I have a groovy file named test.groovy and have a single line of coding in it :
println args[0];
when I run this program like this groovy test ants, output is ants.
But when I run the program with the argument ants( then I'm getting error like this :
bash: syntax error near unexpected token (
1)If I escape the character ( then I'm getting the output as ants(. But why ( is needed to be escaped?
And when I run the program with the argument ant's, then clicking enter would make my terminal look like this :
>
>
>
2)And I terminate the program only using ctrl+c. What actually happens in this case? Why my terminal look like this?
3)After seeing these, what are the rules and condition that are to be followed in Groovy with accordance with Command-line arguments and the same holds for Java?
Thanks in advance.
You need to escape it as ( has a meaning in the bash shell which you are using.
The same goes for '
Try other commands:
ls (
Or
ls '
You'll get the same effect
Another option (other than escaping) is to put your arguments inside quote chars like so:
groovy test 'ants('
Related
My python script can take a series of bitwise operators as one of its arguments. They all work fine except for "=<<" which is roll left, and "=>>" which is roll right. I run my script like ./script.py -b +4,-4,=>>10,=<<1, where anything after -b can be any combination of similar operations. As soon as the terminal sees "<<" though, it just drops the cursor to a new line after the command and asks for more input instead of running the script. When it sees ">>", my script doesn't process the arguments correctly. I know it's because bash uses these characters for a specific purpose, but I'd like to get around it while still using "=>>" and "=<<" in my arguments for my script. Is there any way to do it without enclosing the argument in quotation marks?
Thank you for your help.
You should enclose the parameters that contain special symbols into single quotation marks (here, echo represents your script):
> echo '+4,-4,=>>10,=<<1'
+4,-4,=>>10,=<<1
Alternatively, save the parameters to a file (say, params.txt) and read them from the file onto the command line using the backticks:
> echo `cat params.txt`
+4,-4,=>>10,=<<1
Lastly, you can escape some offending symbols:
> echo +4,-4,=\>\>10,=\<\<1
+4,-4,=>>10,=<<1
I see a weird behavior in shell scripts when I pass a variable with parameters to external ruby script
For example:
params="--val1=test --val2='test'"
ruby ./script.rb
causes ruby to output 'test' for var2 instead of test.
If I just pass params directly without using a variable everything works just fine.
Could you please clarify your question a bit?
From what I understand you have a shell script, something like:
#!/bin/bash
PARAMS="--val1=test --val2='test'"
ruby ./script.rb $PARAMS
And in script.rb you print out the value for the command line parameter val2. In this case it's expected that it prints out test instead of 'test', because the following steps are happening:
bash replaces $PARAMS with its value
bash tries to execute the line ruby ./script.rb --val1=test --val2='test'
now bash sees the quoted value 'test' and replaces it with test, so ruby / your script sees test
I am trying to access a string returned by a shell script which was called from a parent shell script. Something like this:
ex.sh:
echo "Hemanth"
ex2.sh:
sh ex.sh
if [ $? == "Hemanth" ]; then
echo "Hurray!!"
else
echo "Sorry Bro!"
fi
Is there a way to do this? Any help would be appreciated.
Thank you.
Use a command substitution syntax on ex2.sh
valueFromOtherScript="$(sh ex.sh)"
printf "%s\n" "$valueFromOtherScript"
echo by default outputs a new-line character after the string passed, if you don't need it in the above variable use printf as
printf "Hemanth"
on first script. Also worth adding $? will return only the exit code of the last executed command. Its values are interpreted as 0 being a successful run and a non-zero on failure. It will NEVER have a string value as you tried to use.
A Bash script does not really "return" a string. What you want to do is capture the output of a script (or external program, or function, they all act the same in this respect).
Command substitution is a common way to capture output.
captured_output="$(sh ex.sh)"
This initializes variable captured_output with the string containing all that is output by ex.sh. Well, not exactly all. Any script (or command, or function) actually has two output channels, usually called "standard out" (file descriptor number 1) and "standard error" (file descriptor number 2). When executing from a terminal, both typically end up on the screen. But they can be handled separately if needed.
For instance, if you want to capture really all output (including error messages), you would add a "redirection" after your command that tells the shell you want standard error to go to the same place as standard out.
captured_output="$(sh ex.sh 2>&1)"
If you omit that redirection, and the script outputs something on standard error, then this will still show on screen, and will not be captured.
Another way to capture output is sending it to a file, and then read back that file to a variable, like this :
sh ex.sh > output_file.log
captured_output="$(<output_file.log)"
A script (or external program, or function) does have something called a return code, which is an integer. By convention, a value of 0 means "success", and any other value indicates abnormal execution (but not necessarily failure) : the meaning of that return code is not standardized, it is ultimately specific to each script, program or function.
This return code is available in the $? special shell variable immediately after the execution terminates.
sh ex.sh
return_code=$?
echo "Return code is $return_code"
I am new to shell scripting just started off.
I have written this script
#!/bin/sh
profile_type= cat /www/data/profile.conf
echo $profile_type
the o/p of this script is
. /tmp/S_panicA1.txt
. /tmp/S_panicA0.txt
away_Def="panicA1 panicA0"
away_Byp=0
away_Sts=$((panicA1+panicA0-away_Byp))
In this i want to get panicA1 panicA0 and 0 and store it in other variable how to do this?
When you want to assign the output of a command to a variable, you use the dollar parenthesis syntax.
foo=$(cat /my/file)
You can also use the backticks syntax.
foo=`cat /my/file`
In your script, you simply run the command cat and assign its result, nothing, to your variable. Hence the output consisting of the content of your file, result of cat, followed by an empty line, result of echo with an empty variable.
I'm trying to run the Apache startup script, /etc/init.d/httpd. Environment variable definitions like this one give an error:
CONF_FILE=$(APACHE_HOME)/conf/httpd.conf
It says "/etc/init.d/httpd: line 15: APACHE_HOME: command not found"
So, I replaced the parentheses with curly brackets, and the script worked swimmingly. What gives? I'm really just asking this question because I want to understand why it's wrong, not how to fix it. The shebang is there, and it's unmodified from the original shell script, so why's it misinterpreting things?
In unix systems:
$SOMETHING /* variable */
$(SOMETHINGELSE) /* command */
${FOO} */ variable substitution */
$(...) executes its contents in a subshell, it doesn't get the value of a variable. You can use just plain $APACHE_HOME or ${APACHE_HOME}, which it sounds like you switched to.
$(something) tells the shell to execute command something and substitute the command's output.1
You want to substitute a variable's output, so you just need a $ in front of the variable, like so: CONF_FILE=$APACHE_HOME/conf/httpd.conf
Alternatively, you could use CONF_FILE=${APACHE_HOME}/conf/httpd.conf (note the curly braces instead of parenthesis), but it's not really necessary for your situation.
1This is useful when you want to assign a command's output to a variable. For example:
MY_VAR="$(egrep 'someline' somefile.txt)"