Print name if parameter passed to function












4















I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.



function exitIfEmpty()

{

        if [ -z "$1" ]

        then

        echo "Exiting because ${!1} is empty"

        exit 1

        fi

}


when called like so



exitIfEmpty someKey



should print 



Exiting because someKey is empty





bash







share|improve this question









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  • Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

    – choroba
    16 hours ago











  • @choroba I want to print the parameter name if possible not its value.

    – Xerxes
    16 hours ago






  • 1





    What do you mean by the name? Function arguments don't have names.

    – choroba
    16 hours ago
















4















I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.



function exitIfEmpty()

{

        if [ -z "$1" ]

        then

        echo "Exiting because ${!1} is empty"

        exit 1

        fi

}


when called like so



exitIfEmpty someKey



should print 



Exiting because someKey is empty





bash







share|improve this question









New contributor




Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





















  • Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

    – choroba
    16 hours ago











  • @choroba I want to print the parameter name if possible not its value.

    – Xerxes
    16 hours ago






  • 1





    What do you mean by the name? Function arguments don't have names.

    – choroba
    16 hours ago














4












4








4


1






I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.



function exitIfEmpty()

{

        if [ -z "$1" ]

        then

        echo "Exiting because ${!1} is empty"

        exit 1

        fi

}


when called like so



exitIfEmpty someKey



should print 



Exiting because someKey is empty





bash







share|improve this question









New contributor




Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












I have written a little function that exits if the value of the function argument is empty, I would like to be able to also print the name of the parameter (not the value!) if it is possible, my following implementation fails to print the name of the parameter.



function exitIfEmpty()

{

        if [ -z "$1" ]

        then

        echo "Exiting because ${!1} is empty"

        exit 1

        fi

}


when called like so



exitIfEmpty someKey



should print 



Exiting because someKey is empty





bash




bash






share|improve this question









New contributor




Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question




share|improve this question








edited 14 hours ago









GAD3R

27.5k1858114




27.5k1858114






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asked 16 hours ago









XerxesXerxes

1936




1936




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New contributor





Xerxes is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.













  • Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

    – choroba
    16 hours ago











  • @choroba I want to print the parameter name if possible not its value.

    – Xerxes
    16 hours ago






  • 1





    What do you mean by the name? Function arguments don't have names.

    – choroba
    16 hours ago



















  • Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

    – choroba
    16 hours ago











  • @choroba I want to print the parameter name if possible not its value.

    – Xerxes
    16 hours ago






  • 1





    What do you mean by the name? Function arguments don't have names.

    – choroba
    16 hours ago

















Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

– choroba
16 hours ago





Wait, when you give it "someKey", $1 won't be empty. If it's empty, there's nothing to print.

– choroba
16 hours ago













@choroba I want to print the parameter name if possible not its value.

– Xerxes
16 hours ago





@choroba I want to print the parameter name if possible not its value.

– Xerxes
16 hours ago




1




1





What do you mean by the name? Function arguments don't have names.

– choroba
16 hours ago





What do you mean by the name? Function arguments don't have names.

– choroba
16 hours ago










3 Answers
3






active

oldest

votes


















12














What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.



If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference ${!var}. ${!var} expands to the value of the variable named by `var.



So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "${!1}" to get the value of the variable named in $1, and plain "$1" to get the name.



E.g. this will print variable bar is empty, exiting, and exit the shell:



#!/bin/bash

exitIfEmpty() {

    if [ -z "${!1}" ]; then

        echo "variable $1 is empty, exiting"

        exit 1

    fi

}

foo=x

unset bar

exitIfEmpty foo

exitIfEmpty bar





share|improve this answer































    2














    echo "Exiting because $1 is empty"


    should do the trick.






    share|improve this answer



















    • 2





      Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

      – ilkkachu
      16 hours ago











    • ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

      – jimbobmcgee
      12 hours ago



















    1














    Pass the name as second argument



    function exitIfEmpty()
    
{
    
        if [ -z "$1" ]
    
        then
    
        echo "Exiting because ${2} is empty"
    
        exit 1
    
        fi
    
}
    

    
exitIfEmpty $someKey someKey





    share|improve this answer








    New contributor




    JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.





















    • Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

      – αғsнιη
      13 hours ago











    • This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

      – John Kugelman
      9 hours ago











    Your Answer








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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12














    What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.



    If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference ${!var}. ${!var} expands to the value of the variable named by `var.



    So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "${!1}" to get the value of the variable named in $1, and plain "$1" to get the name.



    E.g. this will print variable bar is empty, exiting, and exit the shell:



    #!/bin/bash
    
exitIfEmpty() {
    
    if [ -z "${!1}" ]; then
    
        echo "variable $1 is empty, exiting"
    
        exit 1
    
    fi
    
}
    
foo=x
    
unset bar
    
exitIfEmpty foo
    
exitIfEmpty bar





    share|improve this answer




























      12














      What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.



      If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference ${!var}. ${!var} expands to the value of the variable named by `var.



      So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "${!1}" to get the value of the variable named in $1, and plain "$1" to get the name.



      E.g. this will print variable bar is empty, exiting, and exit the shell:



      #!/bin/bash
      
exitIfEmpty() {
      
    if [ -z "${!1}" ]; then
      
        echo "variable $1 is empty, exiting"
      
        exit 1
      
    fi
      
}
      
foo=x
      
unset bar
      
exitIfEmpty foo
      
exitIfEmpty bar





      share|improve this answer


























        12












        12








        12







        What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.



        If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference ${!var}. ${!var} expands to the value of the variable named by `var.



        So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "${!1}" to get the value of the variable named in $1, and plain "$1" to get the name.



        E.g. this will print variable bar is empty, exiting, and exit the shell:



        #!/bin/bash
        
exitIfEmpty() {
        
    if [ -z "${!1}" ]; then
        
        echo "variable $1 is empty, exiting"
        
        exit 1
        
    fi
        
}
        
foo=x
        
unset bar
        
exitIfEmpty foo
        
exitIfEmpty bar





        share|improve this answer













        What gets passed to the function is just a string. If you run func somevar, what is passed is the string somevar. If you run func $somevar, what is passed is (the word-split) value of the variable somevar. Neither is a variable reference, a pointer or anything like that, they're just strings.



        If you want to pass the name of a variable to a function, and then look at the value of that variable, you'll need to use a nameref (Bash 4.3 or later, IIRC), or an indirect reference ${!var}. ${!var} expands to the value of the variable named by `var.



        So, you just have it the wrong way in the script, if you pass the name of a variable to function, use "${!1}" to get the value of the variable named in $1, and plain "$1" to get the name.



        E.g. this will print variable bar is empty, exiting, and exit the shell:



        #!/bin/bash
        
exitIfEmpty() {
        
    if [ -z "${!1}" ]; then
        
        echo "variable $1 is empty, exiting"
        
        exit 1
        
    fi
        
}
        
foo=x
        
unset bar
        
exitIfEmpty foo
        
exitIfEmpty bar






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 16 hours ago









        ilkkachuilkkachu

        62.7k10103180




        62.7k10103180

























            2














            echo "Exiting because $1 is empty"


            should do the trick.






            share|improve this answer



















            • 2





              Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

              – ilkkachu
              16 hours ago











            • ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

              – jimbobmcgee
              12 hours ago
















            2














            echo "Exiting because $1 is empty"


            should do the trick.






            share|improve this answer



















            • 2





              Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

              – ilkkachu
              16 hours ago











            • ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

              – jimbobmcgee
              12 hours ago














            2












            2








            2







            echo "Exiting because $1 is empty"


            should do the trick.






            share|improve this answer













            echo "Exiting because $1 is empty"


            should do the trick.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 16 hours ago









            PankiPanki

            838412




            838412








            • 2





              Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

              – ilkkachu
              16 hours ago











            • ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

              – jimbobmcgee
              12 hours ago














            • 2





              Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

              – ilkkachu
              16 hours ago











            • ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

              – jimbobmcgee
              12 hours ago








            2




            2





            Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

            – ilkkachu
            16 hours ago





            Wouldn't this always print Exiting because is empty, if the test is against $1 also? That doesn't seem very useful.

