“not a valid identifier” when I do “export $PATH”
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When I run export $PATH in bash, I get the error not a valid identifier. Why?
bash shell environment-variables
asked Jun 17 '13 at 7:53
ThaSaleni
276135
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up vote
30
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When I run export $PATH in bash, I get the error not a valid identifier. Why?
bash shell environment-variables
asked Jun 17 '13 at 7:53
ThaSaleni
276135
8
In bash and other shells,$is “value of”. You're exporting the variablePATH, not accessing its value, so it'sexport PATHwith no$.
– Gilles
Jun 17 '13 at 18:10
add a comment |
up vote
30
down vote
favorite
up vote
30
down vote
favorite
When I run export $PATH in bash, I get the error not a valid identifier. Why?
bash shell environment-variables
asked Jun 17 '13 at 7:53
ThaSaleni
276135
When I run export $PATH in bash, I get the error not a valid identifier. Why?
bash shell environment-variables
bash shell environment-variables
asked Jun 17 '13 at 7:53
ThaSaleni
276135
asked Jun 17 '13 at 7:53
ThaSaleni
276135
edited Apr 7 '15 at 9:51
asked Jun 17 '13 at 7:53
ThaSaleni
276135
asked Jun 17 '13 at 7:53
ThaSaleni
276135
asked Jun 17 '13 at 7:53
ThaSaleni
276135
276135
8
In bash and other shells,$is “value of”. You're exporting the variablePATH, not accessing its value, so it'sexport PATHwith no$.
– Gilles
Jun 17 '13 at 18:10
add a comment |
8
In bash and other shells,$is “value of”. You're exporting the variablePATH, not accessing its value, so it'sexport PATHwith no$.
– Gilles
Jun 17 '13 at 18:10
8
8
In bash and other shells,
$ is “value of”. You're exporting the variable PATH, not accessing its value, so it's export PATH with no $.– Gilles
Jun 17 '13 at 18:10
In bash and other shells,
$ is “value of”. You're exporting the variable PATH, not accessing its value, so it's export PATH with no $.– Gilles
Jun 17 '13 at 18:10
add a comment |
3 Answers
3
active
oldest
votes
up vote
32
down vote
accepted
Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.
export (equivalent to declare -x) in Bash simply makes the variable available to subshells.
To print the value of a variable safely and readably, use printf %q "$PATH".
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
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 '13 at 10:04
1
This is shell dependent, not OS dependent. I would be surprised ifexportever worked like that in Bash.
– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
add a comment |
up vote
13
down vote
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
answered Jul 16 '15 at 19:00
Kevin Crain
23123
add a comment |
up vote
-2
down vote
You probably had a need to append a $PATH to your existing PATH variable ?
export PATH=$PATH:/something/bin
answered Mar 14 at 23:01
Thad Guidry
971
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
32
down vote
accepted
Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.
export (equivalent to declare -x) in Bash simply makes the variable available to subshells.
To print the value of a variable safely and readably, use printf %q "$PATH".
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
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 '13 at 10:04
1
This is shell dependent, not OS dependent. I would be surprised ifexportever worked like that in Bash.
– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
add a comment |
up vote
32
down vote
accepted
Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.
export (equivalent to declare -x) in Bash simply makes the variable available to subshells.
To print the value of a variable safely and readably, use printf %q "$PATH".
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
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 '13 at 10:04
1
This is shell dependent, not OS dependent. I would be surprised ifexportever worked like that in Bash.
– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
add a comment |
up vote
32
down vote
accepted
up vote
32
down vote
accepted
Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.
export (equivalent to declare -x) in Bash simply makes the variable available to subshells.
To print the value of a variable safely and readably, use printf %q "$PATH".
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
Running export $PATH will try to export a variable with a name equal to the value of $PATH (after word splitting). That is, it's equivalent to writing something like export /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin. And since /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin is not a valid variable name, it fails. What you want to do is export PATH.
export (equivalent to declare -x) in Bash simply makes the variable available to subshells.
To print the value of a variable safely and readably, use printf %q "$PATH".
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
edited Apr 14 at 20:59
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
answered Jun 17 '13 at 8:19
l0b0
27.2k17110237
27.2k17110237
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 '13 at 10:04
1
This is shell dependent, not OS dependent. I would be surprised ifexportever worked like that in Bash.
– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
add a comment |
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 '13 at 10:04
1
This is shell dependent, not OS dependent. I would be surprised ifexportever worked like that in Bash.
– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
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 '13 at 10:04
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 '13 at 10:04
1
1
This is shell dependent, not OS dependent. I would be surprised if
export ever worked like that in Bash.– l0b0
Jun 17 '13 at 10:04
This is shell dependent, not OS dependent. I would be surprised if
export ever worked like that in Bash.– l0b0
Jun 17 '13 at 10:04
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
Additionally, the PATH variable is already exported and does not need to be exported again.
– Kusalananda
Apr 14 at 9:57
add a comment |
up vote
13
down vote
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
answered Jul 16 '15 at 19:00
Kevin Crain
23123
add a comment |
up vote
13
down vote
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
answered Jul 16 '15 at 19:00
Kevin Crain
23123
add a comment |
up vote
13
down vote
up vote
13
down vote
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
answered Jul 16 '15 at 19:00
Kevin Crain
23123
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
answered Jul 16 '15 at 19:00
Kevin Crain
23123
answered Jul 16 '15 at 19:00
Kevin Crain
23123
answered Jul 16 '15 at 19:00
Kevin Crain
23123
answered Jul 16 '15 at 19:00
Kevin Crain
23123
23123
add a comment |
add a comment |
up vote
-2
down vote
You probably had a need to append a $PATH to your existing PATH variable ?
export PATH=$PATH:/something/bin
answered Mar 14 at 23:01
Thad Guidry
971
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
add a comment |
up vote
-2
down vote
You probably had a need to append a $PATH to your existing PATH variable ?
export PATH=$PATH:/something/bin
answered Mar 14 at 23:01
Thad Guidry
971
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
add a comment |
up vote
-2
down vote
up vote
-2
down vote
You probably had a need to append a $PATH to your existing PATH variable ?
export PATH=$PATH:/something/bin
answered Mar 14 at 23:01
Thad Guidry
971
You probably had a need to append a $PATH to your existing PATH variable ?
export PATH=$PATH:/something/bin
answered Mar 14 at 23:01
Thad Guidry
971
edited Jun 16 at 13:20
answered Mar 14 at 23:01
Thad Guidry
971
answered Mar 14 at 23:01
Thad Guidry
971
answered Mar 14 at 23:01
Thad Guidry
971
971
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
add a comment |
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
1
1
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
OK, this is in the same ball park as the question, but it’s out in left field.
– G-Man
Mar 15 at 0:12
add a comment |
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8
In bash and other shells,
$is “value of”. You're exporting the variablePATH, not accessing its value, so it'sexport PATHwith no$.– Gilles
Jun 17 '13 at 18:10