关联数组通过字符串初始化且无需声明。

huangapple
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英文:

associative array initialization via string and without declare

问题

I want to have a global associative array, that is filled at several locations and I can not get it to work to initialize the array with the content of a string without using the declare -A -g over and over at said locations(which I don't feel like is the smartest approach). I've extracted the issue to the code below:

  1. #!/bin/bash
  3. declare -A arr
  4. declare inputString
  6. inputString="([key]=val)"
  8. arr="$inputString"
  9. printf "from variable: ${arr[*]}\n"
  10. printf "from variable: ${arr[key]}\n"
  12. arr=([key]=val)
  13. printf "inline: ${arr[*]}\n"
  14. printf "inline: ${arr[key]}\n"
  16. declare -A arr=$inputString
  17. printf "with declare: ${arr[*]}\n"
  18. printf "with declare: ${arr[key]}\n"

the output is:

  1. from variable: ([key]=val)
  2. from variable: 
  3. inline: val
  4. inline: val
  5. with declare: val
  6. with declare: val

whereas I would expect it to be:

  1. from variable: val
  2. from variable: val
  3. inline: val
  4. inline: val
  5. with declare: val
  6. with declare: val

what do I have to do to accomplish this task?

英文:

I want to have a global associative array, that is filled at several locations and I can not get it to work to initialize the array with the content of a string without using the declare -A -g over and over at said locations(which I don't feel like is the smartest approach). I've extracted the issue to the code below:

  1. #!/bin/bash
  3. declare -A arr
  4. declare inputString
  6. inputString="([key]=val)"
  8. arr="$inputString"
  9. printf "from variable: ${arr[*]}\n"
  10. printf "from variable: ${arr[key]}\n"
  12. arr=([key]=val)
  13. printf "inline: ${arr[*]}\n"
  14. printf "inline: ${arr[key]}\n"
  16. declare -A arr=$inputString
  17. printf "with declare: ${arr[*]}\n"
  18. printf "with declare: ${arr[key]}\n"

the output is:

  1. from variable: ([key]=val)
  2. from variable: 
  3. inline: val
  4. inline: val
  5. with declare: val
  6. with declare: val

whereas I would expect it to be:

  1. from variable: val
  2. from variable: val
  3. inline: val
  4. inline: val
  5. with declare: val
  6. with declare: val

what do I have to do to accomplish this task?

答案1

得分: 4

以下是翻译好的内容：

One bash 4.3+ approach using a nameref:

  1. myadd() {                             # $1 == array name; $2 == key name; $3 == value
  2.     declare -n _arr="$1"              # nameref
  3.     _arr+=([$2]="$3")
  4. }
  6. myclear() {
  7.     declare -n _arr="$1"
  8.     _arr=()
  9. }
  11. Taking for a test drive:
  13. unset      arr1 arr2
  14. declare -A arr1 arr2
  16. myadd arr1 keyX 3
  17. myadd arr1 Index_45 16
  18. myadd arr2 long_string_abc long_value_xyz
  20. typeset -p arr1 arr2
  21.     ==> declare -A arr1=([keyX]="3" [Index_45]="16" )
  22.     ==> declare -A arr2=([long_string_abc]="long_value_xyz" )
  24. myclear arr1
  26. myadd arr1 aaa 111
  27. myadd arr1 bbb 222
  29. typeset -p arr1
  30.     ==> declare -A arr1=([bbb]="222" [aaa]="111" )
英文:

One bash 4.3+ approach using a nameref:

  1. myadd() {                             # $1 == array name; $2 == key name; $3 == value
  2.     declare -n _arr="$1"              # nameref
  3.     _arr+=(["$2"]="$3")
  4. }
  6. myclear() {
  7.     declare -n _arr="$1"
  8.     _arr=()
  9. }

Taking for a test drive:

  1. unset      arr1 arr2
  2. declare -A arr1 arr2
  4. myadd arr1 keyX 3
  5. myadd arr1 Index_45 16
  6. myadd arr2 long_string_abc long_value_xyz
  8. typeset -p arr1 arr2
  9.     ==> declare -A arr1=([keyX]="3" [Index_45]="16" )
  10.     ==> declare -A arr2=([long_string_abc]="long_value_xyz" )
  12. myclear arr1
  14. myadd arr1 aaa 111
  15. myadd arr1 bbb 222
  17. typeset -p arr1
  18.     ==> declare -A arr1=([bbb]="222" [aaa]="111" )

huangapple
  • 本文由 发表于 2023年4月13日 21:41:02
  • 转载请务必保留本文链接：https://go.coder-hub.com/76006149.html
  • associative-array
  • bash
  • declare
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