            – ilkkachu
            16 hours ago













            ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

            – jimbobmcgee
            12 hours ago





            ITYM echo 'Exit because $1 is empty' or echo "Exit because $1 is empty" -- single-quote strings don't do variable expansion; double-quoted strings require $ to be escaped

            – jimbobmcgee
            12 hours ago











            1














            Pass the name as second argument



            function exitIfEmpty()
            
{
            
        if [ -z "$1" ]
            
        then
            
        echo "Exiting because ${2} is empty"
            
        exit 1
            
        fi
            
}
            

            
exitIfEmpty $someKey someKey





            share|improve this answer








            New contributor




            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





















            • Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

              – αғsнιη
              13 hours ago











            • This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

              – John Kugelman
              9 hours ago
















            1














            Pass the name as second argument



            function exitIfEmpty()
            
{
            
        if [ -z "$1" ]
            
        then
            
        echo "Exiting because ${2} is empty"
            
        exit 1
            
        fi
            
}
            

            
exitIfEmpty $someKey someKey





            share|improve this answer








            New contributor




            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.





















            • Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

              – αғsнιη
              13 hours ago











            • This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

              – John Kugelman
              9 hours ago














            1












            1








            1







            Pass the name as second argument



            function exitIfEmpty()
            
{
            
        if [ -z "$1" ]
            
        then
            
        echo "Exiting because ${2} is empty"
            
        exit 1
            
        fi
            
}
            

            
exitIfEmpty $someKey someKey





            share|improve this answer








            New contributor




            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.










            Pass the name as second argument



            function exitIfEmpty()
            
{
            
        if [ -z "$1" ]
            
        then
            
        echo "Exiting because ${2} is empty"
            
        exit 1
            
        fi
            
}
            

            
exitIfEmpty $someKey someKey






            share|improve this answer








            New contributor




            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|improve this answer



            share|improve this answer






            New contributor




            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 16 hours ago









            JShorthouseJShorthouse

            37316




            37316




            New contributor




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            New contributor





            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            JShorthouse is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.













            • Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

              – αғsнιη
              13 hours ago











            • This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

              – John Kugelman
              9 hours ago



















            • Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

              – αғsнιη
              13 hours ago











            • This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

              – John Kugelman
              9 hours ago

















            Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

            – αғsнιη
            13 hours ago





            Here in the question, OP needs to check the value of first argument if it's set else exit; also OP may have not only passing single argument to the function.

            – αғsнιη
            13 hours ago













            This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

            – John Kugelman
            9 hours ago





            This doesn't work. If $someKey is empty then the call will be exitIfEmpty someKey and $2 will be unset. Always quote variables: exitIfEmpty "$someKey" someKey.

            – John Kugelman
            9 hours ago










            Xerxes is a new contributor. Be nice, and check out our Code of Conduct.










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            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

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            FæreyjarHestar færeyskahjaltlandshestinumíslenski hesturinnBretlandskolanámumenskuWikipediafæreyskuWikipediaRossið - síða um færeyska hrossið á færeysku Færeyskur hestur Úr Wikipediu, frjálsa alfræðiritinu Jump to navigation Jump to search Rani og Grani. Grani í vetrarfeldi. Færeyski hesturinn (færeyska: Føroyska rossið ) er hestur sem hefur lifað í Færeyjum í hundruði ára. Hann er smágerður, 120-132 cm á hæð, og helst skyldur hjaltlandshestinum. Hlutverk hans var að draga vagna og plóg og bera klyfjar. Hann hefur fjórar gangtegundir eins og íslenski hesturinn (þar með talið tölt) og er með fjölda litaafbrigði. Nú til dags er hann notaður til frístunda. Í lok 19. aldar voru til um 800 hross. Mörg þeirra voru seld til Bretlands til að vinna í kolanámum og um 1960 voru aðeins nokkrir hestar eftir. Átak var gert í að varðveita hestinn og eru nú til 70-80 hross í Færeyjum. [1] Heimild |   Wikimedia Commons er með margmiðlunarefni sem te
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