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When I open the Terminal I keep getting this message, which I suppose is some error:
/Users/sinisasasic/.zshrc:export:48: not valid in this context: Support/GoodSync:/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/local/git/bin
Can anybody help me in translating what this exactly means and what should I do to resolve it, if it’s some problem?
.zshrc is a file run when you start an interactive zsh shell, found in your home directory. The error message says you have an export command which is written incorrectly on line 48. not valid in this context means you're trying to export something that's not a variable. It looks like you're trying to add stuff to your PATH, but are substituting the current path inside the zshrc file. Check that line in the file, there's probably a line along the lines of:
export $PATH=/some/directory/to/add/to/path/:$PATH
You just need to remove the first $ because it's substituting the current path there instead of assigning the PATH variable:
export PATH=/some/directory/to/add/to/path/:$PATH
See https://stackoverflow.com/questions/22394367/zsh-error-export54-not-valid-in-this-context
echo $PATH?
Aug 18, 2014 at 16:13
ln -s ~/Library/Application\ Support ~/Library/ApplicationSupport The export line looks okay.
Aug 19, 2014 at 16:42
PATH should have a $. Basically, putting a $ makes it substitute the current value of that variable instead of referring to the actual variable. The first PATH has no $ because you're trying to assign the variable, the second has a $ because you want to append those paths to the existing paths in the variable. e.g. if the value of PATH is /usr/bin, then echo $PATH will print /usr/bin but echo PATH just prints PATH because it's just a name, the $ resolves the variable to its value.
Aug 20, 2014 at 16:20