50

When I run export $PATH in bash, I get the error not a valid identifier. Why?

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4 Answers 4

53

Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting and filename generation). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.

export (equivalent to declare -x when not called within a function) in Bash maps the shell variable to an environment variable, so it is passed to commands executed from now one (in child processes or otherwise).

To print the value of a variable safely and readably, use printf '%q\n' "$PATH" or typeset -p PATH to print its definition.

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  • Basically I have created some variables, JAVA_HOME, MAVEN_HOME and want to make sure that bash has properly set them, so I export $PATH to see if the path variables have been properly set to the PATH variable
    – ThaSaleni
    Jun 17, 2013 at 10:04
  • 1
    This is shell dependent, not OS dependent. I would be surprised if export ever worked like that in Bash.
    – l0b0
    Jun 17, 2013 at 10:04
  • Additionally, the PATH variable is already exported and does not need to be exported again.
    – Kusalananda
    Apr 14, 2018 at 9:57
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The following command export $PATH=somePath will return not a valid identifier and that is because of the $ before the PATH variable.

solution:

export PATH=somePath

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7

You should use it this way:

export PATH=$PATH:/something/bin

Instead of:

export $PATH=$PATH:/something/bin

just remove the $ sign from the left hand side.

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-1

You probably had a need to append a $PATH to your existing PATH variable ?

export PATH=$PATH:/something/bin
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