How to find the location(s) of specific characters in a string - string

With the aid of this question, I can find out if a string holds a specific character. I want to be able to find out where the character actually is. For example for the string banana, how would I be able to determine the letter n is the 3rd and 5th letter, or for the letter a is the 2nd,4th and 6th letter. and b is the first letter.
Q: For a given string, how can I find the location of a given character in that string?

You can do it with a for loop.
char=a
string=banana
len=${#string}
for (( i=0; i < len; i++ )); do
if [[ $char == ${string:$i:1} ]]
then echo $i
fi
done
The positions printed are zero-based. You could echo $((i+1)) to get 1-based positions instead.
${string:$i:1} extracts the ith character of the string, using bash's substring operator, as explained in Shell Parameter Expansion:
${parameter:offset:length}
This is referred to as Substring Expansion. It expands to up to length characters of the value of parameter starting at the character specified by offset.

Here's a fancy way to do it:
#!/usr/bin/env bash
findChar(){
string="${1}"
char="${2}"
length=${#string}
offset=0
r=()
while true; do
string="${string#*${char}}"
length_new="${#string}"
if [[ "${length}" == "${length_new}" ]]; then
echo "${r[#]}"
return
fi
offset=($(( $offset + $length - $length_new )))
r+=("${offset}")
length="${length_new}"
done
}
findChar banana b
findChar banana a

Here's my take on this:
#!/usr/bin/env bash
[[ ${BASH_VERSINFO[0]} < 4 ]] && { echo "Requires bash 4."; exit 1; }
string="${1:-banana}"
declare -A result=()
for ((i=0; i<${#string}; i++)); do
result[${string:$i:1}]="${result[${string:$i:1}]} $i"
done
declare -p result
The idea is that we walk through the string, adding character positions to strings that are values in an array whose subscripts are the letters you're interested in. It's quick & easy, and gives you a result set you can manipulate afterwards, rather than just sending things to stdout.
My result with this is:
$ ./foo
declare -A result='([a]=" 1 3 5" [b]=" 0" [n]=" 2 4" )'
$ ./foo barber
declare -A result='([a]=" 1" [b]=" 0 3" [e]=" 4" [r]=" 2 5" )'
Results are zero-based (i.e. "b" is in position 0).
Note an interesting side-effect of this method is that every position is preceded by a space, so if you want to count the number of occurrences of a character, you can just count the spaces:
$ declare -A result
$ result[a]=" 1 3 5"
$ count="${result[a]//[0-9]/}"
$ echo "${#count}"
3
$
I don't know what you're planning to do with this data, but if you like, you could easily turn these string results into arrays of their own for easier handling within bash.
Note that associative arrays were introduced with bash version 4.

Related

How To Parameter Array To The Function In Bash [duplicate]

