I have to replace blank spaces (" ") of all the values from an array in shell script. So, I need to make an array like this:
$ array[0]="one"
$ array[1]="two three"
$ array[2]="four five"
looks like this:
$ array[0]="one"
$ array[1]="two!three"
$ array[2]="four!five"
Replacing every blank space to another character with a loop or something, not changing value by value.
array=('one' 'two three' 'four five') # initial assignment
array=( "${array[@]// /_}" ) # perform expansion on all array members at once
printf '%s\n' "${array[@]}" # print result
Bash shell supports a find and replace via substitution for string manipulation operation. The syntax is as follows:
${varName//Pattern/Replacement}
Replace all matches of Pattern with Replacement.
x=" This is a test "
## replace all spaces with * ####
echo "${x// /*}"
You should now be able tp simply loop through the array and replace the spaces woth whatever you want.
$!/bin/sh
array=('one' 'two three' 'four five')
for i in "${!array[@]}"
do
array[$i]=${array[$i]/ /_}
done
echo ${array[@]}
echo is unfortunate, especially with unquoted arguments. printf '%q\n' "${array[@]}" would be a more effective demonstration of where the boundaries between array elements are, since echo doesn't visually distinguish between an argument with a space and two separate arguments (and leaving out your quotes means that something that would otherwise have been one argument with a space gets transformed into multiple arguments during string-splitting phase).
array[0]=one, maybe?array[0] = oneisn't an assignment.