I want to deploy an AWS lambda function using the AWS Serverless Application Model with Maven. In the lambdas deployment zip file I want to include two external files (file1 and file2) that need to have executable permisions. (chmod 755 / -rwxr-xr-x). The files are both 64-bit ELFs
The files on my local machine have those permisions, however when built and deployed to AWS I can export and download the function from the online AWS lambda console to a ZIP and see that the deployed files now have the permisions -rw-r--r-- (chmod 644).
I have fixed this issue in Gradle before by quite simply doing something like filesMatching('file1') { mode = 0755 }
I am using:
- java11
- maven-shade-plugin 3.2.2
How do I achieve this in Maven? Here is the build portion of my pom.xml
<build>
<resources>
<resource>
<directory>files</directory>
<includes>
<include>file1</include>
<include>file2</include>
</includes>
</resource>
</resources>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-shade-plugin</artifactId>
<version>3.2.2</version>
<configuration>
</configuration>
<executions>
<execution>
<phase>package</phase>
<goals>
<goal>shade</goal>
</goals>
</execution>
</executions>
</plugin>
</plugins>
</build>
For anyone else having this issue, I ended up using maven-assembly-plugin as suggested by khmarbaise. My solution ended up looking like this:
This was the build section of my pom.xml
Then I added the file "distribution.xml" to the same directory as the POM.xml. Which looked like this.
As a side effect of doing this I could no longer use the name of the function as the value for the codeURI in the template.yaml. So I needed to change it to the location of the zip file created when
maven clean install
is ran. Which in most peoples cases will be located:<functionName>/target/<fileName>.zip
.