Accessing array index variable from bash shell script loop?












7















I want to access the array index variable while looping thru an array in my bash shell script.



myscript.sh

#!/bin/bash

AR=('foo' 'bar' 'baz' 'bat')

for i in ${AR[*]}; do

  echo $i

done


The result of the above script is:



foo

bar

baz

bat


The result I seek is:



0

1

2

3


How do I alter my script to achieve this?






bash shell shell-script







share|improve this question


















  • 3





    Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

    – jordanm
    Apr 23 '16 at 2:40
















7















I want to access the array index variable while looping thru an array in my bash shell script.



myscript.sh

#!/bin/bash

AR=('foo' 'bar' 'baz' 'bat')

for i in ${AR[*]}; do

  echo $i

done


The result of the above script is:



foo

bar

baz

bat


The result I seek is:



0

1

2

3


How do I alter my script to achieve this?






bash shell shell-script







share|improve this question


















  • 3





    Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

    – jordanm
    Apr 23 '16 at 2:40














7












7








7


2






I want to access the array index variable while looping thru an array in my bash shell script.



myscript.sh

#!/bin/bash

AR=('foo' 'bar' 'baz' 'bat')

for i in ${AR[*]}; do

  echo $i

done


The result of the above script is:



foo

bar

baz

bat


The result I seek is:



0

1

2

3


How do I alter my script to achieve this?






bash shell shell-script







share|improve this question














I want to access the array index variable while looping thru an array in my bash shell script.



myscript.sh

#!/bin/bash

AR=('foo' 'bar' 'baz' 'bat')

for i in ${AR[*]}; do

  echo $i

done


The result of the above script is:



foo

bar

baz

bat


The result I seek is:



0

1

2

3


How do I alter my script to achieve this?






bash shell shell-script




bash shell shell-script






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 23 '16 at 2:31









MowzerMowzer

179227




179227








  • 3





    Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

    – jordanm
    Apr 23 '16 at 2:40














  • 3





    Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

    – jordanm
    Apr 23 '16 at 2:40








3




3





Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

– jordanm
Apr 23 '16 at 2:40





Also note that you basically never want "${array[*]}" instead of "${array[@]}". Using * instead of @ more or less treats it as a string instead of an array.

– jordanm
Apr 23 '16 at 2:40










2 Answers
2






active

oldest

votes


















8














You can do this using indirection. From the bash manpage:




If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.




For your example:



#!/bin/bash

AR=('foo' 'bar' 'baz' 'bat')

for i in "${!AR[@]}"; do

  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"

done


This results in:



${AR[0]}=foo

${AR[1]}=bar

${AR[2]}=baz

${AR[3]}=bat


Note that this also work for non-succesive indexes:



#!/bin/bash

AR=([3]='foo' [5]='bar' [25]='baz' [7]='bat')

for i in "${!AR[@]}"; do

  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"

done


This results in:



${AR[3]}=foo

${AR[5]}=bar

${AR[7]}=bat

${AR[25]}=baz





share|improve this answer





















  • 2





    I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

    – Alexandros
    Mar 17 '18 at 16:05



















4














Additional to jordanm's answer you can also do a C like loop in bash:



for ((idx=0; idx<${#array[@]}; ++idx)); do

    echo "$idx" "${array[idx]}"

done





share|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    8














    You can do this using indirection. From the bash manpage:




    If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.




    For your example:



    #!/bin/bash
    
AR=('foo' 'bar' 'baz' 'bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[0]}=foo
    
${AR[1]}=bar
    
${AR[2]}=baz
    
${AR[3]}=bat


    Note that this also work for non-succesive indexes:



    #!/bin/bash
    
AR=([3]='foo' [5]='bar' [25]='baz' [7]='bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[3]}=foo
    
${AR[5]}=bar
    
${AR[7]}=bat
    
${AR[25]}=baz





    share|improve this answer





















    • 2





      I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

      – Alexandros
      Mar 17 '18 at 16:05
















    8














    You can do this using indirection. From the bash manpage:




    If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.




