trying to use variable in bash-curl, getting curl: (3) <url> malformed - bash

Question

we get error on trying to use variable in curl url block, how i can fix it?
variable is
NEXUS_UPLOAD='http://mynexus.com/nexus/content/repositories/releases/ru/123/project/1.0/1.0.289/'
#!/bin/bash -v
for i in $(cat upload.list); do
content=$(curl -v -u "$NEXUS_USER":"$NEXUS_USER_PASSWORD" --upload-file "$i" "${NEXUS_UPLOAD}")
echo $content
done
error is:
* <url> malformed
* Closing connection -1
curl: (3) <url> malformed
if we have direct link in this script, curl worked correctly
#!/bin/bash -v
for i in $(cat upload.list); do
content=$(curl -v -u "$NEXUS_USER":"$NEXUS_USER_PASSWORD" --upload-file "$i" http://mynexus.com/nexus/content/repositories/releases/ru/123/project/1.0/1.0.289/)
echo $content
done
using just $NEXUS_UPLOAD without quotes getting - curl: no URL specified!
also I try to use "$NEXUS_UPLOAD" in script, but still get malformed error

Answer 1

Remove the curly brackets around NEXUS_UPLOAD and use -x to enable the trace option, and don't forget to "export" the variables.
You can see a list of exported variables with declare -p
#!/bin/bash -x
for i in $(cat upload.list); do
content=$(curl -v -u "$NEXUS_USER":"$NEXUS_USER_PASSWORD" --upload-file "$i" "$NEXUS_UPLOAD")
echo $content
done
Here's a link explaining when they are useful.
For example:
i=image.jpg
convert $i ${i%jpg}png