I am trying to sort all values of a list and add them, sorted, into another list . The sort -nu works fine but he doesn't want to add $i to the new SortedList. What could be the problem here?
function Sort() {
SortedList=""
for i in $list;
do
echo $i
$SortedList= "$SortedList $i"
done | sort -nu
}
echo " This is the sorted list : $SortedList"
I think you can do something like the following.
#!/bin/bash
list="7 4 2 5 3"
function Sort() {
SortedList=$(echo $list | tr " " "\n" | sort -nu)
}
Sort
echo $SortedList
It may fail because we still don't know how your list looks like.
SortedList=$(tr " " "\n" <<<"${list}" | sort -nu).Your solution tries to store the list items to the SortedList before the sort itself. The echo $i part sends the list to the pipe and the sort -nu then prints it (sorted) to the STDOUT. If you want to store the sorted list to the variable, just try this:
function Sort() {
SortedList=`for i in $list; do echo $i; done | sort -nu`
}
If you're wondering why the SortedList variable is empty after the function call, the problem is the pipe to the sort command. Without the pipe it works fine. From wiki:
In the most commonly used simple pipelines the shell connects a series of sub-processes via pipes, and executes external commands within each sub-process. Thus the shell itself is doing no direct processing of the data flowing through the pipeline.
That means the for part of the pipe starts a new sub-process (with the sorted var initialized), that changes the sorted variable as it is supposed to. But, when the process ends, you are back in the environment with the initial value of the variable.
Also, consult this post (Set a parent shell's variable from a subshell)
$SortedList= "$SortedList $i"should beSortedList="$SortedList $i". A $ and a space less. But I guess that wouldn't be enough.SortedListcould be sorted at the end... The pipe won't change the iteration order.sortcommand. Without the pipe it works fine. I suspect the for command is due to the pipe executed in different shell.