As we know, in bash programming the way to pass arguments is $1, ..., $N. However, I found it not easy to pass an array as an argument to a function which receives more than one argument. Here is one example:
f(){
x=($1)
y=$2
for i in "${x[#]}"
do
echo $i
done
....
}
a=("jfaldsj jflajds" "LAST")
b=NOEFLDJF
f "${a[#]}" $b
f "${a[*]}" $b
As described, function freceives two arguments: the first is assigned to x which is an array, the second to y.
f can be called in two ways. The first way use the "${a[#]}" as the first argument, and the result is:
jfaldsj
jflajds
The second way use the "${a[*]}" as the first argument, and the result is:
jfaldsj
jflajds
LAST
Neither result is as I wished. So, is there anyone having any idea about how to pass array between functions correctly?
You cannot pass an array, you can only pass its elements (i.e. the expanded array).
#!/bin/bash
function f() {
a=("$#")
((last_idx=${#a[#]} - 1))
b=${a[last_idx]}
unset a[last_idx]
for i in "${a[#]}" ; do
echo "$i"
done
echo "b: $b"
}
x=("one two" "LAST")
b='even more'
f "${x[#]}" "$b"
echo ===============
f "${x[*]}" "$b"
The other possibility would be to pass the array by name:
#!/bin/bash
function f() {
name=$1[#]
b=$2
a=("${!name}")
for i in "${a[#]}" ; do
echo "$i"
done
echo "b: $b"
}
x=("one two" "LAST")
b='even more'
f x "$b"
You can pass an array by name reference to a function in bash (since version 4.3+), by setting the -n attribute:
show_value () # array index
{
local -n myarray=$1
local idx=$2
echo "${myarray[$idx]}"
}
This works for indexed arrays:
$ shadock=(ga bu zo meu)
$ show_value shadock 2
zo
It also works for associative arrays:
$ declare -A days=([monday]=eggs [tuesday]=bread [sunday]=jam)
$ show_value days sunday
jam
See also nameref or declare -n in the man page.
You could pass the "scalar" value first. That would simplify things:
f(){
b=$1
shift
a=("$#")
for i in "${a[#]}"
do
echo $i
done
....
}
a=("jfaldsj jflajds" "LAST")
b=NOEFLDJF
f "$b" "${a[#]}"
At this point, you might as well use the array-ish positional params directly
f(){
b=$1
shift
for i in "$#" # or simply "for i; do"
do
echo $i
done
....
}
f "$b" "${a[#]}"
This will solve the issue of passing array to function:
#!/bin/bash
foo() {
string=$1
array=($#)
echo "array is ${array[#]}"
echo "array is ${array[1]}"
return
}
array=( one two three )
foo ${array[#]}
colors=( red green blue )
foo ${colors[#]}
Try like this
function parseArray {
array=("$#")
for data in "${array[#]}"
do
echo ${data}
done
}
array=("value" "value1")
parseArray "${array[#]}"
Pass the array as a function
array() {
echo "apple pear"
}
printArray() {
local argArray="${1}"
local array=($($argArray)) # where the magic happens. careful of the surrounding brackets.
for arrElement in "${array[#]}"; do
echo "${arrElement}"
done
}
printArray array
Here is an example where I receive 2 bash arrays into a function, as well as additional arguments after them. This pattern can be continued indefinitely for any number of bash arrays and any number of additional arguments, accommodating any input argument order, so long as the length of each bash array comes just before the elements of that array.
Function definition for print_two_arrays_plus_extra_args:
# Print all elements of a bash array.
# General form:
# print_one_array array1
# Example usage:
# print_one_array "${array1[#]}"
print_one_array() {
for element in "$#"; do
printf " %s\n" "$element"
done
}
# Print all elements of two bash arrays, plus two extra args at the end.
# General form (notice length MUST come before the array in order
# to be able to parse the args!):
# print_two_arrays_plus_extra_args array1_len array1 array2_len array2 \
# extra_arg1 extra_arg2
# Example usage:
# print_two_arrays_plus_extra_args "${#array1[#]}" "${array1[#]}" \
# "${#array2[#]}" "${array2[#]}" "hello" "world"
print_two_arrays_plus_extra_args() {
i=1
# Read array1_len into a variable
array1_len="${#:$i:1}"
((i++))
# Read array1 into a new array
array1=("${#:$i:$array1_len}")
((i += $array1_len))
# Read array2_len into a variable
array2_len="${#:$i:1}"
((i++))
# Read array2 into a new array
array2=("${#:$i:$array2_len}")
((i += $array2_len))
# You can now read the extra arguments all at once and gather them into a
# new array like this:
extra_args_array=("${#:$i}")
# OR you can read the extra arguments individually into their own variables
# one-by-one like this
extra_arg1="${#:$i:1}"
((i++))
extra_arg2="${#:$i:1}"
((i++))
# Print the output
echo "array1:"
print_one_array "${array1[#]}"
echo "array2:"
print_one_array "${array2[#]}"
echo "extra_arg1 = $extra_arg1"
echo "extra_arg2 = $extra_arg2"
echo "extra_args_array:"
print_one_array "${extra_args_array[#]}"
}
Example usage:
array1=()
array1+=("one")
array1+=("two")
array1+=("three")
array2=("four" "five" "six" "seven" "eight")
echo "Printing array1 and array2 plus some extra args"
# Note that `"${#array1[#]}"` is the array length (number of elements
# in the array), and `"${array1[#]}"` is the array (all of the elements
# in the array)
print_two_arrays_plus_extra_args "${#array1[#]}" "${array1[#]}" \
"${#array2[#]}" "${array2[#]}" "hello" "world"
Example Output:
Printing array1 and array2 plus some extra args
array1:
one
two
three
array2:
four
five
six
seven
eight
extra_arg1 = hello
extra_arg2 = world
extra_args_array:
hello
world
For further examples and detailed explanations of how this works, see my longer answer on this topic here: Passing arrays as parameters in bash
You can also create a json file with an array, and then parse that json file with jq
For example:
my-array.json:
{
"array": ["item1","item2"]
}
script.sh:
ARRAY=$(jq -r '."array"' $1 | tr -d '[],"')
And then call the script like:
script.sh ./path-to-json/my-array.json

