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1

0

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

share|improve this question

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 like http://...?password=something&... where something is 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 -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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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

1

0

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

share|improve this question

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 like http://...?password=something&... where something is 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 -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
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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

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

share|improve this question

edited Mar 14 '17 at 16:10

ilkkachu

63.5k10104181

asked Mar 14 '17 at 16:00

readerreader

155

password1=$(.....)

wget "http://.........?something&thecommandiscontinues"

share|improve this question

edited Mar 14 '17 at 16:10

ilkkachu

63.5k10104181

asked Mar 14 '17 at 16:00

readerreader

155

share|improve this question

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

asked Mar 14 '17 at 16:00

readerreader

155

asked Mar 14 '17 at 16:00

readerreader

155

2 Answers
2

active

oldest

votes

3

Replace something with $password1

share|improve this answer

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 . Try password="mypass" - remove $( )
  
  – George Vasiliou
  Mar 14 '17 at 21:35
  
  

  
  
  
  

add a comment | 

1

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.

share|improve this answer

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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3

Replace something with $password1

share|improve this answer

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 . Try password="mypass" - remove $( )
  
  – George Vasiliou
  Mar 14 '17 at 21:35
  
  

  
  
  
  

add a comment | 

3

Replace something with $password1

share|improve this answer

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 . Try password="mypass" - remove $( )
  
  – George Vasiliou
  Mar 14 '17 at 21:35
  
  

  
  
  
  

add a comment | 

Replace something with $password1

share|improve this answer

edited Mar 14 '17 at 18:52

Michael Mrozek♦

62.5k29194214

answered Mar 14 '17 at 16:40

Tom ReingoldTom Reingold

311

share|improve this answer

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

answered Mar 14 '17 at 16:40

Tom ReingoldTom Reingold

311

answered Mar 14 '17 at 16:40

Tom ReingoldTom Reingold

311

1

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.

share|improve this answer

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 | 

1

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.

share|improve this answer

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.

share|improve this answer

edited 9 hours ago

Rui F Ribeiro

42.1k1484142

answered Mar 14 '17 at 16:04

KingofkechKingofkech

4881620

myFunction () 
 wget -arg "something[...]$password1"

myVariable=$(myFunction);

share|improve this answer

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

answered Mar 14 '17 at 16:04

KingofkechKingofkech

4881620

answered Mar 14 '17 at 16:04

KingofkechKingofkech

4881620