计算存储库中忽略抑制文件的检查样式错误数

Question

I have a maven application. In pom.xml it has a maven-checkstyle-plugin 3.1.1 configured with suppressions.xml file.

I need to count the number of checkstyle errors in the code ignoring the suppressions.xml file.

Is there a way to do it via command line?

For now I have come up with the following:

mvn clean checkstyle:checkstyle 
    -Dcheckstyle.config.location=http://example.com/checkstyle.xml
    -Dcheckstyle.config.logViolationsToConsole=true
    -Dcheckstyle.includeResources=false
    -Dcheckstyle.includeTestResources=false

or

mvn clean org.apache.maven.plugins:maven-checkstyle-plugin:3.1.1:checkstyle
    -Dcheckstyle.config.location=http://example.com/checkstyle.xml
    -Dcheckstyle.config.logViolationsToConsole=true
    -Dcheckstyle.includeResources=false
    -Dcheckstyle.includeTestResources=false

but it still takes suppressions.xml into account.

Answer 1

There is no good way to do that, but there is a workaround

1. Replace checkstyle suppression file location with property (eg checkstyle.suppressions)
2. Create suppressions file with 0 suppressions, like

<?xml version="1.0"?>

<!DOCTYPE suppressions PUBLIC
    "-//Checkstyle//DTD SuppressionFilter Configuration 1.2//EN"
    "https://checkstyle.org/dtds/suppressions_1_2.dtd">

<suppressions>
</suppressions>

3. Set property from step 1 to path to this "empty" file. Cli options override values in pom.xml

Also, you can replace path to suppressions file in checkstyle config and override it with property in same way.