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How to pass a variable in a command in UNIX?



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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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1















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



















  • 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

















1















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



















  • 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













1












1








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
















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 14 '17 at 16:10









ilkkachu

63.5k10104181




63.5k10104181










asked Mar 14 '17 at 16:00









readerreader

155




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












  • 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







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










2 Answers
2






active

oldest

votes


















3














Replace something with $password1






share|improve this answer




















  • 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



















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

























  • 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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Replace something with $password1






share|improve this answer




















  • 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
















3














Replace something with $password1






share|improve this answer




















  • 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














3












3








3







Replace something with $password1






share|improve this answer















Replace something with $password1







share|improve this answer














share|improve this answer



share|improve this answer








edited Mar 14 '17 at 18:52









Michael Mrozek♦

62.5k29194214




62.5k29194214










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 . Try password="mypass" - remove $( )

    – George Vasiliou
    Mar 14 '17 at 21:35













  • 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








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














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

























  • 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















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

























  • 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













1












1








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















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














share|improve this answer



share|improve this answer








edited 9 hours ago









Rui F Ribeiro

42.1k1484142




42.1k1484142










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

















  • 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

















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