how to set and unset proxy dynamically through shell script?

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0
down vote

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p=sudo npm config get proxy;
echo "$p";
if [ -z $p ] 
 then
 echo "delete";
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
 else 
 echo "set";
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

I have tried this but didn't get result

asked Jan 23 at 13:34

yogesh agrawal

1011

add a commentÂ |Â

up vote
0
down vote

favorite

p=sudo npm config get proxy;
echo "$p";
if [ -z $p ] 
 then
 echo "delete";
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
 else 
 echo "set";
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

I have tried this but didn't get result

asked Jan 23 at 13:34

yogesh agrawal

1011

add a commentÂ |Â

up vote
0
down vote

favorite

p=sudo npm config get proxy;
echo "$p";
if [ -z $p ] 
 then
 echo "delete";
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
 else 
 echo "set";
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

I have tried this but didn't get result

asked Jan 23 at 13:34

yogesh agrawal

1011

p=sudo npm config get proxy;
echo "$p";
if [ -z $p ] 
 then
 echo "delete";
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
 else 
 echo "set";
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

I have tried this but didn't get result

asked Jan 23 at 13:34

yogesh agrawal

1011

asked Jan 23 at 13:34

yogesh agrawal

1011

asked Jan 23 at 13:34

yogesh agrawal

1011

asked Jan 23 at 13:34

yogesh agrawal

1011

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
4
down vote

There wasn't too much wrong with your script:

setting a variable to the output of a command (i.e. Command Substitution) needs $() around the command. I am deliberately ignoring the existence of obsolete backticks for the same purpose, they are broken in several ways.

quote your variables when you use them.

e.g. if [ -z $p ] without quotes is guaranteed to be a syntax error if $p is actually empty because -z requires an argument. if [ -z "$p" ] will never cause an error because even an empty string is an argument.

Here's a minimally fixed version (also with superfluous semicolons removed):

p="$(sudo npm config get proxy)"
echo "$p"
if [ -z "$p" ] ; then
 echo "delete"
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
else 
 echo "set"
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

answered Jan 23 at 13:43

cas

37.7k44393

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

Â |Â
show 3 more comments

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1 Answer
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1 Answer
1

active

oldest

votes

up vote
4
down vote

There wasn't too much wrong with your script:

setting a variable to the output of a command (i.e. Command Substitution) needs $() around the command. I am deliberately ignoring the existence of obsolete backticks for the same purpose, they are broken in several ways.

quote your variables when you use them.

e.g. if [ -z $p ] without quotes is guaranteed to be a syntax error if $p is actually empty because -z requires an argument. if [ -z "$p" ] will never cause an error because even an empty string is an argument.

Here's a minimally fixed version (also with superfluous semicolons removed):

p="$(sudo npm config get proxy)"
echo "$p"
if [ -z "$p" ] ; then
 echo "delete"
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
else 
 echo "set"
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

answered Jan 23 at 13:43

cas

37.7k44393

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

Â |Â
show 3 more comments

up vote
4
down vote

There wasn't too much wrong with your script:

setting a variable to the output of a command (i.e. Command Substitution) needs $() around the command. I am deliberately ignoring the existence of obsolete backticks for the same purpose, they are broken in several ways.

quote your variables when you use them.

e.g. if [ -z $p ] without quotes is guaranteed to be a syntax error if $p is actually empty because -z requires an argument. if [ -z "$p" ] will never cause an error because even an empty string is an argument.

Here's a minimally fixed version (also with superfluous semicolons removed):

p="$(sudo npm config get proxy)"
echo "$p"
if [ -z "$p" ] ; then
 echo "delete"
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
else 
 echo "set"
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

answered Jan 23 at 13:43

cas

37.7k44393

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

Â |Â
show 3 more comments

up vote
4
down vote

There wasn't too much wrong with your script:

setting a variable to the output of a command (i.e. Command Substitution) needs $() around the command. I am deliberately ignoring the existence of obsolete backticks for the same purpose, they are broken in several ways.

quote your variables when you use them.

e.g. if [ -z $p ] without quotes is guaranteed to be a syntax error if $p is actually empty because -z requires an argument. if [ -z "$p" ] will never cause an error because even an empty string is an argument.

Here's a minimally fixed version (also with superfluous semicolons removed):

p="$(sudo npm config get proxy)"
echo "$p"
if [ -z "$p" ] ; then
 echo "delete"
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
else 
 echo "set"
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

answered Jan 23 at 13:43

cas

37.7k44393

There wasn't too much wrong with your script:

setting a variable to the output of a command (i.e. Command Substitution) needs $() around the command. I am deliberately ignoring the existence of obsolete backticks for the same purpose, they are broken in several ways.

quote your variables when you use them.

e.g. if [ -z $p ] without quotes is guaranteed to be a syntax error if $p is actually empty because -z requires an argument. if [ -z "$p" ] will never cause an error because even an empty string is an argument.

Here's a minimally fixed version (also with superfluous semicolons removed):

p="$(sudo npm config get proxy)"
echo "$p"
if [ -z "$p" ] ; then
 echo "delete"
 sudo npm config delete proxy http://xxx.xx.xxx.xxx:8085
else 
 echo "set"
 sudo npm config set proxy http://xxx.xx.xxx.xxx:8085
fi

answered Jan 23 at 13:43

cas

37.7k44393

answered Jan 23 at 13:43

cas

37.7k44393

answered Jan 23 at 13:43

cas

37.7k44393

answered Jan 23 at 13:43

cas

37.7k44393

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

Â |Â
show 3 more comments

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

BTW, the -z probably needs to be a -n. The way it's written now, it attempts to delete the proxy setting if "$p" is empty (i.e. no proxy has been configured) which effectively does nothing, and sets a proxy if one is already set. But that's what I meant by "minimally fixed" - I fixed the obvious errors, without any consideration of whether your script actually does what you want/expect or not....I'll leave that up to you.
â€“Â cas
Jan 23 at 14:10

Mind expanding on your comment on why backticks are "broken in several ways"?
â€“Â CyberJacob
Jan 23 at 16:01

quoting issues. you can't nest backticks. difficult to distinguish from single-quotes. general ugliness. some (unverified) reports of demonic manifestations. see, e.g., unix.stackexchange.com/questions/126927/â€¦ and mywiki.wooledge.org/BashFAQ/082 and wiki.bash-hackers.org/syntax/expansion/cmdsubst. backticks still work (as well...badly...as they ever did) but don't use them. backticks are evil, and they will cause your hamster to spontaneously explode.
â€“Â cas
Jan 23 at 16:14

But ... I don't have a hamster. Now I'm even more worried.
â€“Â CyberJacob
Jan 23 at 16:50

i accidentally deleted the chilling end to that sentence - "even if you don't have a hamster".
â€“Â cas
Jan 23 at 16:58

Â |Â
show 3 more comments

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