Bash shell how to remove string from array with -= operator [duplicate]

I need to remove an element from an array in bash shell.
Generally I'd simply do:
array=("${(#)array:#<element to remove>}")
Unfortunately the element I want to remove is a variable so I can't use the previous command.
Down here an example:
array+=(pluto)
array+=(pippo)
delete=(pluto)
array( ${array[#]/$delete} ) -> but clearly doesn't work because of {}
Any idea?
The following works as you would like in bash and zsh:
$ array=(pluto pippo)
$ delete=pluto
$ echo ${array[#]/$delete}
pippo
$ array=( "${array[#]/$delete}" ) #Quotes when working with strings
If need to delete more than one element:
...
$ delete=(pluto pippo)
for del in ${delete[#]}
do
array=("${array[#]/$del}") #Quotes when working with strings
done
Caveat
This technique actually removes prefixes matching $delete from the elements, not necessarily whole elements.
Update
To really remove an exact item, you need to walk through the array, comparing the target to each element, and using unset to delete an exact match.
array=(pluto pippo bob)
delete=(pippo)
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
Note that if you do this, and one or more elements is removed, the indices will no longer be a continuous sequence of integers.
$ declare -p array
declare -a array=([0]="pluto" [2]="bob")
The simple fact is, arrays were not designed for use as mutable data structures. They are primarily used for storing lists of items in a single variable without needing to waste a character as a delimiter (e.g., to store a list of strings which can contain whitespace).
If gaps are a problem, then you need to rebuild the array to fill the gaps:
for i in "${!array[#]}"; do
new_array+=( "${array[i]}" )
done
array=("${new_array[#]}")
unset new_array
You could build up a new array without the undesired element, then assign it back to the old array. This works in bash:
array=(pluto pippo)
new_array=()
for value in "${array[#]}"
do
[[ $value != pluto ]] && new_array+=($value)
done
array=("${new_array[#]}")
unset new_array
This yields:
echo "${array[#]}"
pippo
This is the most direct way to unset a value if you know it's position.
$ array=(one two three)
$ echo ${#array[#]}
3
$ unset 'array[1]'
$ echo ${array[#]}
one three
$ echo ${#array[#]}
2
This answer is specific to the case of deleting multiple values from large arrays, where performance is important.
The most voted solutions are (1) pattern substitution on an array, or (2) iterating over the array elements. The first is fast, but can only deal with elements that have distinct prefix, the second has O(n*k), n=array size, k=elements to remove. Associative array are relative new feature, and might not have been common when the question was originally posted.
For the exact match case, with large n and k, possible to improve performance from O(nk) to O(n+klog(k)). In practice, O(n) assuming k much lower than n. Most of the speed up is based on using associative array to identify items to be removed.
Performance (n-array size, k-values to delete). Performance measure seconds of user time
N K New(seconds) Current(seconds) Speedup
1000 10 0.005 0.033 6X
10000 10 0.070 0.348 5X
10000 20 0.070 0.656 9X
10000 1 0.043 0.050 -7%
As expected, the current solution is linear to N*K, and the fast solution is practically linear to K, with much lower constant. The fast solution is slightly slower vs the current solution when k=1, due to additional setup.
The 'Fast' solution: array=list of input, delete=list of values to remove.
declare -A delk
for del in "${delete[#]}" ; do delk[$del]=1 ; done
# Tag items to remove, based on
for k in "${!array[#]}" ; do
[ "${delk[${array[$k]}]-}" ] && unset 'array[k]'
done
# Compaction
array=("${array[#]}")
Benchmarked against current solution, from the most-voted answer.
