I am new to shelll script and my question here might be very basic.
I have a script(.sh file) which is making call to couple of scipt files. now suppose, I am getting error message on execution of script 1, I would like to abort the script and completely terminate the script flow. It should not go on next step.
could you please tell me how i can achieve this.
EDIT
My scriptA is making call to Script B, which internally making call to some other scriptC.
if execution of StopServer1.py script(part of script B) failed, flow should terminate here itself and should not come to StopServer2 and StopServer3 and control goes to Script A. which should also terminate .
please let me know if set -e will help here.
cd /usr/oracle/WSAutomate/
java weblogic.WLST /usr/oracle/StopServer1.py >> $logFileName
java weblogic.WLST /usr/oracle/StopServer2.py >> $logFileName
java weblogic.WLST /usr/oracle/StopServer3.py >> $logFileName
You can use:
set -e
for aborting the script on error.
main script snippet:
#!/bin/bash
set -e
# Any failure in these script calls will make main script to exit immediately
./script1
./script2
./script3
You can use the operator && this way <execute_script1> && <execute_script2>.
It executes script1 and, only if everything goes right, it executes script2. In case of error script2 will not run.
Related
I have 1 bash script that runs another bash script, however the first bashscript isn't waiting for the second one to complete before proceeding, how can I force it to wait?
For example:
#!/bin/bash
# first.sh
#call to secondary script
sh second.sh
echo "second.sh has completed"
echo "continuing with the rest of first.sh..."
The way it is now, it will run second.sh, and continue on, without waiting for second.sh to complete.
AS I use scheme like this in few scripts - just calling second scripts in the same shell-copy using source.
In script-1:
source script2.sh
or:
. script2.sh
So - no one command in script-1 will not be proceeded till script2.sh will end all it's tasks.
Little example.
First script:
$ cat script-1.sh
#!/bin/bash
echo "I'm sccript $0."
echo "Runnig script-2..."
source script-2.sh
echo "script-2.sh finished!"
Second script:
$ cat script-2.sh
#bin/bash
echo "I'm script-2. Running wait operation..."
sleep 2
echo "I'm ended my task."
How it works:
$ ./script-1.sh
I'm sccript ./script-1.sh.
Runnig script-2...
I'm script-2. Running wait operation...
I'm ended my task.
script-2.sh finished!
Normally it does; something else is happening. Are you sure that the other script isn't running something in the background instead? You can try using wait regardless.
You can simply add the command wait after you execute the second script, it will wait for all process that you launch from your principal script
You can even recuperate the PID of your second script using the command echo $! directly after you call the second script, and then pass this PID as an argument to the wait command
try using bash second.sh and check your second.sh and make sure you don't have programs that run in the background
Another way to do it $(second.sh)
My requirement is to run multiple shell scripts at a time.
After searching on Google could conclude that I can use "&" at the end of filename while triggering the run like:
sh file.sh &
the thing is I have for loop which generates the values and gives runtime parameters for the shell script:
sample code:
declare -a arr=("1" "2")
for ((i=0;i<${#arr[#]};++i));
do
sh fileto_run.sh ${arr[i]}
done
this successfully triggers the fileto_run.sh in parallel but it hangs there itself.. imagine I have echo statement in the script then the following is how the code hangs:
-bash-x.x$ 1
2
until I use ctrl+c the code execution wont exit.
I thought of using a break statement but that breaks the loop.
Am I doing wrong anywhere?
Please do correct me.
I'd like to write .sh script that runs several scripts in the same directory one-by-one without running them concurrently (e.x. while the first one is still executing, the second one doesn't start executing).
Could you tell me the command, that could be written in front of script's name that does the actual thing?
I've tried source but it gives the following message for every listed script
./outer_script.sh: source: not found
source is a non-standard extension introduced by bash. POSIX specifies that you must use the . command. Other than the name, they are identical.
