In array operator in bash - arrays

Is there a way to test whether an array contains a specified element?
e.g., something like:
array=(one two three)
if [ "one" in ${array} ]; then
...
fi

A for loop will do the trick.
array=(one two three)
for i in "${array[#]}"; do
if [[ "$i" = "one" ]]; then
...
break
fi
done

Try this:
array=(one two three)
if [[ "${array[*]}" =~ "one" ]]; then
echo "'one' is found"
fi

I got an function 'contains' in my .bashrc-file:
contains ()
{
param=$1;
shift;
for elem in "$#";
do
[[ "$param" = "$elem" ]] && return 0;
done;
return 1
}
It works well with an array:
contains on $array && echo hit || echo miss
miss
contains one $array && echo hit || echo miss
hit
contains onex $array && echo hit || echo miss
miss
But doesn't need an array:
contains one four two one zero && echo hit || echo miss
hit

I like using grep for this:
if echo ${array[#]} | grep -qw one; then
# "one" is in the array
...
fi
(Note that both -q and -w are non-standard options to grep: -w tells it to work on whole words only, and -q ("quiet") suppresses all output.)

array="one two three"
if [ $(echo "$array" | grep one | wc -l) -gt 0 ] ;
then echo yes;
fi
If that's ugly, you could hide it away in a function.

if you just want to check whether an element is in array, another approach
case "${array[#]/one/}" in
"${array[#]}" ) echo "not in there";;
*) echo "found ";;
esac

In_array() {
local NEEDLE="$1"
local ELEMENT
shift
for ELEMENT; do
if [ "$ELEMENT" == "$NEEDLE" ]; then
return 0
fi
done
return 1
}
declare -a ARRAY=( "elem1" "elem2" "elem3" )
if In_array "elem1" "${ARRAY[#]}"; then
...
A nice and elegant version of the above.

in_array() {
local needle=$1 el
shift
for el in "$#"; do
if [ "$el" = "$needle" ]; then
return 0
fi
done
return 1
}
if in_array 1 1 2 3; then
echo true
else
echo false
fi
# alternatively
a=(1 2 3)
if in_array 1 "${a[#]}"; then
...

OPTIONS=('-q','-Q','-s','-S')
find="$(grep "\-q" <<< "${OPTIONS[#]}")"
if [ "$find" = "${OPTIONS[#]}" ];
then
echo "arr contains -q"
fi

Related

Bash - How to check if value doesn't exist in array?

Folks,
I have an array, ex,
declare -a arr=("crm" "hr" "pos")
I need to output error if the passed value doesn't exist in this array
I'm trying this use below snippet but it prints "No match found" for any value
match=0
for i in "${arr[#]}"; do
if ! [[ $2 == "$i" ]]; then
match=1
break
fi
done
if [[ $match = 1 ]]; then
echo "No match found"
fi
Any idea how to loop in array and popup error if value doesn't exist ?
There are already answers about your question see check value is in an array, but a fix/idea for your specific approach is something like.
#!/usr/bin/env bash
declare -a arr=("crm" "hr" "pos")
match=0
for i in "${arr[#]}"; do
if ! [[ $2 == "$i" ]]; then
((match++))
fi
done
if (( match == ${#arr[*]} )); then
printf >&2 "No match found\n"
fi
The above script increments match every time the test inside the for loop is true, and in the end compare match with the length of the array ${#arr[*]}. A more verbose output is to put set -x on after the shebang and add a $ sign to the variable match (which is not required) inside the (( )).
if (( $match == ${#arr[*]} )); then
Your original approach always breaks the loop when it does not have a match, doing so will not continue the loop.
You reversed the logic. Just leave out the ! and swap the final test:
match=0
for i in "${arr[#]}"; do
if [[ $2 == "$i" ]]; then
match=1
break
fi
done
if [[ $match = 0 ]]; then
echo "No match found"
fi
This doesn't require a loop.
c.f. if a 2 strings BOTH present in 1 array in a bash script
$: x=bar
$: [[ " ${arr[*]} " =~ " $x " ]] && echo found || echo nope
nope
$: x=hr
$: [[ " ${arr[*]} " =~ " $x " ]] && echo found || echo nope
found
or
$: match=0
$: set -- foo bar
$: [[ " ${arr[*]} " =~ " $2 " ]] && match=1
$: echo match
0
$: set -- foo crm
$: [[ " ${arr[*]} " =~ " $2 " ]] && match=1
$: echo match
1
The logic seems incorrect.
If it does not match, you break the loop.
If it does match, the loop goes to the next iteration. Then it becomes unmatched and finally breaks the loop
declare -a arr=("crm" "hr" "pos")
not_match=0
for i in "${arr[#]}"; do
if [ "$2" = "$i" ]; then
not_match=1
break
fi
done
if [ $not_match -eq 0 ]; then
echo "No match found"
fi