for target in "${delete[#]}"; do
for i in "${!array[#]}"; do
if [[ ${array[i]} = $target ]]; then
unset 'array[i]'
fi
done
done
array=("${array[#]}")
Here's a one-line solution with mapfile:
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "<regexp>")
Example:
$ arr=("Adam" "Bob" "Claire"$'\n'"Smith" "David" "Eve" "Fred")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 6 Contents: Adam Bob Claire
Smith David Eve Fred
$ mapfile -d $'\0' -t arr < <(printf '%s\0' "${arr[#]}" | grep -Pzv "^Claire\nSmith$")
$ echo "Size: ${#arr[*]} Contents: ${arr[*]}"
Size: 5 Contents: Adam Bob David Eve Fred
This method allows for great flexibility by modifying/exchanging the grep command and doesn't leave any empty strings in the array.
Partial answer only
To delete the first item in the array
unset 'array[0]'
To delete the last item in the array
unset 'array[-1]'
To expand on the above answers, the following can be used to remove multiple elements from an array, without partial matching:
ARRAY=(one two onetwo three four threefour "one six")
TO_REMOVE=(one four)
TEMP_ARRAY=()
for pkg in "${ARRAY[#]}"; do
for remove in "${TO_REMOVE[#]}"; do
KEEP=true
if [[ ${pkg} == ${remove} ]]; then
KEEP=false
break
fi
done
if ${KEEP}; then
TEMP_ARRAY+=(${pkg})
fi
done
ARRAY=("${TEMP_ARRAY[#]}")
unset TEMP_ARRAY
This will result in an array containing:
(two onetwo three threefour "one six")
Here's a (probably very bash-specific) little function involving bash variable indirection and unset; it's a general solution that does not involve text substitution or discarding empty elements and has no problems with quoting/whitespace etc.
delete_ary_elmt() {
local word=$1 # the element to search for & delete
local aryref="$2[#]" # a necessary step since '${!$2[#]}' is a syntax error
local arycopy=("${!aryref}") # create a copy of the input array
local status=1
for (( i = ${#arycopy[#]} - 1; i >= 0; i-- )); do # iterate over indices backwards
elmt=${arycopy[$i]}
[[ $elmt == $word ]] && unset "$2[$i]" && status=0 # unset matching elmts in orig. ary
done
return $status # return 0 if something was deleted; 1 if not
}
array=(a 0 0 b 0 0 0 c 0 d e 0 0 0)
delete_ary_elmt 0 array
for e in "${array[#]}"; do
echo "$e"
done
# prints "a" "b" "c" "d" in lines
Use it like delete_ary_elmt ELEMENT ARRAYNAME without any $ sigil. Switch the == $word for == $word* for prefix matches; use ${elmt,,} == ${word,,} for case-insensitive matches; etc., whatever bash [[ supports.
It works by determining the indices of the input array and iterating over them backwards (so deleting elements doesn't screw up iteration order). To get the indices you need to access the input array by name, which can be done via bash variable indirection x=1; varname=x; echo ${!varname} # prints "1".
You can't access arrays by name like aryname=a; echo "${$aryname[#]}, this gives you an error. You can't do aryname=a; echo "${!aryname[#]}", this gives you the indices of the variable aryname (although it is not an array). What DOES work is aryref="a[#]"; echo "${!aryref}", which will print the elements of the array a, preserving shell-word quoting and whitespace exactly like echo "${a[#]}". But this only works for printing the elements of an array, not for printing its length or indices (aryref="!a[#]" or aryref="#a[#]" or "${!!aryref}" or "${#!aryref}", they all fail).
So I copy the original array by its name via bash indirection and get the indices from the copy. To iterate over the indices in reverse I use a C-style for loop. I could also do it by accessing the indices via ${!arycopy[#]} and reversing them with tac, which is a cat that turns around the input line order.
A function solution without variable indirection would probably have to involve eval, which may or may not be safe to use in that situation (I can't tell).
Using unset
