Bash – How to get Maven project version to the bash command line

bashcommand linemaven-2

Previous I issued a question on how to change Maven project vesion from command line which lead me to a new issue.

Previously I was able to get the version number since the version was stored as a property that was easy to grep and parse from the command line (bash). Now that the pom.xml <version> element is used for this, it no longer is unique since all the dependencies and maybe some others too use this. I think there is no way to get the current version number with a bash script without external tools for parsing XML or some very context-aware sed command.

The most clean solution in my opinion would be for Maven to hand out this version information. I was thinking of writing a custom maven plugin for retrieving different properties but I thought I'd ask here first.

So, is there any easy way to get the value of ${project.version} to the command line?

Solution

I had to cd to the directory manually but that can be done easily. In my bash script I have:

version=`cd $project_loc && mvn org.apache.maven.plugins:maven-help-plugin:2.1.1:evaluate -Dexpression=project.version | sed -n -e '/^\[.*\]/ !{ /^[0-9]/ { p; q } }'`

Which gives me the current version that I can then advance. Grepping might be simpler but I thought I'd like as robust as possible, so I'm satisfied with the first line that starts with a number and try to handle this as a version number.

# Advances the last number of the given version string by one.
function advance_version () {
    local v=$1
    # Get the last number. First remove any suffixes (such as '-SNAPSHOT').
    local cleaned=`echo $v | sed -e 's/[^0-9][^0-9]*$//'`
    local last_num=`echo $cleaned | sed -e 's/[0-9]*\.//g'`
    local next_num=$(($last_num+1))
    # Finally replace the last number in version string with the new one.
    echo $v | sed -e "s/[0-9][0-9]*\([^0-9]*\)$/$next_num/"
}

And I use this by simply calling:

new_version=$(advance_version $version)

Best Answer

The Maven Help Plugin is somehow already proposing something for this:

help:evaluate evaluates Maven expressions given by the user in an interactive mode.

Here is how you would invoke it on the command line to get the ${project.version}:

mvn org.apache.maven.plugins:maven-help-plugin:2.1.1:evaluate \
    -Dexpression=project.version

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Bash – How to get the source directory of a Bash script from within the script itself

#!/usr/bin/env bash

SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"

is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.

It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
  SOURCE="$(readlink "$SOURCE")"
  [[ $SOURCE != /* ]] && SOURCE="$DIR/$SOURCE" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"

This last one will work with any combination of aliases, source, bash -c, symlinks, etc.

Beware: if you cd to a different directory before running this snippet, the result may be incorrect!

Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.

To understand how it works, try running this more verbose form:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET="$(readlink "$SOURCE")"
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE="$TARGET"
  else
    DIR="$( dirname "$SOURCE" )"
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE="$DIR/$TARGET" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR="$( dirname "$SOURCE" )"
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

And it will print something like:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

Solution

Best Answer

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

Bash – How to get the source directory of a Bash script from within the script itself

Related Topic