Compare array elements with an element

I am trying to compare an array with a value and print an error statement if the match is not found.
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
For example, this stores values such as small, large and medium which are the files present in new-dir.
The value to be compared with will be entered by the user viz. var
I have tried something like following:
(for i in "${arraylist[#]}"; do [[ "$i" == "$var"]] && exit 0; done) && echo "found" || echo "not found"
Also tried, however, doesn't work:
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
count=0
for((i=0; i<${#arraylist[#]}; i++)); do
if [ "$arraylist[$i]" == "$var" ] ; then
count=1
fi
done
if [ $count -eq 0 ]; then
echo "Not found"
fi
Is there any other way to do this comparison?
This code works. Just replace arraylist with your array.
arraylist=('hello' 'world')
var='hello'
count=0
for i in "${arraylist[#]}"
do
if [ "$i" == "$var" ]; then
count=1
fi
done
if [ $count -eq 0 ]; then
echo "Not found"
fi
To extract the first "word" from each filename, I'd do
declare -A prefixes
cd new-dir
for file in *; do
read -r prefix _ <<< "$file"
prefixes["$prefix"]=1
done
cd -
then to look for a match:
# $var is somehow provided
result=""
for p in "${!prefixes[#]}"; do
if [[ $p == "$var" ]]; then
result=found
break
fi
done
echo "${result:-not found}"
This requires bash v4.0+ for the use of an associative array.
If you have bash v4.3+ you can do
[[ -v prefixes[$var] ]] && echo found || echo not found
Alternately:
shopt -s nullglob
files=( new-dir/"$var "* )
(( ${#files[#]} > 0 )) && echo found || echo not found
This is what shell programming is about. Keep it simple :
Array1=( "item1" "item2" "item3" "item-4" )
var="item3"
count=$(echo ${Array1[#]} | tr ' ' '\n' | awk '$1 == "'"$var"'"{print $0}' | wc -l)
[ $count -eq 0 ] && echo "Not found" || echo "found"
The arraylist might be an array with just one index.
The command:
arraylist="$(ls new-dir/ | cut -d' ' -f1)"
might just assign whole "ls new-dir" output to ${arraylist[0]}
You can check the index value in an array by echo ${arraylist[1]} (to confirm).
Try changing the command to generate an array for a list of files/dirs to:
arraylist=($(ls new-dir/))
The complete script:
arraylist=($(ls new-dir/))
(for i in "${arraylist[#]}"; do [[ "$i" == "$var" ]] && exit 0; done) && echo "found" || echo "not found"
You can also ignore the whole array assignment and just use subshell-command in for loop, eg:
(for i in $(ls -1 new-dir/); do [[ "$i" == "$var" ]] && exit 0; done) && echo "found" || echo "not found"
Remember, no quotes " around $(ls -1 new-dir/)
Another one-liner without any array or loops:
ls -1 new-dir/ | grep -q "$var" && echo "found" || echo "not found"
Edit:
As #M. Nejat Aydin suggested, use [[ -e new-dir/$var ]] && echo "found" || echo "Not found" and do not use ls in the script.

