Move related pipeline class to new pipeline module

4 files changed, 147 insertions, 145 deletions

diff --git a/app/models/ci/pipeline.rb b/app/models/ci/pipeline.rb
index acaa028eaa2..3d5acc00f8f 100644
--- a/app/models/ci/pipeline.rb
+++ b/app/models/ci/pipeline.rb
@@ -434,7 +434,7 @@ module Ci
     def update_duration
       return unless started_at
 
-      self.duration = Gitlab::Ci::PipelineDuration.from_pipeline(self)
+      self.duration = Gitlab::Ci::Pipeline::Duration.from_pipeline(self)
     end
 
     def execute_hooks
diff --git a/lib/gitlab/ci/pipeline/duration.rb b/lib/gitlab/ci/pipeline/duration.rb
new file mode 100644
index 00000000000..469fc094cc8
--- /dev/null
+++ b/lib/gitlab/ci/pipeline/duration.rb
@@ -0,0 +1,143 @@
+module Gitlab
+  module Ci
+    module Pipeline
+      # # Introduction - total running time
+      #
+      # The problem this module is trying to solve is finding the total running
+      # time amongst all the jobs, excluding retries and pending (queue) time.
+      # We could reduce this problem down to finding the union of periods.
+      #
+      # So each job would be represented as a `Period`, which consists of
+      # `Period#first` as when the job started and `Period#last` as when the
+      # job was finished. A simple example here would be:
+      #
+      # * A (1, 3)
+      # * B (2, 4)
+      # * C (6, 7)
+      #
+      # Here A begins from 1, and ends to 3. B begins from 2, and ends to 4.
+      # C begins from 6, and ends to 7. Visually it could be viewed as:
+      #
+      #     0  1  2  3  4  5  6  7
+      #        AAAAAAA
+      #           BBBBBBB
+      #                       CCCC
+      #
+      # The union of A, B, and C would be (1, 4) and (6, 7), therefore the
+      # total running time should be:
+      #
+      #     (4 - 1) + (7 - 6) => 4
+      #
+      # # The Algorithm
+      #
+      # The algorithm used here for union would be described as follow.
+      # First we make sure that all periods are sorted by `Period#first`.
+      # Then we try to merge periods by iterating through the first period
+      # to the last period. The goal would be merging all overlapped periods
+      # so that in the end all the periods are discrete. When all periods
+      # are discrete, we're free to just sum all the periods to get real
+      # running time.
+      #
+      # Here we begin from A, and compare it to B. We could find that
+      # before A ends, B already started. That is `B.first <= A.last`
+      # that is `2 <= 3` which means A and B are overlapping!
+      #
+      # When we found that two periods are overlapping, we would need to merge
+      # them into a new period and disregard the old periods. To make a new
+      # period, we take `A.first` as the new first because remember? we sorted
+      # them, so `A.first` must be smaller or equal to `B.first`. And we take
+      # `[A.last, B.last].max` as the new last because we want whoever ended
+      # later. This could be broken into two cases:
+      #
+      #     0  1  2  3  4
+      #        AAAAAAA
+      #           BBBBBBB
+      #
+      # Or:
+      #
+      #     0  1  2  3  4
+      #        AAAAAAAAAA
+      #           BBBB
+      #
+      # So that we need to take whoever ends later. Back to our example,
+      # after merging and discard A and B it could be visually viewed as:
+      #
+      #     0  1  2  3  4  5  6  7
+      #        DDDDDDDDDD
+      #                       CCCC
+      #
+      # Now we could go on and compare the newly created D and the old C.
+      # We could figure out that D and C are not overlapping by checking
+      # `C.first <= D.last` is `false`. Therefore we need to keep both C
+      # and D. The example would end here because there are no more jobs.
+      #
+      # After having the union of all periods, we just need to sum the length
+      # of all periods to get total time.
+      #
+      #     (4 - 1) + (7 - 6) => 4
+      #
+      # That is 4 is the answer in the example.
+      module Duration
+        extend self
+
+        Period = Struct.new(:first, :last) do
+          def duration
+            last - first
+          end
+        end
+
+        def from_pipeline(pipeline)
+          status = %w[success failed running canceled]
+          builds = pipeline.builds.latest
+            .where(status: status).where.not(started_at: nil).order(:started_at)
+
+          from_builds(builds)
+        end
+
+        def from_builds(builds)
+          now = Time.now
+
+          periods = builds.map do |b|
+            Period.new(b.started_at, b.finished_at || now)
+          end
+
+          from_periods(periods)
+        end
+
+        # periods should be sorted by `first`
+        def from_periods(periods)
+          process_duration(process_periods(periods))
+        end
+
+        private
+
+        def process_periods(periods)
+          return periods if periods.empty?
+
+          periods.drop(1).inject([periods.first]) do |result, current|
+            previous = result.last
+
+            if overlap?(previous, current)
+              result[-1] = merge(previous, current)
+              result
+            else
+              result << current
+            end
+          end
+        end
+
+        def overlap?(previous, current)
+          current.first <= previous.last
+        end
+
+        def merge(previous, current)
+          Period.new(previous.first, [previous.last, current.last].max)
+        end
+
+        def process_duration(periods)
+          periods.sum(&:duration)
+        end
+      end
+    end
+  end
+end