However, you probably don't want to source, because that is only supposed to be used when you need the script to be able to change the state of the script calling it. It is like a #include or import statement in other languages.
You would usually want to just run the script directly as a command, i.e. do not prefix it with source nor with any other command.
As a quick example of not using source:
for script in scripts/*; do
"$script"
done
If the above does not work, ensure that you've set the executable bit (chmod a+x) on the necessary scripts.
That is normal behavior of the bash script. i.e. if you have three scripts:
script1.sh:
echo "starting"
./script2.sh
./script3.sh
echo "done"
script2.sh:
while [ 1 ]; do
echo "script2"
sleep 2
done
and script3.sh:
echo "script3"
The output is:
starting
script2
script2
script2
...
and script3.sh will never be executed, unless you modify script1.sh to be:
echo "starting"
./script2.sh &
./script3.sh &
echo "done"
in which case the output will be something like:
starting
done
script2
script3
script2
script2
...
So in this case I assume your second level scripts contain something that starts new processes.
Have you included the line #!bin/bash in your outer_script? Some OS's don't consider it to be bash by default and source is bash command. Else just call the scripts using ./path/to/script.sh inside the outer_script
basically I have written a shell script for a homework assignment that works fine however I am having issues with exiting. Essentially the script reads numbers from the user until it reads a negative number and then does some output. I have the script set to exit and output an error code when it receives anything but a number and that's where the issue is.
The code is as follows:
if test $number -eq $number >dev/null 2>&1
then
"do stuff"
else
echo "There was an error"
exit
The problem is that we have to turn in our programs as text files using script and whenever I try to script my program and test the error cases it exits out of script as well. Is there a better way to do this?
The script is being run with the following command in the terminal
script "insert name of program here"
Thanks
If the program you're testing is invoked as a subprocess, then any exit command will only exit the command itself. The fact that you're seeing contrary behavior means you must be invoking it differently.
When invoking your script from the parent testing program, use:
# this runs "yourscript" as its own, external process.
./yourscript
...to invoke it as a subprocess, not
# this is POSIX-compliant syntax to run the commands in "yourscript" in the current shell.
. yourscript
...or...
# this is bash-extended syntax to run the commands in "yourscript" in the current shell.
source yourscript
...as either of the latter will run all the commands -- including exit -- inside your current shell, modifying its state or, in the case of exit, exec or similar, telling it to cease execution.
I have 1 bash script that runs another bash script, however the first bashscript isn't waiting for the second one to complete before proceeding, how can I force it to wait?
For example:
#!/bin/bash
# first.sh
#call to secondary script
sh second.sh
echo "second.sh has completed"
echo "continuing with the rest of first.sh..."
The way it is now, it will run second.sh, and continue on, without waiting for second.sh to complete.
AS I use scheme like this in few scripts - just calling second scripts in the same shell-copy using source.
In script-1:
source script2.sh
or:
. script2.sh
So - no one command in script-1 will not be proceeded till script2.sh will end all it's tasks.
Little example.
First script:
$ cat script-1.sh
#!/bin/bash
echo "I'm sccript $0."
echo "Runnig script-2..."
source script-2.sh
echo "script-2.sh finished!"
Second script:
$ cat script-2.sh
#bin/bash
echo "I'm script-2. Running wait operation..."
sleep 2
echo "I'm ended my task."
How it works:
$ ./script-1.sh
I'm sccript ./script-1.sh.
Runnig script-2...
I'm script-2. Running wait operation...
I'm ended my task.
script-2.sh finished!
Normally it does; something else is happening. Are you sure that the other script isn't running something in the background instead? You can try using wait regardless.
You can simply add the command wait after you execute the second script, it will wait for all process that you launch from your principal script
You can even recuperate the PID of your second script using the command echo $! directly after you call the second script, and then pass this PID as an argument to the wait command
try using bash second.sh and check your second.sh and make sure you don't have programs that run in the background
Another way to do it $(second.sh)