Check multiple var if exists in array with grep

I am using this code to check one $var if exists in array :
if echo ${myArr[#]} | grep -qw $myVar; then echo "Var exists on array" fi
How could I combine more than one $vars to my check? Something like grep -qw $var1,$var2; then ... fi
Thank you in Advance.
if echo ${myArr[#]} | grep -qw -e "$myVar" -e "$otherVar"
then
echo "Var exists on array"
fi
From the man-page:
-e PATTERN, --regexp=PATTERN
Use PATTERN as the pattern.
This can be used to specify multiple search patterns, or to protect a pattern beginning with a hyphen (-). (-e is specified by POSIX.)
But if you want to use arrays like this you might as well use the bash built-in associative arrays.
To implement and logic:
myVar1=home1
myVar2=home2
myArr[0]=home1
myArr[1]=home2
if echo ${myArr[#]} | grep -qw -e "$myVar1.*$myVar2" -e "$myVar2.*$myVar1"
then
echo "Var exists on array"
fi
# using associative arrays
declare -A assoc
assoc[home1]=1
assoc[home2]=1
if [[ ${assoc[$myVar1]} && ${assoc[$myVar2]} ]]; then
echo "Var exists on array"
fi
Actually you don't need grep for this, Bash is perfectly capable of doing Extended Regular Expressions itself (Bash 3.0 or later).
pattern="$var1|$var2|$var3"
for element in "${myArr[#]}"
do
if [[ $element =~ $pattern ]]
then
echo "$pattern exists in array"
break
fi
done
Something quadratic, but aware of spaces:
myArr=(aa "bb c" ddd)
has_values(){
for e in "${myArr[#]}" ; do
for f ; do
if [ "$e" = "$f" ]; then return 0 ; fi
done
done
return 1
}
if has_values "ee" "bb c" ; then echo yes ; else echo "no" ; fi
this example will print no because "bb c" != "bb c"

Check if an element is present in a Bash array [duplicate]

This question already has answers here:
Check if a Bash array contains a value
(41 answers)
Closed 2 years ago.
I was wondering if there is an efficient way to check if an element is present within an array in Bash? I am looking for something similar to what I can do in Python, like:
arr = ['a','b','c','d']
if 'd' in arr:
do your thing
else:
do something
I've seen solutions using associative array for bash for Bash 4+, but I am wondering if there is another solution out there.
Please understand that I know the trivial solution is to iterate in the array, but I don't want that.
You could do:
if [[ " ${arr[*]} " == *" d "* ]]; then
echo "arr contains d"
fi
This will give false positives for example if you look for "a b" -- that substring is in the joined string but not as an array element. This dilemma will occur for whatever delimiter you choose.
The safest way is to loop over the array until you find the element:
array_contains () {
local seeking=$1; shift
local in=1
for element; do
if [[ $element == "$seeking" ]]; then
in=0
break
fi
done
return $in
}
arr=(a b c "d e" f g)
array_contains "a b" "${arr[#]}" && echo yes || echo no # no
array_contains "d e" "${arr[#]}" && echo yes || echo no # yes
Here's a "cleaner" version where you just pass the array name, not all its elements
array_contains2 () {
local array="$1[#]"
local seeking=$2
local in=1
for element in "${!array}"; do
if [[ $element == "$seeking" ]]; then
in=0
break
fi
done
return $in
}
array_contains2 arr "a b" && echo yes || echo no # no
array_contains2 arr "d e" && echo yes || echo no # yes
For associative arrays, there's a very tidy way to test if the array contains a given key: The -v operator
$ declare -A arr=( [foo]=bar [baz]=qux )
$ [[ -v arr[foo] ]] && echo yes || echo no
yes
$ [[ -v arr[bar] ]] && echo yes || echo no
no
See 6.4 Bash Conditional Expressions in the manual.
Obvious caveats aside, if your array was actually like the one above, you could do
if [[ ${arr[*]} =~ d ]]
then
do your thing
else
do something
fi
Initialize array arr and add elements
set variable to search for SEARCH_STRING
check if your array contains element
arr=()
arr+=('a')
arr+=('b')
arr+=('c')
SEARCH_STRING='b'
if [[ " ${arr[*]} " == *"$SEARCH_STRING"* ]];
then
echo "YES, your arr contains $SEARCH_STRING"
else
echo "NO, your arr does not contain $SEARCH_STRING"
fi
If array elements don't contain spaces, another (perhaps more readable) solution would be:
if echo ${arr[#]} | grep -q -w "d"; then
echo "is in array"
else
echo "is not in array"
fi
array=("word" "two words") # let's look for "two words"
using grep and printf:
(printf '%s\n' "${array[#]}" | grep -x -q "two words") && <run_your_if_found_command_here>
using for:
(for e in "${array[#]}"; do [[ "$e" == "two words" ]] && exit 0; done; exit 1) && <run_your_if_found_command_here>
For not_found results add || <run_your_if_notfound_command_here>
As bash does not have a built-in value in array operator and the =~ operator or the [[ "${array[#]" == *"${item}"* ]] notation keep confusing me, I usually combine grep with a here-string:
colors=('black' 'blue' 'light green')
if grep -q 'black' <<< "${colors[#]}"
then
echo 'match'
fi
Beware however that this suffers from the same false positives issue as many of the other answers that occurs when the item to search for is fully contained, but is not equal to another item:
if grep -q 'green' <<< "${colors[#]}"
then
echo 'should not match, but does'
fi
If that is an issue for your use case, you probably won't get around looping over the array:
for color in "${colors[#]}"
do
if [ "${color}" = 'green' ]
then
echo "should not match and won't"
break
fi
done
for color in "${colors[#]}"
do
if [ "${color}" = 'light green' ]
then
echo 'match'
break
fi
done
Here's another way that might be faster, in terms of compute time, than iterating. Not sure. The idea is to convert the array to a string, truncate it, and get the size of the new array.
For example, to find the index of 'd':
arr=(a b c d)
temp=`echo ${arr[#]}`
temp=( ${temp%%d*} )
index=${#temp[#]}
You could turn this into a function like:
get-index() {
Item=$1
Array="$2[#]"
ArgArray=( ${!Array} )
NewArray=( ${!Array%%${Item}*} )
Index=${#NewArray[#]}
[[ ${#ArgArray[#]} == ${#NewArray[#]} ]] && echo -1 || echo $Index
}
You could then call:
get-index d arr
and it would echo back 3, which would be assignable with:
index=`get-index d arr`
FWIW, here's what I used:
expr "${arr[*]}" : ".*\<$item\>"
This works where there are no delimiters in any of the array items or in the search target. I didn't need to solve the general case for my applicaiton.

