Get latest version of Maven with curl - bash

Question

My task is to grab latest version of maven in bash script using curl https://apache.osuosl.org/maven/maven-3/
Output should be: 3.8.5
curl -s "https://apache.osuosl.org/maven/maven-3/" | grep -o "[0-9].[0-9].[0-9]" | tail -1
works for me ty

Answer 1

One way is to
Use grep to only get the lines containing the folder link
Use sed with an regex to get only the version number
Use sort to sort the lines
Use tail -n1 to get the last line
curl -s https://apache.osuosl.org/maven/maven-3/ \
| grep folder \
| gsed -E 's/.*([[:digit:]]\.[[:digit:]]\.[[:digit:]]).*/\1/' \
| sort \
| tail -n1
Output:
3.8.5

Answer 2

The url you're mentioning is an HTML-document and curl is not an HTML-parser, so I'd look into an HTML-parser, like xidel, if I were you:
$ xidel -s "https://apache.osuosl.org/maven/maven-3/" -e '//pre/a[last()]'
3.8.5/
$ xidel -s "https://apache.osuosl.org/maven/maven-3/" -e 'substring-before(//pre/a[last()],"/")'
3.8.5

Answer 3

I would suggest to use the central repository instead one of the apache maven sites because the information on central more maded for machine consumption.
It will be taken into account that the file maven-metadata.xml contains the appropriate information.
There are two options using the <version>..</version> tag and the <latest>..</latest> tag.
MVNVERSION=$(curl https://repo1.maven.org/maven2/org/apache/maven/maven/maven-metadata.xml \
| grep "<version>.*</version>" \
| tr -s " " \
| cut -d ">" -f2 \
| cut -d "<" -f1 \
| sort -V -r \
| head -1)
The second option:
MVNVERSION=$(curl https://repo1.maven.org/maven2/org/apache/maven/maven/maven-metadata.xml \
| grep "<latest>.*</latest>" \
| cut -d ">" -f2 \
| cut -d "<" -f1)