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Tweak sort order of tags in get-version.

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We want to sort such that the most recent/relevant tag is first
and gets used to set the compiled-in version.

The solution is far from general, but works for the tag formats
used by dnsmasq. v2.84 sorts before v2.83, but v2.83 sorts
before v2.83rc1 and 2.83rc1 sorts before v2.83test1
This commit is contained in:
Simon Kelley
2021-01-29 23:20:06 +00:00
parent 20295012b8
commit f1204a875e
1 changed files with 5 additions and 2 deletions
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7
bld/get-version
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@@ -9,7 +9,10 @@
# If we can find one which matches $v[0-9].* then we assume it's # If we can find one which matches $v[0-9].* then we assume it's
# a version-number tag, else we just use the whole string. # a version-number tag, else we just use the whole string.
# If there is more than one v[0-9].* tag, sort them and use the # If there is more than one v[0-9].* tag, sort them and use the
# first. This favours, eg v2.63 over 2.63rc6. # first. The insane arguments to the sort command are to ensure
# that, eg v2.64 comes before v2.63, but v2.63 comes before v2.63rc1
# and v2.63rc1 comes before v2.63test1
# Change directory to the toplevel source directory. # Change directory to the toplevel source directory.
if test -z "$1" || ! test -d "$1" || ! cd "$1"; then if test -z "$1" || ! test -d "$1" || ! cd "$1"; then
@@ -28,7 +31,7 @@ else
vers=`cat $1/VERSION | sed 's/[(), ]/,/ g' | tr ',' '\n' | grep ^v[0-9]` vers=`cat $1/VERSION | sed 's/[(), ]/,/ g' | tr ',' '\n' | grep ^v[0-9]`
if [ $? -eq 0 ]; then if [ $? -eq 0 ]; then
echo "${vers}" | sort -r | head -n 1 | sed 's/^v//' echo "${vers}" | sort -k1.2,1.5r -k 1.6,1.6 -k1.8,1.9r -k1.10,1.11r | head -n 1 | sed 's/^v//'
else else
cat $1/VERSION cat $1/VERSION
fi fi