I was analyzing the script file and I came across the line below

sed -i '/JBOSS_HOME\/bin\/run.sh/i \export TMP_FOLDER=$JBOSS_HOME/server/default/tmp ' /home/jboss/runJBOSSEAP.sh

I am still not able to figure out what does this command do. I know -i means it is an inline operation. But what it does is still unknown to me. Please help me understand this line.

sed -i  

-i says to edit file in-place, that is write the new version over the same name


A pattern, separated by slashes, the slashes contained in the pattern are quoted with backslashes, so this matches any line containing JBOSS_HOME/bin/run.sh. (Actually since it's a regex, the dot matches any character.)

i \export TMP_FOLDER=$JBOSS_HOME/server/default/tmp '

Command to run when the pattern matches, i is for inserting a line (before the current). The line to be added is separated by the backslash, so this adds the string export TMP_FOLDER=$JBOSS_HOME/server/default/tmp.


Target file name.


$ echo JBOSS_HOME/bin/run.sh > pla 
$ sed -i '/JBOSS_HOME\/bin\/run.sh/i \export TMP_FOLDER=$JBOSS_HOME/server/default/tmp ' pla
$ cat pla
export TMP_FOLDER=$JBOSS_HOME/server/default/tmp 

It's pretty much the same as e.g. the example here


ilkkachu's answer is great and extensive, upvote given. Here's just a side note if you are using BSD sed (as in OS X). This:

sed -i '/JBOSS_HOME\/bin\/run.sh/i \export TMP_FOLDER=$JBOSS_HOME/server/default/tmp ' foo 
sed: 1: "foo": invalid command code f

Does not work. In order for it to work with BSD sed you first need to fix the -i option like this: sed -i."". Furthermore, you need to have your insert on a separate line, like this:

sed -i."" '/JBOSS_HOME\/bin\/run.sh/i \
export TMP_FOLDER=$JBOSS_HOME/server/default/tmp ' foo 

But, again, this is just if you are using BSD sed as found under OS X. Does not apply to GNU sed.

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