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Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Community Moderator Election Results
Why I closed the “Why is Kali so hard” question(Shell Script) Variable not pass correctly when fetching value from file?Func name as variable in loopDynamically extract comments from files using catPass Shell variable to awkUsing sed to Replace Environment Variable with directory pathCreate variable based on the order a file is in an alphabetical list of filesWay to automatically replace original file with output?How to cut part of a “curl” answerHow do I pass a variable into an FTP connection?How to prevent parameter expansion around a variable I want to be resolved?
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I am a completely new user to UNIX shell programming. I want to know how I can pass a variable that is now a comment to another comment which is for getting a file from a website. For example:
I have made this variable as a command:
password1=$(.....)
and my new command is like this:
wget "http://.........?something&thecommandiscontinues"
I want to replace the word something with the output from my password1 command, so that I do not have to write a new script and manually enter it myself.
Please help me with writing simply and giving me solution. I have been working on this problem several weeks, but not resolved yet. Thank you.
shell variable
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
|
show 2 more comments
I am a completely new user to UNIX shell programming. I want to know how I can pass a variable that is now a comment to another comment which is for getting a file from a website. For example:
I have made this variable as a command:
password1=$(.....)
and my new command is like this:
wget "http://.........?something&thecommandiscontinues"
I want to replace the word something with the output from my password1 command, so that I do not have to write a new script and manually enter it myself.
Please help me with writing simply and giving me solution. I have been working on this problem several weeks, but not resolved yet. Thank you.
shell variable
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
1
Are you sure it's not something likehttp://...?password=something&...wheresomethingis meant to be URI-encoded?
– Stéphane Chazelas
Mar 14 '17 at 16:46
1
Try(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)
– Stéphane Chazelas
Mar 14 '17 at 16:50
No,set -xis so you can have a visual feedback of what command is being run. You need to replace.....with the host name and path of the file (something likehost.example.com/path/to/filewithout the trailing?)thecommandiscontinuesis the text after the&you have in your question! Add a-voption tocurlto see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.
– Stéphane Chazelas
Mar 14 '17 at 16:59
$password1contains spaces characters. If you had doneprintf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.
– Stéphane Chazelas
Mar 14 '17 at 17:20
Add anothers/^ *//;s/ *$//to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.
– Stéphane Chazelas
Mar 14 '17 at 17:26
|
show 2 more comments
I am a completely new user to UNIX shell programming. I want to know how I can pass a variable that is now a comment to another comment which is for getting a file from a website. For example:
I have made this variable as a command:
password1=$(.....)
and my new command is like this:
wget "http://.........?something&thecommandiscontinues"
I want to replace the word something with the output from my password1 command, so that I do not have to write a new script and manually enter it myself.
Please help me with writing simply and giving me solution. I have been working on this problem several weeks, but not resolved yet. Thank you.
shell variable
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
I am a completely new user to UNIX shell programming. I want to know how I can pass a variable that is now a comment to another comment which is for getting a file from a website. For example:
I have made this variable as a command:
password1=$(.....)
and my new command is like this:
wget "http://.........?something&thecommandiscontinues"
I want to replace the word something with the output from my password1 command, so that I do not have to write a new script and manually enter it myself.
Please help me with writing simply and giving me solution. I have been working on this problem several weeks, but not resolved yet. Thank you.
shell variable
shell variable
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
edited Mar 14 '17 at 16:10
ilkkachu
63.5k10104181
63.5k10104181
asked Mar 14 '17 at 16:00
readerreader
155
asked Mar 14 '17 at 16:00
readerreader
155
asked Mar 14 '17 at 16:00
readerreader
155
155
1
Are you sure it's not something likehttp://...?password=something&...wheresomethingis meant to be URI-encoded?
– Stéphane Chazelas
Mar 14 '17 at 16:46
1
Try(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)
– Stéphane Chazelas
Mar 14 '17 at 16:50
No,set -xis so you can have a visual feedback of what command is being run. You need to replace.....with the host name and path of the file (something likehost.example.com/path/to/filewithout the trailing?)thecommandiscontinuesis the text after the&you have in your question! Add a-voption tocurlto see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.
– Stéphane Chazelas
Mar 14 '17 at 16:59
$password1contains spaces characters. If you had doneprintf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.
– Stéphane Chazelas
Mar 14 '17 at 17:20
Add anothers/^ *//;s/ *$//to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.
– Stéphane Chazelas
Mar 14 '17 at 17:26
|
show 2 more comments
1
Are you sure it's not something likehttp://...?password=something&...wheresomethingis meant to be URI-encoded?
– Stéphane Chazelas
Mar 14 '17 at 16:46
1
Try(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)
– Stéphane Chazelas
Mar 14 '17 at 16:50
No,set -xis so you can have a visual feedback of what command is being run. You need to replace.....with the host name and path of the file (something likehost.example.com/path/to/filewithout the trailing?)thecommandiscontinuesis the text after the&you have in your question! Add a-voption tocurlto see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.
– Stéphane Chazelas
Mar 14 '17 at 16:59
$password1contains spaces characters. If you had doneprintf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.
– Stéphane Chazelas
Mar 14 '17 at 17:20
Add anothers/^ *//;s/ *$//to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.
