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Your code is correct. You have declared your variable as an array, and you are successfully appending to it. To display all of the elements of your variable, try echo "${my_array[@]}". (Another an...
Answer
#4: Post edited
by
r~~
·
2021-03-22T20:22:05Z (over 3 years ago)
Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array, and you are successfully appending to it.To display all of the elements of your variable, try `echo "${my_array[@]}"`.- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
- `echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
- ---
- Note also that in Bash versions >= 4, the `readarray` builtin will do all of this work for you:
- `readarray -t my_array < my_file.txt`
- You don't even need to declare `my_array` beforehand if you use this builtin. Bash 4 was released in 2009, so unless you're working on a pretty old or stripped-down system, you are probably safe using that.
- Your code is correct. You have declared your variable as an array, and you are successfully appending to it.
- To display all of the elements of your variable, try `echo "${my_array[@]}"`. (Another answer suggests `declare -p` for this, but `declare -p` will possibly give you more information than you wanted. Also, be aware that `declare -p` isn't intended to *substitute* for your `declare -a`; it's an additional command to show the status of your variable.)
- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
- `echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
- ---
- Note also that in Bash versions >= 4, the `readarray` builtin will do all of this work for you:
- `readarray -t my_array < my_file.txt`
- You don't even need to declare `my_array` beforehand if you use this builtin. Bash 4 was released in 2009, so unless you're working on a pretty old or stripped-down system, you are probably safe using that.
#3: Post edited
by
r~~
·
2021-03-22T20:17:59Z (over 3 years ago)
- Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array, and you are successfully appending to it.
- To display all of the elements of your variable, try `echo "${my_array[@]}"`.
- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
`echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
- Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array, and you are successfully appending to it.
- To display all of the elements of your variable, try `echo "${my_array[@]}"`.
- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
- `echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
- ---
- Note also that in Bash versions >= 4, the `readarray` builtin will do all of this work for you:
- `readarray -t my_array < my_file.txt`
- You don't even need to declare `my_array` beforehand if you use this builtin. Bash 4 was released in 2009, so unless you're working on a pretty old or stripped-down system, you are probably safe using that.
#2: Post edited
by
r~~
·
2021-03-22T20:10:20Z (over 3 years ago)
Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array type, and you are successfully appending to it.- To display all of the elements of your variable, try `echo "${my_array[@]}"`.
- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
- `echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
- Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array, and you are successfully appending to it.
- To display all of the elements of your variable, try `echo "${my_array[@]}"`.
- To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
- To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
- `echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.
#1: Initial revision
by
r~~
·
2021-03-22T20:07:28Z (over 3 years ago)
Your code is correct; another answer suggests switching to `declare -p my_array`, but this is not necessary or useful (all it does is print the attributes of the variable, which implicitly means that `my_array` defaults to being a normal string variable—the same effect is achieved by omitting the `declare` altogether). You have declared your variable as an array type, and you are successfully appending to it.
To display all of the elements of your variable, try `echo "${my_array[@]}"`.
To display the current number of elements in your variable, try `echo "${#my_array[@]}"`.
To display the third element in the array, try `echo "${my_array[2]}"` (arrays in Bash are zero-indexed).
`echo $my_array` will only display the first element of the array, for what I presume are historical reasons, but that's the documented behavior. I would avoid using this ‘feature’ with array variables; it only leads to confusion.