How to get the correct number of background jobs running, f.ex. $(jobs | wc -l | xargs) returns 1 instead of 0 - bash

Question

I have a jenkins build job that starts processes in the background. I need to write a function that checks wether there are still background processes running. To test it I came up with this:
#!/bin/bash -e
function waitForUploadFinish() {
runningJobs=$(jobs | wc -l | xargs)
echo "Waiting for ${runningJobs} background upload jobs to finish"
while [ "$(jobs | wc -l | xargs)" -ne 0 ];do
echo "$(jobs | wc -l | xargs)"
echo -n "." # no trailing newline
sleep 1
done
echo ""
}
for i in {1..3}
do
sleep $i &
done
waitForUploadFinish
The problem is it never comes down to 0. Even when the last sleep is done, there is still one job running?
mles-MBP:ionic mles$ ./jobs.sh
Waiting for 3 background upload jobs to finish
3
.2
.1
.1
.1
.1
.1
.1
Why I don't want to use wait here
In the Jenkins build job script where this snippet is for, i'm starting background upload processes for huge files. They don't run for 3 seconds like in the example here with sleep. They can take up to 30 minutes to proceed. If I use wait here, the user would see something like this in the log:
upload huge_file1.ipa &
upload huge_file2.ipa &
upload huge_file3.ipa &
wait
They would wonder why is nothing going on?
Instead I want to implement something like this:
upload huge_file1.ipa &
upload huge_file2.ipa &
upload huge_file3.ipa &
Waiting for 3 background upload jobs to finish
............
Waiting for 2 background upload jobs to finish
.................
Waiting for 1 background upload jobs to finish
.........
Upload done
That's why I need the loop with the current running background jobs.

Answer 1

This fixes it:
function waitForUploadFinish() {
runningJobs=$(jobs | wc -l | xargs)
echo "Waiting for ${runningJobs} background upload jobs to finish"
while [ `jobs -r | wc -l | tr -d " "` != 0 ]; do
jobs -r | wc -l | tr -d " "
echo -n "." # no trailing newline
sleep 1
done
echo ""
}
Note: you will only count the background processes that are started by this bash script, you will not see the background processes from the starting shell.
As the gniourf_gniourf commented: if you only need to wait and don't need to output then a simple wait after the sleeps is much simpler.
for i in {1..3}; do
sleep $i &
done
wait

Answer 2

Please consider comments made by gniourf_gniourf, as your design is not good to start with.
However, despite a much simpler and more efficient solution being possible, there is the question of why you are seeing what you are seeing.
I modified the first line of your loop, like so :
while [ "$(jobs | tee >(cat >&2) | wc -l | xargs)" -ne 0 ];do
The tee command takes its input and sends it to both standard out and to the file passed as argument. >(cat >&2) is syntax that, to explain it simply, provides a file to the tee command, but that file really is a named FIFO and anything written to that file will be sent to standard error, bypassing the pipeline and allowing you to see what jobs is spitting out, all while allowing the rest of the pipeline to operate normally.
If you do that, you will notice that the "job" that jobs keeps on returning is not a job, but a message stating some other job has finished. I do not know why it keeps on repeating that, but this is the cause of the problem.
You could replace :
while [ "$(jobs | wc -l | xargs)" -ne 0 ];do
With :
while [ "$(jobs -p | grep "^[0-9]" | wc -l | xargs)" -ne 0 ];do
This will cause jobs to echo PIDs, and filter out any line that does not begin with a number, so messages will not be counted.