To remove an element at particular index, we can use unset and then do copy to another array. Only just unset is not required in this case. Because unset does not remove the element it just sets null string to the particular index in array.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
unset 'arr[1]'
declare -a arr2=()
i=0
for element in "${arr[#]}"
do
arr2[$i]=$element
((++i))
done
echo "${arr[#]}"
echo "1st val is ${arr[1]}, 2nd val is ${arr[2]}"
echo "${arr2[#]}"
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"
Output is
aa cc dd ee
1st val is , 2nd val is cc
aa cc dd ee
1st val is cc, 2nd val is dd
Using :<idx>
We can remove some set of elements using :<idx> also. For example if we want to remove 1st element we can use :1 as mentioned below.
declare -a arr=('aa' 'bb' 'cc' 'dd' 'ee')
arr2=("${arr[#]:1}")
echo "${arr2[#]}"
echo "1st val is ${arr2[1]}, 2nd val is ${arr2[2]}"
Output is
bb cc dd ee
1st val is cc, 2nd val is dd
http://wiki.bash-hackers.org/syntax/pe#substring_removal
${PARAMETER#PATTERN} # remove from beginning
${PARAMETER##PATTERN} # remove from the beginning, greedy match
${PARAMETER%PATTERN} # remove from the end
${PARAMETER%%PATTERN} # remove from the end, greedy match
In order to do a full remove element, you have to do an unset command with an if statement. If you don't care about removing prefixes from other variables or about supporting whitespace in the array, then you can just drop the quotes and forget about for loops.
See example below for a few different ways to clean up an array.
options=("foo" "bar" "foo" "foobar" "foo bar" "bars" "bar")
# remove bar from the start of each element
options=("${options[#]/#"bar"}")
# options=("foo" "" "foo" "foobar" "foo bar" "s" "")
# remove the complete string "foo" in a for loop
count=${#options[#]}
for ((i = 0; i < count; i++)); do
if [ "${options[i]}" = "foo" ] ; then
unset 'options[i]'
fi
done
# options=( "" "foobar" "foo bar" "s" "")
# remove empty options
# note the count variable can't be recalculated easily on a sparse array
for ((i = 0; i < count; i++)); do
# echo "Element $i: '${options[i]}'"
if [ -z "${options[i]}" ] ; then
unset 'options[i]'
fi
done
# options=("foobar" "foo bar" "s")
# list them with select
echo "Choose an option:"
PS3='Option? '
select i in "${options[#]}" Quit
do
case $i in
Quit) break ;;
*) echo "You selected \"$i\"" ;;
esac
done
Output
Choose an option:
1) foobar
2) foo bar
3) s
4) Quit
Option?
Hope that helps.
There is also this syntax, e.g. if you want to delete the 2nd element :
array=("${array[#]:0:1}" "${array[#]:2}")
which is in fact the concatenation of 2 tabs. The first from the index 0 to the index 1 (exclusive) and the 2nd from the index 2 to the end.
POSIX shell script does not have arrays.
So most probably you are using a specific dialect such as bash, korn shells or zsh.
Therefore, your question as of now cannot be answered.
Maybe this works for you:
unset array[$delete]
What I do is:
array="$(echo $array | tr ' ' '\n' | sed "/itemtodelete/d")"
BAM, that item is removed.
This is a quick-and-dirty solution that will work in simple cases but will break if (a) there are regex special characters in $delete, or (b) there are any spaces at all in any items. Starting with:
array+=(pluto)
array+=(pippo)
delete=(pluto)
Delete all entries exactly matching $delete:
array=(`echo $array | fmt -1 | grep -v "^${delete}$" | fmt -999999`)
resulting in
echo $array -> pippo, and making sure it's an array:
echo $array[1] -> pippo
fmt is a little obscure: fmt -1 wraps at the first column (to put each item on its own line. That's where the problem arises with items in spaces.) fmt -999999 unwraps it back to one line, putting back the spaces between items. There are other ways to do that, such as xargs.
Addendum: If you want to delete just the first match, use sed, as described here:
array=(`echo $array | fmt -1 | sed "0,/^${delete}$/{//d;}" | fmt -999999`)
Actually, I just noticed that the shell syntax somewhat has a behavior built-in that allows for easy reconstruction of the array when, as posed in the question, an item should be removed.
# let's set up an array of items to consume:
x=()
for (( i=0; i<10; i++ )); do
x+=("$i")
done
# here, we consume that array:
while (( ${#x[#]} )); do
i=$(( $RANDOM % ${#x[#]} ))
echo "${x[i]} / ${x[#]}"
x=("${x[#]:0:i}" "${x[#]:i+1}")
done
Notice how we constructed the array using bash's x+=() syntax?
You could actually add more than one item with that, the content of a whole other array at once.
In ZSH this is dead easy (note this uses more bash compatible syntax than necessary where possible for ease of understanding):
# I always include an edge case to make sure each element
# is not being word split.
start=(one two three 'four 4' five)
work=(${(#)start})
idx=2
val=${work[idx]}
# How to remove a single element easily.
# Also works for associative arrays (at least in zsh)
work[$idx]=()
echo "Array size went down by one: "
[[ $#work -eq $(($#start - 1)) ]] && echo "OK"
echo "Array item "$val" is now gone: "
[[ -z ${work[(r)$val]} ]] && echo OK
echo "Array contents are as expected: "
wanted=("${start[#]:0:1}" "${start[#]:2}")
[[ "${(j.:.)wanted[#]}" == "${(j.:.)work[#]}" ]] && echo "OK"
echo "-- array contents: start --"
print -l -r -- "-- $#start elements" ${(#)start}
echo "-- array contents: work --"
print -l -r -- "-- $#work elements" "${work[#]}"
Results:
Array size went down by one:
OK
Array item two is now gone:
OK
Array contents are as expected:
OK
-- array contents: start --
-- 5 elements
one
two
three
four 4
five
-- array contents: work --
-- 4 elements
one
three
four 4
five
To avoid conflicts with array index using unset - see https://stackoverflow.com/a/49626928/3223785 and https://stackoverflow.com/a/47798640/3223785 for more information - reassign the array to itself: ARRAY_VAR=(${ARRAY_VAR[#]}).
#!/bin/bash
ARRAY_VAR=(0 1 2 3 4 5 6 7 8 9)
unset ARRAY_VAR[5]
unset ARRAY_VAR[4]
ARRAY_VAR=(${ARRAY_VAR[#]})
echo ${ARRAY_VAR[#]}
A_LENGTH=${#ARRAY_VAR[*]}
for (( i=0; i<=$(( $A_LENGTH -1 )); i++ )) ; do
echo ""
echo "INDEX - $i"
echo "VALUE - ${ARRAY_VAR[$i]}"
done
exit 0
[Ref.: https://tecadmin.net/working-with-array-bash-script/ ]
How about something like:
array=(one two three)
array_t=" ${array[#]} "
delete=one
array=(${array_t// $delete / })
unset array_t
#/bin/bash
echo "# define array with six elements"
arr=(zero one two three 'four 4' five)
echo "# unset by index: 0"
unset -v 'arr[0]'
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
arr_delete_by_content() { # value to delete
for i in ${!arr[*]}; do
[ "${arr[$i]}" = "$1" ] && unset -v 'arr[$i]'
done
}
echo "# unset in global variable where value: three"
arr_delete_by_content three
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
echo "# rearrange indices"
arr=( "${arr[#]}" )
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_value() { # value arrayelements..., returns array decl.
local e val=$1; new=(); shift
for e in "${#}"; do [ "$val" != "$e" ] && new+=("$e"); done
declare -p new|sed 's,^[^=]*=,,'
}
echo "# new array without value: two"
declare -a arr="$(delete_value two "${arr[#]}")"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
delete_values() { # arraydecl values..., returns array decl. (keeps indices)
declare -a arr="$1"; local i v; shift
for v in "${#}"; do
for i in ${!arr[*]}; do
[ "$v" = "${arr[$i]}" ] && unset -v 'arr[$i]'
done
done
declare -p arr|sed 's,^[^=]*=,,'
}
echo "# new array without values: one five (keep indices)"
declare -a arr="$(delete_values "$(declare -p arr|sed 's,^[^=]*=,,')" one five)"
for i in ${!arr[*]}; do echo "arr[$i]=${arr[$i]}"; done
# new array without multiple values and rearranged indices is left to the reader

