有效检索包含提交的版本 [英] Efficient retrieval of releases that contain a commit
问题描述
在命令行中，如果我输入
git tag --contains {commit}
获取包含给定提交的发布列表，每个发布需要大约11到20秒承诺。由于目标代码库存在超过300,000次提交，所以需要大量的检索所有提交的信息。
然而， gitk 显然设法检索这些数据做得很好。从我搜索的内容来看，它使用了缓存。
我有两个问题：
- 如何解释缓存格式？
- 有没有办法从
git命令行获取转储工具来生成相同的信息？
您可以直接从 git rev-list 。
latest.awk ：
BEGIN {thiscommit =; }
$ 1 ==commit{
if（thiscommit！=）
print thiscommit，tags [thiscommit]
thiscommit = $ 2
line [$ 2] = NR
latest = 0;如果（行[$ i]>最近）{
latest =行[$ i];如果（行[$ i]>最近）为$（i = 3; i <= NF; ++ i）
标签[$ 2] =标签[$ i];
}
next;
}
$ 1！=commit{tags [thiscommit] = $ 0; }
END {if（thiscommit！=）print thiscommit，tags [thiscommit]; }
示例命令：
git rev-list --date-order --children --format =％d --all | awk -f latest.awk
你也可以使用 - topo-order ，你可能必须清除 $ 1！=commit逻辑中的不需要的引用。
根据您想要的传递性以及清单的明确程度，累积标签可能需要字典。这里有一个明确列出所有提交的所有ref：
all.awk ：
BEGIN {
thiscommit =;
}
$ 1 ==commit{
if（thiscommit！=）
print thiscommit，tags [thiscommit]
thiscommit = $ 2
行[$ 2] = NR
split（，看过）;
for（i = 3; i <= NF; ++ i）{
nnew = split（tags [$ i]，new）;
for（n = 1; n <= nnew; ++ n）{
if（！seen [new [n]]）{
tags [$ 2] = tags [$ 2] new [n]
[new [n]] = 1
}
}
}
next;
}
$ 1！=commit{
nnew = split（$ 0，new，，）;
new [1] = substr（new [1]，3）;
new [nnew] = substr（new [nnew]，1，length（new [nnew]） - 1）;
for（n = 1; n <= nnew; ++ n）
tags [thiscommit] = tags [thiscommit]new [n]
}
END {if（thiscommit！=）print thiscommit，tags [thiscommit]; }
all.awk 花了几分钟要做322K的Linux内核仓库提交，大约每秒1000次或类似的事情（大量重复的字符串和冗余处理），所以如果你真的在完整的交叉产品之后，你可能想用C ++重写。 ..但我不认为gitk显示，只有最近的邻居，对不对？
In the command line, if I type
git tag --contains {commit}
to obtain a list of releases that contain a given commit, it takes around 11 to 20 seconds for each commit. Since the target code base there exists more than 300,000 commits, it would take a lot to retrieve this information for all commits.
However, gitk apparently manages to do a good job retrieving this data. From what I searched, it uses a cache for that purpose.
I have two questions:
- How can I interpret that cache format?
- Is there a way to obtain a dump from the
gitcommand line tool to generate that same information?
You can get this almost directly from git rev-list.
latest.awk:
BEGIN { thiscommit=""; }
$1 == "commit" {
if ( thiscommit != "" )
print thiscommit, tags[thiscommit]
thiscommit=$2
line[$2]=NR
latest = 0;
for ( i = 3 ; i <= NF ; ++i ) if ( line[$i] > latest ) {
latest = line[$i];
tags[$2] = tags[$i];
}
next;
}
$1 != "commit" { tags[thiscommit] = $0; }
END { if ( thiscommit != "" ) print thiscommit, tags[thiscommit]; }
a sample command:
git rev-list --date-order --children --format=%d --all | awk -f latest.awk
you can also use --topo-order, and you'll probably have to weed out unwanted refs in the $1!="commit" logic.
Depending on what kind of transitivity you want and how explicit the listing has to be, accumulating the tags might need a dictionary. Here's one that gets an explicit listing of all refs for all commits:
all.awk:
BEGIN {
thiscommit="";
}
$1 == "commit" {
if ( thiscommit != "" )
print thiscommit, tags[thiscommit]
thiscommit=$2
line[$2]=NR
split("",seen);
for ( i = 3 ; i <= NF ; ++i ) {
nnew=split(tags[$i],new);
for ( n = 1 ; n <= nnew ; ++n ) {
if ( !seen[new[n]] ) {
tags[$2]= tags[$2]" "new[n]
seen[new[n]] = 1
}
}
}
next;
}
$1 != "commit" {
nnew=split($0,new,", ");
new[1]=substr(new[1],3);
new[nnew]=substr(new[nnew],1,length(new[nnew])-1);
for ( n = 1; n <= nnew ; ++n )
tags[thiscommit] = tags[thiscommit]" "new[n]
}
END { if ( thiscommit != "" ) print thiscommit, tags[thiscommit]; }
all.awk took a few minutes to do the 322K linux kernel repo commits, about a thousand a second or something like that (lots of duplicate strings and redundant processing) so you'd probably want to rewrite that in C++ if you're really after the complete cross-product ... but I don't think gitk shows that, only the nearest neighbors, right?
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