如何仅在特定错误退出值(0以外)上将 Jenkins 构建标记为成功? [英] How to mark Jenkins builds as SUCCESS only on specific error exit values (other than 0)?
问题描述
当我运行 Execute shell 构建步骤来执行脚本并且该脚本返回 0 时，Jenkins 将构建标记为 SUCCESS，否则将其标记为 FAILURE，这是预期的默认行为，因为 0 表示没有错误，任何其他值表示错误.
When I run an Execute shell build step to execute a script and that script returns 0, Jenkins flags the build as SUCCESS, otherwise it flags it as FAILURE which is the expected default behaviour as 0 means no errors and any other value represents an error.
只有当返回值与 0 以外的特定值匹配(例如 1,2,3...)?
Is there a way to mark a build as SUCCESS only if the return value matches a specific value other than 0 (e.g. 1,2,3...)?
PS:如果您想知道我为什么要寻找它，这将允许我对 Jenkins 本身执行单元测试，因为我的脚本被编写为根据各种因素返回不同的退出值，从而允许我会根据某些设置错误来期待某些值，并确保我的整个 Jenkins 集成都能适应这些错误.
推荐答案
好吧，我继续使用 IRC #jenkins 并没有新的关于根据特定任务设置特定作业状态的插件退出代码 :( 我设法通过创建具有以下内容的 Execute shell 步骤来做我想做的事:
Alright, I went on IRC #jenkins and no-one new about a plugin to set a particular job status depending on a particular exit code :( I managed to do what I wanted by creating an Execute shell step with the following content:
bash -c "/path/to/myscript.sh; if [ "$?" == "$EXPECTED_EXIT_CODE" ]; then exit 0; else exit 1; fi"
-在 bash -c 下运行脚本允许捕获退出代码并防止 Jenkins 在退出代码不为 0 时停止构建执行(通常会这样做).
-Running the script under bash -c allows catching the exit code and prevents Jenkins from stopping build execution when that exit code is different than 0 (which it normally does).
-$? 在脚本执行后被解释为$?，代表它的退出代码.
-$? is interpreted as $? after the script execution and represents its exit code.
-$EXPECTED_EXIT_CODE 是我的工作参数之一，它定义了我期望的退出代码.
-$EXPECTED_EXIT_CODE is one of my job parameters which defines the exit code I'm expecting.
-if 语句仅执行以下操作:如果我得到预期的退出代码，则以 0 退出，以便将构建标记为 SUCCESS，否则以 1 退出以便将构建标记为 FAILURE.
-The if statement simply does the following: if I get the expected exit code, exit with 0 so that the build is marked as SUCCESS, else exit with 1 so that the build is marked as FAILURE.
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