In Bash, how to convert number list into ranges of numbers?

Currently I have a sorted output of numbers from a command:
18,19,62,161,162,163,165
I would like to condense these number lists into a list of single numbers or ranges of numbers
18-19,62,161-163,165
I thought about trying to sort through the array in bash and read the next number to see if it is +1... I have a PHP function that does essentially the same thing, but I'm having trouble transposing it to Bash:
foreach ($missing as $key => $tag) {
$next = $missing[$key+1];
if (!isset($first)) {
$first = $tag;
}
if($next != $tag + 1) {
if($first == $tag) {
echo '<tr><td>'.$tag.'</td></tr>';
} else {
echo '<tr><td>'.$first.'-'.$tag.'</td></tr>';
}
unset($first);
}
}
I'm thinking there's probably a one-liner in bash that could do this but my Googling is coming up short....
UPDATE:
Thank you #Karoly Horvath for a quick answer which I used to finish my project. I'd sure be interested in any simpler solutions out there.
Yes, shell does variable substitution, if prev is not set, that line becomes:
if [ -ne $n+1]
Here is a working version:
numbers="18,19,62,161,162,163,165"
echo $numbers, | sed "s/,/\n/g" | while read num; do
if [[ -z $first ]]; then
first=$num; last=$num; continue;
fi
if [[ num -ne $((last + 1)) ]]; then
if [[ first -eq last ]]; then echo $first; else echo $first-$last; fi
first=$num; last=$num
else
: $((last++))
fi
done | paste -sd ","
18-19,62,161-163,165
Only with a function in bash:
#!/bin/bash
list2range() {
set -- ${#//,/ } # convert string to parameters
local first a b string IFS
local -a array
local endofrange=0
while [[ $# -ge 1 ]]; do
a=$1; shift; b=$1
if [[ $a+1 -eq $b ]]; then
if [[ $endofrange -eq 0 ]]; then
first=$a
endofrange=1
fi
else
if [[ $endofrange -eq 1 ]]; then
array+=($first-$a)
else
array+=($a)
fi
endofrange=0
fi
done
IFS=","; echo "${array[*]}"
}
list2range 18,19,62,161,162,163,165
Output:
18-19,62,161-163,165

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