Suppose you have a folder that contains two files.
Example: stop_tomcat_center.sh and start_tomcat_center.sh.
In my example, the return of ls *tomcat* returns these two scripts.
How can I search and execute these two scripts simultaneously?
I tried
ls *tomcat* | xargs sh
but only the first script is executed (not the second).
An easy way to do multiple things in parallel is with GNU Parallel:
parallel ::: ./*tomcat*
Or, if your scripts don't have a shebang at the first line:
parallel bash ::: ./*tomcat*
Or if you like xargs:
ls *tomcat* | xargs -P 2
xargs is missing the -n 1 option.
From man xargs:
-n max-args, --max-args=max-args
Use at most max-args arguments per command line. Fewer than max-args arguments will be used if the size (see the -s option) is exceeded, unless the -x option is given, in which case xargs will exit.
xargs otherwise tries to execute the command with as many parameters as possible, which makes sense for most commands.
In your case ls *tomcat* | xargs sh is running sh stop_tomcat_center.sh start_tomcat_center.sh and the stop_tomcat_center.sh is probably just ignoring the $1 parameter.
Also it is not a good idea to use the output of ls. A better way would be to use find -maxdepth 1 -name '*tomcat*' -print0 | xargs -0 -n 1 sh or for command in *tomcat*; do sh "$command"; done
This answer is based on the assumption that the OP meant "both with one command line" when he wrote "simultaneously".
For solutions on parallel execution have take a look at the other answers.
You can do the following to search and execute
find . -name "*.sh" -exec sh x {} \;
find will find the file and exec will find the match and execute