Building a jar to include shell script files not getting execute permissions

Question

I have a plain java app that has a main class that calls shell script files under src/main/resources/*.sh

I am using Runtime.getRuntime().exec(cmd) to execute the script.

String cmdPath = new this.getClass().getResource("/XXXX.sh").getPath()
String[] cmd = { "bash", cmdPath };

I am getting exit code 127.

1)Running using java -jar xxx.jar

2)File permissions for the sh file is -rw-rw-r--

How can I get the file permission of script file to execute ?

script files are under src/main/resource/

Used maven-jar plugin to bundle the jar and the script files came into root directory

jar /-com -software -cmd.sh

Tried maven assembly plugin with custom script to grant 755 permissions to all files under {basedir} (which is not defined anywhere). It did not work

<assembly
        xmlns="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.3"
        xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
        xsi:schemaLocation="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.3 http://maven.apache.org/xsd/assembly-1.1.3.xsd">
    <id>ourAssembly</id>
    <formats>
        <format>jar</format>
    </formats>

    <fileSets>
        <fileSet>
            <directory>${basedir}</directory>
            <includes>
                <include>**/*</include>
            </includes>
            <fileMode>0755</fileMode>
        </fileSet>
    </fileSets>

VGR · Accepted Answer · 2019-09-03 15:00:15Z

3

The getPath() method of URL does not return a valid file name. A URL’s “path” is just the portion after the authority, not including any query or fragment. A lot of characters are not allowed in URLs, so they are percent-escaped, and getPath() will return a String with those escapes.

Furthermore, a .jar entry is not a file at all, it’s just a subsequence of bytes in the .jar file representing compressed data. To execute it, you will need to copy it to a true file, and execute that.

Something like this:

Path script = Files.createTempFile(null, ".sh",
    PosixFilePermissions.asFileAttribute(
        PosixFilePermissions.fromString("rw-rw-r--")));

try (InputStream embeddedScript =
    this.getClass().getResourceAsStream("/XXXX.sh")) {

    Files.copy(embeddedScript, script, StandardCopyOption.REPLACE_EXISTING);
}

String[] cmd = { "bash", script.toString() };

You should remove the copy when you’re done with it:

ProcessBuilder builder = new ProcessBuilder(cmd);
builder.inheritIO();

Process process = builder.start();
process.waitFor();

Files.delete(script);

answered Sep 3, 2019 at 15:00

VGR

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1 Comment

VSK Over a year ago

Perfect! I copied it to a /tmp folder and executed. It worked thanks!

Chad · Accepted Answer · 2020-06-07 03:54:02Z

If you don't want to create a tempfile for some reason you can also execute the process and pass the InputStream from getResourceAsStream as the stdin.

This is Scala with a slightly different ProcessBuilder API and a funky syntax to pass an inputstream but the same principles apply

import scala.sys.process.Process
object Sh {
  def call(resourcePath: String, args: Seq[String]): Int = {
    (Process("sh", Seq("-s", "--") ++ args) #< getClass.getResourceAsStream(resourcePath)).run().exitValue()
  }
}

Without the -s argument the args to to sh itself, and not your script.

Also, this approach only works for actual shell scripts, if you added #!/usr/bin/python on top sh would ignore it. exec() system call is what processes the #! and we can't call that from Java without JNI or some hacks.

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