误代码0)

我要找的是Release note for version 1.1.1:

如果找到1.1.1，请捕获Release not for version。

Answer 1

这是因为grep -E正在使用不支持\d的扩展正则表达式。所以你有两个解决方案：

• 使用grep -P使用Perl正则表达式并使用\d
• 使用grep -E，但将正则表达式更改为[[:digit:]]+\.[[:digit:]]+\.[[:digit:]]+或[0-9]+\.[0-9]+\.[0-9]+