    For your example:



    #!/bin/bash
    
AR=('foo' 'bar' 'baz' 'bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[0]}=foo
    
${AR[1]}=bar
    
${AR[2]}=baz
    
${AR[3]}=bat


    Note that this also work for non-succesive indexes:



    #!/bin/bash
    
AR=([3]='foo' [5]='bar' [25]='baz' [7]='bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[3]}=foo
    
${AR[5]}=bar
    
${AR[7]}=bat
    
${AR[25]}=baz





    share|improve this answer





















    • 2





      I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

      – Alexandros
      Mar 17 '18 at 16:05














    8












    8








    8







    You can do this using indirection. From the bash manpage:




    If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.




    For your example:



    #!/bin/bash
    
AR=('foo' 'bar' 'baz' 'bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[0]}=foo
    
${AR[1]}=bar
    
${AR[2]}=baz
    
${AR[3]}=bat


    Note that this also work for non-succesive indexes:



    #!/bin/bash
    
AR=([3]='foo' [5]='bar' [25]='baz' [7]='bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[3]}=foo
    
${AR[5]}=bar
    
${AR[7]}=bat
    
${AR[25]}=baz





    share|improve this answer















    You can do this using indirection. From the bash manpage:




    If the first character of parameter is an exclamation point (!), it introduces a level of variable indirection. Bash uses the value of the variable formed from the rest of parameter as the name of the variable; this variable is then expanded and that value is used in the rest of the substitution, rather than the value of parameter itself. This is known as indirect expansion. The exceptions to this are the expansions of ${!prefix*} and ${!name[@]} described below. The exclamation point must immediately follow the left brace in order to introduce indirection.




    For your example:



    #!/bin/bash
    
AR=('foo' 'bar' 'baz' 'bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[0]}=foo
    
${AR[1]}=bar
    
${AR[2]}=baz
    
${AR[3]}=bat


    Note that this also work for non-succesive indexes:



    #!/bin/bash
    
AR=([3]='foo' [5]='bar' [25]='baz' [7]='bat')
    
for i in "${!AR[@]}"; do
    
  printf '${AR[%s]}=%sn' "$i" "${AR[i]}"
    
done


    This results in:



    ${AR[3]}=foo
    
${AR[5]}=bar
    
${AR[7]}=bat
    
${AR[25]}=baz






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Feb 8 at 14:48









    Kusalananda

    131k17250409




    131k17250409










    answered Apr 23 '16 at 2:38









    jordanmjordanm

    30.8k38695




    30.8k38695








    • 2





      I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

      – Alexandros
      Mar 17 '18 at 16:05














    • 2





      I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

      – Alexandros
      Mar 17 '18 at 16:05








    2




    2





    I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

    – Alexandros
    Mar 17 '18 at 16:05





    I think in the semicolon was meant to be outside the double quote in: for i in "${!AR[@]};" do

    – Alexandros
    Mar 17 '18 at 16:05













    4














    Additional to jordanm's answer you can also do a C like loop in bash:



    for ((idx=0; idx<${#array[@]}; ++idx)); do
    
    echo "$idx" "${array[idx]}"
    
done





    share|improve this answer




























      4














      Additional to jordanm's answer you can also do a C like loop in bash:



      for ((idx=0; idx<${#array[@]}; ++idx)); do
      
    echo "$idx" "${array[idx]}"
      
done





      share|improve this answer


























        4












        4








        4







        Additional to jordanm's answer you can also do a C like loop in bash:



        for ((idx=0; idx<${#array[@]}; ++idx)); do
        
    echo "$idx" "${array[idx]}"
        
done





        share|improve this answer













        Additional to jordanm's answer you can also do a C like loop in bash:



        for ((idx=0; idx<${#array[@]}; ++idx)); do
        
    echo "$idx" "${array[idx]}"
        
done






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Apr 23 '16 at 2:41









        pfnueselpfnuesel

        2,71941939




        2,71941939






























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