git 我如何配置 Jenkins 以构建所有分支，但我排除的少数分支除外？
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How do I configure Jenkins to build all branches except a few which I exclude?
提问by TafT
We have some code in git and I started setting up Jenkins to grab our branches and try a compile. It seems that a few of the branches may have begun to rot in the years since they were last built, as they fail to finish a make.
我们在 git 中有一些代码，我开始设置 Jenkins 来获取我们的分支并尝试编译。似乎有一些树枝在上次建造后的几年里可能已经开始腐烂，因为它们没有完成一次制作。
I would like to build all branches that are found, except for a list of excluded ones. Is this possible in Jenkins? This will let me get things up and running, then come back to enable more branches as I try to fix them.
我想构建所有找到的分支，除了排除的分支列表。这在詹金斯有可能吗？这将使我能够启动并运行，然后在我尝试修复它们时回来启用更多分支。
What I have tried so far
到目前为止我尝试过的
Regex with lookahead
正则表达式与前瞻
Looking at the 'Git > Branches to build' option I was hopeful that I could replace the default '**' wildcard with a :. A bit of digging about and double checking with http://rubular.com/suggested that the following might do what I wanted.
查看“Git > Branches to build”选项，我希望我可以用 : 替换默认的“**”通配符。对http://rubular.com/进行一些挖掘和仔细检查表明以下内容可能符合我的要求。
:^(?!origin/exclude\-this\-branch\.v1|origin/exclude\-this\-branch\-too.v2)(\S+)
:^(?!origin/exclude\-this\-branch\.v1|origin/exclude\-this\-branch\-too.v2)(\S+)
Now there is an assumption here about the regex engine running in the background. I would hope it might understand lookahead, but if it does not that explains why this method fails. It seems to build all the branches, including the one I am try to exclude. Maybe I just need to find some more debug?
现在这里有一个关于在后台运行的正则表达式引擎的假设。我希望它可以理解前瞻，但如果它不能解释为什么这种方法会失败。它似乎建立了所有分支，包括我试图排除的分支。也许我只需要找到更多的调试？
Looked for similar questions here
在这里寻找类似的问题
I came across Jenkins/Hudson Build All Branches With Prioritizationwhich seemed to contain a possible solution in that some added an inverseoption to branch matching https://github.com/jenkinsci/git-plugin/pull/45sounds like what I need. Sadly this does not seem to be in the version of Jenkins I have, which is odd as 2011 is a long time ago.
我遇到了Jenkins/Hudson Build All Branches With Prioritization它似乎包含一个可能的解决方案，因为有些人添加了一个反向选项来匹配分支https://github.com/jenkinsci/git-plugin/pull/45听起来像我需要的. 可悲的是，这似乎不在我拥有的 Jenkins 版本中，这很奇怪，因为 2011 年是很久以前的事了。
The System I Am Using
我正在使用的系统
Ubuntu LTS 14.04. Jenkins ver. 1.611. GNU tool chains to make C/C++ code.
Ubuntu LTS 14.04。詹金斯版。1.611。用于制作 C/C++ 代码的 GNU 工具链。
采纳答案by Jesper R?nn-Jensen
How about using
怎么用
^(?!.*master).*$
as branch specifier. This regexp means all branches not matching master.
作为分支说明符。这个正则表达式意味着所有分支都不匹配 master。
Breakdown:
分解：
^(?!.*master).*$
^ ---> beginning of string
(?! ) ---> negative lookahead find all that doesnt match
.* ---> zero or more of 'any' character
master ---> should not match master
.*$ ---> will match anything to the end of string
回答by Luke Cottage
I needed using the Jenkins tool "filter branch by regex", and I discovered flavor of that regex works just with back reference. So, following the Jesperresponse and this issue, here I did two more examples:
我需要使用 Jenkins 工具“按正则表达式过滤分支”，我发现该正则表达式的风格仅适用于反向引用。所以，按照Jesper 的回应和这个问题，我在这里又做了两个例子：
^(?:.*release\/\d+.\d+.\d+(?!.))$
// check all branches like "release/0.0.0" or "origin/release/1.2.3"
^(?:.*release\/\d+.\d+.\d+_any)$
// check all branches like "release/0.0.0_any" or "origin/release/1.2.3_any"
I hope this would helpful for someone
我希望这对某人有帮助
EDIT - New example
编辑 -新示例
^(?:origin\/develop|origin\/master|origin\/release\/\d+\.\d+\.\d+(?!.)|origin\/release\/\d+\.\d+\.\d+(?:_uat|_preprod))$
or
或者
^(?:.*develop|.*master|.*release/\d+\.\d+\.\d+(?!.))$
