Merge pull request #14518 from Bo98/api-license

Fix license handling for API formulae

Library/Homebrew/formulary.rb

@@ -151,7 +151,7 @@ module Formulary
     klass = Class.new(::Formula) do
       desc json_formula["desc"]
       homepage json_formula["homepage"]
-      license json_formula["license"]
+      license SPDX.string_to_license_expression(json_formula["license"])
       revision json_formula["revision"]
       version_scheme json_formula["version_scheme"]
 

Library/Homebrew/test/utils/spdx_spec.rb

@@ -217,6 +217,31 @@ describe SPDX do
     end
   end
 
+  describe ".string_to_license_expression" do
+    it "returns the correct result for 'and', 'or' and 'with'" do
+      expr_string = "Apache-2.0 and (Apache-2.0 with LLVM-exception) and (MIT or NCSA)"
+      expect(described_class.string_to_license_expression(expr_string)).to eq({
+        all_of: [
+          "Apache-2.0",
+          { "Apache-2.0" => { with: "LLVM-exception" } },
+          { any_of: ["MIT", "NCSA"] },
+        ],
+      })
+    end
+
+    it "handles nested brackets" do
+      expect(described_class.string_to_license_expression("A and (B or (C and D))")).to eq({
+        all_of: [
+          "A",
+          any_of: [
+            "B",
+            all_of: ["C", "D"],
+          ],
+        ],
+      })
+    end
+  end
+
   describe ".license_version_info" do
     it "returns license without version" do
       expect(described_class.license_version_info("MIT")).to eq ["MIT"]

Library/Homebrew/utils/spdx.rb

@@ -129,6 +129,40 @@ module SPDX
     end
   end
 
+  def string_to_license_expression(string)
+    return if string.blank?
+
+    result = string
+    result_type = nil
+
+    and_parts = string.split(/ and (?![^(]*\))/)
+    if and_parts.length > 1
+      result = and_parts
+      result_type = :all_of
+    else
+      or_parts = string.split(/ or (?![^(]*\))/)
+      if or_parts.length > 1
+        result = or_parts
+        result_type = :any_of
+      end
+    end
+
+    if result_type
+      result.map! do |part|
+        part = part[1..-2] if part[0] == "(" && part[-1] == ")"
+        string_to_license_expression(part)
+      end
+      { result_type => result }
+    else
+      with_parts = string.split(" with ", 2)
+      if with_parts.length > 1
+        { with_parts.first => { with: with_parts.second } }
+      else
+        result
+      end
+    end
+  end
+
   def license_version_info(license)
     return [license] if ALLOWED_LICENSE_SYMBOLS.include? license