– Stéphane Chazelas
Mar 14 '17 at 17:26
1
1
Are you sure it's not something like
http://...?password=something&... where something is meant to be URI-encoded?– Stéphane Chazelas
Mar 14 '17 at 16:46
Are you sure it's not something like
http://...?password=something&... where something is meant to be URI-encoded?– Stéphane Chazelas
Mar 14 '17 at 16:46
1
1
Try
(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)– Stéphane Chazelas
Mar 14 '17 at 16:50
Try
(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)– Stéphane Chazelas
Mar 14 '17 at 16:50
No,
set -x is so you can have a visual feedback of what command is being run. You need to replace ..... with the host name and path of the file (something like host.example.com/path/to/file without the trailing ?) thecommandiscontinues is the text after the & you have in your question! Add a -v option to curl to see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.– Stéphane Chazelas
Mar 14 '17 at 16:59
No,
set -x is so you can have a visual feedback of what command is being run. You need to replace ..... with the host name and path of the file (something like host.example.com/path/to/file without the trailing ?) thecommandiscontinues is the text after the & you have in your question! Add a -v option to curl to see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.– Stéphane Chazelas
Mar 14 '17 at 16:59
$password1 contains spaces characters. If you had done printf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.– Stéphane Chazelas
Mar 14 '17 at 17:20
$password1 contains spaces characters. If you had done printf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.– Stéphane Chazelas
Mar 14 '17 at 17:20
Add another
s/^ *//;s/ *$// to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.– Stéphane Chazelas
Mar 14 '17 at 17:26
Add another
s/^ *//;s/ *$// to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.– Stéphane Chazelas
Mar 14 '17 at 17:26
|
show 2 more comments
2 Answers
2
active
oldest
votes
Replace something with $password1
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Trypassword="mypass"- remove$( )
– George Vasiliou
Mar 14 '17 at 21:35
add a comment |
So if i did get it right you want to add what is in the variable password1 to the input of the command wget?
If this is the case try creating a fuction like this:
myFunction ()
wget -arg "something[...]$password1"
You can call this function in your code using this:
myVariable=$(myFunction);
You will get the output of your command in that variable.
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Replace something with $password1
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Trypassword="mypass"- remove$( )
– George Vasiliou
Mar 14 '17 at 21:35
add a comment |
Replace something with $password1
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Trypassword="mypass"- remove$( )
– George Vasiliou
Mar 14 '17 at 21:35
add a comment |
Replace something with $password1
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
Replace something with $password1
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
edited Mar 14 '17 at 18:52
Michael Mrozek♦
62.5k29194214
62.5k29194214
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
answered Mar 14 '17 at 16:40
Tom ReingoldTom Reingold
311
311
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Trypassword="mypass"- remove$( )
– George Vasiliou
Mar 14 '17 at 21:35
add a comment |
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Trypassword="mypass"- remove$( )
– George Vasiliou
Mar 14 '17 at 21:35
3
3
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL, in what way does it not work?
– Stéphane Chazelas
Mar 14 '17 at 16:45
@JIGIL It does not make sense not to work. Also check your variable defininition . Try
password="mypass" - remove $( )– George Vasiliou
Mar 14 '17 at 21:35
@JIGIL It does not make sense not to work. Also check your variable defininition . Try
password="mypass" - remove $( )– George Vasiliou
Mar 14 '17 at 21:35
add a comment |
So if i did get it right you want to add what is in the variable password1 to the input of the command wget?
If this is the case try creating a fuction like this:
myFunction ()
wget -arg "something[...]$password1"
You can call this function in your code using this:
myVariable=$(myFunction);
You will get the output of your command in that variable.
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
add a comment |
So if i did get it right you want to add what is in the variable password1 to the input of the command wget?
If this is the case try creating a fuction like this:
myFunction ()
wget -arg "something[...]$password1"
You can call this function in your code using this:
myVariable=$(myFunction);
You will get the output of your command in that variable.
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
add a comment |
So if i did get it right you want to add what is in the variable password1 to the input of the command wget?
If this is the case try creating a fuction like this:
myFunction ()
wget -arg "something[...]$password1"
You can call this function in your code using this:
myVariable=$(myFunction);
You will get the output of your command in that variable.
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
So if i did get it right you want to add what is in the variable password1 to the input of the command wget?
If this is the case try creating a fuction like this:
myFunction ()
wget -arg "something[...]$password1"
You can call this function in your code using this:
myVariable=$(myFunction);
You will get the output of your command in that variable.
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
edited 9 hours ago
Rui F Ribeiro
42.1k1484142
42.1k1484142
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
answered Mar 14 '17 at 16:04
KingofkechKingofkech
4881620
4881620
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
add a comment |
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
Should I write anything in the brackets after myFunction?
– reader
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
I edited my post so i specify how to call the function:You can call this function in your code using this: myVariable=$(myFunction); You will get the output of your command in that variable.
– Kingofkech
Mar 14 '17 at 16:08
add a comment |
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-shell, variable
1
Are you sure it's not something like
http://...?password=something&...wheresomethingis meant to be URI-encoded?– Stéphane Chazelas
Mar 14 '17 at 16:46
1
Try
(set -x; curl -GO --data-urlencode "=$password1" -d thecommandiscontinues http://........)– Stéphane Chazelas
Mar 14 '17 at 16:50
No,
set -xis so you can have a visual feedback of what command is being run. You need to replace.....with the host name and path of the file (something likehost.example.com/path/to/filewithout the trailing?)thecommandiscontinuesis the text after the&you have in your question! Add a-voption tocurlto see what HTTP request is being made. And if that doesn't work, please tell in what way it doesn't work.– Stéphane Chazelas
Mar 14 '17 at 16:59
$password1contains spaces characters. If you had doneprintf '"%s"n' "$password1", you would have seen them. You probably need to change the command that retrieves the password.– Stéphane Chazelas
Mar 14 '17 at 17:20
Add another
s/^ *//;s/ *$//to your sed command to remove leading an trailing space characters. Now what if the password contains space characters? It's common to have spaces in passwords.– Stéphane Chazelas
Mar 14 '17 at 17:26