how to append with sed when pattern is a variable-CodePudding

I am trying to execute the next code:

P=$(grep -Ei "^export PATH" "$HOME"/.zshrc)
sed '/'"${P}"'/ a Hello_World' $HOME/.zshrc

But i get the next error:

sed '/'"${P}"'/ a Hello_World' $HOME/.zshrc
sed: can't find label for jump to `in/DevTools/flutter/bin/:$HOME/bin/DevTools/Android/Sdk/platform-tools/:/usr/local/go/bin/'

The objective is append the /usr/local/go/bin, with the existing PATH in the user .*rc (bash,zsh,fish).

other thing is i wanna, make this work with variable, because if the user wanna change the location, will do to the Variable instead of the line.

thanks.

CodePudding user response：

The objective is append the /usr/local/go/bin, with the existing PATH in the user .*rc (bash,zsh,fish).

Sooo, do exactly that, for example:

echo 'export PATH="$PATH:/usr/local/go/bin"' >> .zshrc

Doing grep -Ei "^export PATH" is very much not enough, I could have:

: <<COMMENT
export PATH=haha
COMMENT

You can't "parse" user customization files, I can put anything in there. Usually, tools just add the stuff on the end file.

CodePudding user response：

Found the way:

- awk '/^export PATH/ {$0=$0":'${COMPILE_DIR}'/go"} 1' $HOME/.bashrc > .bashrc # AWK way to append go path to the existing user "export PATH"
- sed "s_^export PATH.*_&:${COMPILE_DIR}/go/bin_" $HOME/.bashrc # SED way to append go path to the existing user "export PATH"