Bash for loop on variable numbers

I have a situation where I have large number of numbered variables. I want to evaluate each variable and set variable to a specific string if the condition is matched.
#!/bin/bash
var1=""
var2="1233123213"
var3="22332323222324242"
var4=""
var5=""
for i in 1 2 3 4 5
do
if [ -z "$var{$}i" ]
then
var{$}i="None"
fi
echo "var{$}i \r"
done
but the problem is when I run the script I get following.
{1} \r
{2} \r
{3} \r
{4} \r
{5} \r
How I can fix this.
Use indirect variable expansion in bash with syntax {!var}.
From the man bash page,
If the first character of parameter is an exclamation point (!), a level of variable indirection is introduced. 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 exclamation point must immediately follow the left brace in order to introduce indirection.
Modify your code to something like below,
for i in 1 2 3 4 5
do
var="var$i"
[ -z "${!var}" ] && declare "var$i"="none"
done
printf "var1=%s\n" "$var1"
printf "var2=%s\n" "$var2"
printf "var3=%s\n" "$var3"
printf "var4=%s\n" "$var4"
printf "var5=%s\n" "$var5"
The syntax "${!var}" in this case evaluates the value of the variable within the string var which is var1, var2, var3... and the declare syntax sets the variable value at run-time, only for those variables that are empty.
Now on printing those variables produces,
var1=none
var2=1233123213
var3=22332323222324242
var4=none
var5=none
Indirect assignment will work here, but in this specific case arrays seem like a good fit :
#!/bin/bash
declare -a var=()
var+=("")
var+=(1233123213)
var+=(22332323222324242)
var+=("")
var+=("")
for i in "${!var[#]}"
do
[[ "${var[$i]}" ]] || var[$i]="None"
echo "Index: $i - Value: ${var[$i]}"
done
Consider using an array instead of numbered variables:
#!/bin/bash
var[1]=""
var[2]="1233123213"
var[3]="22332323222324242"
var[4]=""
var[5]=""
for i in 1 2 3 4 5
do
if [ -z "${var[i]}" ]
then
var[i]="None"
fi
echo "${var[i]} \r"
done

Reversing a bash for loop

I have this:
for (( count= "$WP_RANGE_START"; count< "$WP_RANGE_STOP"+1; count=count+1 ));
Where WP_RANGE_STARTis a number like 1 and WP_RANGE_STOPis a number like 10.
Right now this will step though going 1,2,...10
How can I do so that it counts backwards?(10,9,...1)
I guess the mirror image of what you have would be
for (( count="$WP_RANGE_STOP"; count >= "$WP_RANGE_START"; count=count-1 ));
But a less cumbersome way to write it would be
for (( count=WP_RANGE_STOP; count >= WP_RANGE_START; count-- ));
The $ is unnecessary in arithmetic context.
When dealing with literals, bash has a range expansion feature using brace expansion:
for i in {0..10}; # or {10..0} or what have you
But it's cumbersome to use with variables, as the brace expansion happens before parameter expansion. It's usually easier to use arithmetic for loops in those cases.
Your incrementing code can be "simplified" as:
for count in $(eval echo {$WP_RANGE_START..$WP_RANGE_STOP});
So, to decrement you can just reverse the parameters"
for count in $(eval echo {$WP_RANGE_STOP..$WP_RANGE_START});
Assuming you've got a bash version of 3 or higher, you can specify an increment or decrement by appending it to the range, like so:
CHANGE=1
for count in $(eval echo {$WP_RANGE_STOP..$WP_RANGE_START..$CHANGE});
The for loop is your problem.
i=11 ; until [ $((i=i-1)) -lt 1 ] ; do echo $i ; done
OUTPUT
10
9
8
7
6
5
4
3
2
1
You don't need any bashisms at all.

Check and modify format of variable in expect script

I am trying to verify that the format of a variable is a number and is at least 10 digits long with leading zeros, inside of an expect script.
In a bash script it would look something like this:
[[ "$var" != +([0-9]) ]] && echo "bad input" && exit
while [[ $(echo -n ${var} | wc -c) -lt 10 ]] ; do var="0${var}" ; done
For the following input:
16
I am trying to achieve the following output:
0000000016
The simplest way to check whether a variable has just digits is to use a regular expression. Expect's regular expressions are entirely up to the task:
if {![regexp {^\d+$} $var]} {
puts "bad input"
exit
}
Padding with zeroes is best done by formatting the value; if you know C's printf(), you'll recognize the format:
set var [format "%010d" $var]
Expect is actually just an extension of TCL, so you can use any facility that TCL provides. TCL is an unusual language, but it's not hard to do what you want.
# Set a test string.
set testvar 1234567890
# Store the match (if any) in matchvar.
regexp {\d{10,}} $testvar matchvar
puts $matchvar
# Test that matchvar holds an integer.
string is integer $matchvar
The string is command is relatively new, so you might have to rely on the return value of regexp if your TCL interpreter doesn't support it.

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