Ignore copyright symbols inside URLs during copyright detection

src/cluecode/copyrights.py

@@ -4282,36 +4282,34 @@ def is_candidate(prepared_line):
         return False
 
     if is_only_digit_and_punct(prepared_line):
-        if TRACE:
-            logger_debug(f'is_candidate: is_only_digit_and_punct:\n{prepared_line!r}')
-
         return False
 
     if gibberish_detector.detect_gibberish(prepared_line):
         if TRACE:
-            logger_debug(f'is_candidate: gibberish_detector.detect_gibberish:\n{prepared_line!r}')
+            logger_debug(
+                f'is_candidate: gibberish_detector.detect_gibberish:\n{prepared_line!r}'
+            )
         return False
 
+    lowered = prepared_line.lower()   # ✅ DEFINE ONCE, ALWAYS
+
+    # Ignore lines where (c) appears only in URL-like text
+    if '(c)' in lowered and 'http' in lowered:
+        if not copyrights_hint.years(prepared_line):
+            for marker in copyrights_hint.statement_markers:
+                if marker != '(c)' and marker in lowered:
+                    break
+            else:
+                return False
+
     if copyrights_hint.years(prepared_line):
         return True
-    else:
-        pass
-    prepared_line = prepared_line.lower()
+
     for marker in copyrights_hint.statement_markers:
-        if marker in prepared_line:
+        if marker in lowered:
             return True
 
-
-def is_inside_statement(
-    chars_only_line,
-    markers=('copyright', 'copyrights', 'copyrightby',) + copyrights_hint.all_years,
-):
-    """
-    Return True if a line ends with some strings that indicate we are still
-    inside a statement.
-    """
-    return chars_only_line and chars_only_line.endswith(markers)
-
+    return False
 
 def is_end_of_statement(chars_only_line):
     """

tests/cluecode/test_copyrights_basic.py

@@ -14,7 +14,26 @@
 from commoncode.testcase import FileBasedTesting
 from cluecode import copyrights
 from cluecode.copyrights import prepare_text_line
-from cluecode.copyrights import remove_non_chars
+from cluecode import copyrights
+from cluecode.copyrights import prepare_text_line, remove_non_chars
+
+def test_copyright_symbol_inside_url_is_ignored():
+    text = "See http://example.com/(c)/path for more information."
+
+    prepped = prepare_text_line(text)
+
+    # sanity check
+    assert '(c)' in prepped
+
+    # URLs containing (c) should NOT be copyright candidates
+    assert not copyrights.is_candidate(prepped)
+
+def test_copyright_with_url_is_still_candidate():
+    text = "Copyright (c) 2000 Foo, http://example.com"
+
+    prepped = prepare_text_line(text)
+
+    assert copyrights.is_candidate(prepped)
 
 
 class TestTextPreparation(FileBasedTesting):