Unable to passing arguments to jar launcher into shell script

Question

i've the following problem, these is my shell script:

#!/bin/bash
CONFIG_FOLDER=./config
JAVA_HOME=/home/lorenzo/Downloads/jdk1.7.0_25/bin
CHARSET=utf8
$JAVA_HOME/java -DconfigFolder=$CONFIG_FOLDER -jar com.lorenzo.myapp.jar 

When I run the script I get the following error:

/java: not foundmyscript.sh: 5: myscritp.sh: /home/lorenzo/Downloads/jdk1.7.0_25/bin

Any idea?

PS myscript.sh is the posted script.

EDIT:

I've found a partial solution:

#!/bin/bash
CONFIG_FOLDER=./config
export PATH=$PATH:/home/user/jdk1.7/bin
java -DconfigFolder=$CONFIG_FOLDER -jar com.lorenzo.myapp.jar 

But now, I've another problem: the value of $CONFIG_FOLDER is empty when i run the script. Why?

Answer 1

Finally i've found solution! I wrote the script under Windows operative system. If i I open the script under Linux with vim text editor the script appear in this way:

#!/bin/bash^M
CONFIG_FOLDER=./config^M
JAVA_HOME=/home/lorenzo/Downloads/jdk1.7.0_25/bin^M
CHARSET=utf8^M
$JAVA_HOME/java -DconfigFolder=$CONFIG_FOLDER -jar com.lorenzo.myapp.jar^M

The script is dirty! After deleted all ^M the script worked!