diff --git a/lib/gitlab/ci/pipeline_duration.rb b/lib/gitlab/ci/pipeline_duration.rb
deleted file mode 100644
index 3208cc2bef6..00000000000
--- a/lib/gitlab/ci/pipeline_duration.rb
+++ /dev/null
@@ -1,141 +0,0 @@
-module Gitlab
-  module Ci
-    # # Introduction - total running time
-    #
-    # The problem this module is trying to solve is finding the total running
-    # time amongst all the jobs, excluding retries and pending (queue) time.
-    # We could reduce this problem down to finding the union of periods.
-    #
-    # So each job would be represented as a `Period`, which consists of
-    # `Period#first` as when the job started and `Period#last` as when the
-    # job was finished. A simple example here would be:
-    #
-    # * A (1, 3)
-    # * B (2, 4)
-    # * C (6, 7)
-    #
-    # Here A begins from 1, and ends to 3. B begins from 2, and ends to 4.
-    # C begins from 6, and ends to 7. Visually it could be viewed as:
-    #
-    #     0  1  2  3  4  5  6  7
-    #        AAAAAAA
-    #           BBBBBBB
-    #                       CCCC
-    #
-    # The union of A, B, and C would be (1, 4) and (6, 7), therefore the
-    # total running time should be:
-    #
-    #     (4 - 1) + (7 - 6) => 4
-    #
-    # # The Algorithm
-    #
-    # The algorithm used here for union would be described as follow.
-    # First we make sure that all periods are sorted by `Period#first`.
-    # Then we try to merge periods by iterating through the first period
-    # to the last period. The goal would be merging all overlapped periods
-    # so that in the end all the periods are discrete. When all periods
-    # are discrete, we're free to just sum all the periods to get real
-    # running time.
-    #
-    # Here we begin from A, and compare it to B. We could find that
-    # before A ends, B already started. That is `B.first <= A.last`
-    # that is `2 <= 3` which means A and B are overlapping!
-    #
-    # When we found that two periods are overlapping, we would need to merge
-    # them into a new period and disregard the old periods. To make a new
-    # period, we take `A.first` as the new first because remember? we sorted
-    # them, so `A.first` must be smaller or equal to `B.first`. And we take
-    # `[A.last, B.last].max` as the new last because we want whoever ended
-    # later. This could be broken into two cases:
-    #
-    #     0  1  2  3  4
-    #        AAAAAAA
-    #           BBBBBBB
-    #
-    # Or:
-    #
-    #     0  1  2  3  4
-    #        AAAAAAAAAA
-    #           BBBB
-    #
-    # So that we need to take whoever ends later. Back to our example,
-    # after merging and discard A and B it could be visually viewed as:
-    #
-    #     0  1  2  3  4  5  6  7
-    #        DDDDDDDDDD
-    #                       CCCC
-    #
-    # Now we could go on and compare the newly created D and the old C.
-    # We could figure out that D and C are not overlapping by checking
-    # `C.first <= D.last` is `false`. Therefore we need to keep both C
-    # and D. The example would end here because there are no more jobs.
-    #
-    # After having the union of all periods, we just need to sum the length
-    # of all periods to get total time.
-    #
-    #     (4 - 1) + (7 - 6) => 4
-    #
-    # That is 4 is the answer in the example.
-    module PipelineDuration
-      extend self
-
-      Period = Struct.new(:first, :last) do
-        def duration
-          last - first
-        end
-      end
-
-      def from_pipeline(pipeline)
-        status = %w[success failed running canceled]
-        builds = pipeline.builds.latest
-          .where(status: status).where.not(started_at: nil).order(:started_at)
-
-        from_builds(builds)
-      end
-
-      def from_builds(builds)
-        now = Time.now
-
-        periods = builds.map do |b|
-          Period.new(b.started_at, b.finished_at || now)
-        end
-
-        from_periods(periods)
-      end
-
-      # periods should be sorted by `first`
-      def from_periods(periods)
-        process_duration(process_periods(periods))
-      end
-
-      private
-
-      def process_periods(periods)
-        return periods if periods.empty?
-
-        periods.drop(1).inject([periods.first]) do |result, current|
-          previous = result.last
-
-          if overlap?(previous, current)
-            result[-1] = merge(previous, current)
-            result
-          else
-            result << current
-          end
-        end
-      end
-
-      def overlap?(previous, current)
-        current.first <= previous.last
-      end
-
-      def merge(previous, current)
-        Period.new(previous.first, [previous.last, current.last].max)
-      end
-
-      def process_duration(periods)
-        periods.sum(&:duration)
-      end
-    end
-  end
-end
diff --git a/spec/lib/gitlab/ci/pipeline_duration_spec.rb b/spec/lib/gitlab/ci/pipeline/duration_spec.rb
index b26728a843c..7c9836e2da6 100644
--- a/spec/lib/gitlab/ci/pipeline_duration_spec.rb
+++ b/spec/lib/gitlab/ci/pipeline/duration_spec.rb
@@ -1,6 +1,6 @@
 require 'spec_helper'
 
-describe Gitlab::Ci::PipelineDuration do
+describe Gitlab::Ci::Pipeline::Duration do
   let(:calculated_duration) { calculate(data) }
 
   shared_examples 'calculating duration' do
@@ -107,9 +107,9 @@ describe Gitlab::Ci::PipelineDuration do
 
   def calculate(data)
     periods = data.shuffle.map do |(first, last)|
-      Gitlab::Ci::PipelineDuration::Period.new(first, last)
+      described_class::Period.new(first, last)
     end
 
-    Gitlab::Ci::PipelineDuration.from_periods(periods.sort_by(&:first))
+    described_class.from_periods(periods.sort_by(&:first))
   end
 end