处理只更改空格的块

Question

在我维护的代码中，我有时会收到pull请求，提交者只是在没有明显原因的情况下重排了一个段落。下面是一个示例：

diff --git a/knuth.tex b/knuth.tex
index 2f6a2f8..7b0827d 100644
--- a/knuth.tex
+++ b/knuth.tex
@@ -1,6 +1,6 @@
 Thus, I came to the conclusion that the designer of a new
 system must not only be the implementer and first
-large||scale user; the designer should also write the first
+large-scale user; the designer should also write the first
 user manual.

 The separation of any of these four components would have
@@ -9,8 +9,7 @@ all these activities, literally hundreds of improvements
 would never have been made, because I would never have
 thought of them or perceived why they were important.

-But a system cannot be successful if it is too strongly
-influenced by a single person. Once the initial design is
-complete and fairly robust, the real test begins as people
-with many different viewpoints undertake their own
-experiments.
+But a system cannot be successful if it is too strongly influenced by
+a single person. Once the initial design is complete and fairly
+robust, the real test begins as people with many different viewpoints
+undertake their own experiments.

正如您所看到的，第一个块引入了一个实际的更改，用-替换了||，而第二个块除了换行符和空格之外没有更改任何内容。事实上，第二个块的word-diff将是空的。

是否可以设置一个策略(例如在GitHub上或在我的CI中)来拒绝包含这种“空”块的提交，或者重新格式化补丁以完全忽略这些块，以便我可以在没有它们的情况下git apply它？

相关：How to git-apply a git word diff

Answer 1

如果你正在寻找一个内置的解决方案，我不知道。然而，这并不意味着这不能相对容易地构建到CI系统中。

您可以将适当的git diff命令的输出通过管道传输到如下脚本中，如果输入包含如上第二个块的补丁，则该脚本将退出1。

#!/usr/bin/env ruby

def filter(arr)
  arr.join.split("\n\n").map { |x| x.gsub(/\s+/, ' ') }.join("\n\n")
end

def should_reject(before, after)
  return false if before.empty? && after.empty?
  before = filter(before)
  after = filter(after)
  return true if before == after
  false
end

chunk = nil
before = []
after = []
while (line = gets)
  trimmed = line[1..-1]
  case line
  when /^(\+\+\+|---)/
    # Do nothing.
  when /^@@ /
    if should_reject(before, after)
      warn "Useless change to hunk #{chunk}"
      exit 1
    end
    chunk = line
    before = []
    after = []
  when /^ /
    before << trimmed
    after << trimmed
  when /^\+/
    after << trimmed
  when /^-/
    before << trimmed
  end
end

if should_reject(before, after)
  warn "Useless change to hunk #{chunk}"
  exit 1
end

它基本上将每个块拆分为块，块之间有一个空行，将所有空格转换为空格，然后对它们进行比较。如果它们相等，它会报错并退出非零。您可能希望将其修改为更健壮，就像处理CRLF结尾之类的，但这种方法是可行的。

顺便说一句，让这类更改更容易发现的一种方法是使用逐行语句样式。每个句子，无论长度如何，都在一行上，并且每行只有一个句子。这使得区分任何类型的更改变得容易得多，并完全避免了包装问题。