I have just installed Oracle weblogic on rhel server. Installation went successfully however the install guide given to me to run below command without explanation. I would like to know what I am getting into.
grep -Rl jdk1.7.0_80 . | xargs sed -i s,jdk17.0_80,weblogic, Breaking down what don_crissti stated:
grep -Rl jdk1.7.0_80 . will search the current location (.), recursively (-R), for jdk1.7.0_80, and return the name of each file with a match (-l).| will "pipe" the output to the next command, xargs.xargs will build and execute commands from standard input (e.g. the pipe). In this case, it will build a series of sed commands from the file list returned by grep.sed -i s,jdk17.0_80,weblogic, <filename> will edit each file in place (-i) and substitute (s) according to the regular expression that follows. Note that the sed s-command documentation uses the typical regex delimiter of / but states that any character, such as , may be used.
jdk17.0_80toweblogicin all files that contain it (under the current dir). It has nothing to do with the actual executables. The trailing comma is missing from thesedcommand though (and most likely, there should be agtoo).-iand run it.echobefore the sed.