Ensure there won't be any regression with where(nil) calls

Consider this scenario:

    if params[:foo]
      conditions = { foo: true }
    end

    foos = Foo.where(conditions).order(:id)

When params[:foo] is nil, this would call:

    foos = Foo.where(nil).order(:id)

In this scenario, we want Foo.where(conditions) to be the same as calling
Foo.all, otherwise we'd get a "NoMethodError order for WhereChain".

Related to rails#8332.

activerecord/lib/active_record/relation/query_methods.rb

@@ -452,8 +452,8 @@ def bind!(value) # :nodoc:
     #
     # If the condition is any other blank-ish object than nil, then where is a # no-op and returns
     # the current relation.
-    def where(opts = nil, *rest)
-      if opts.nil?
+    def where(opts = :chain, *rest)
+      if opts == :chain
         WhereChain.new(spawn)
       elsif opts.blank?
         self
@@ -464,8 +464,8 @@ def where(opts = nil, *rest)
 
     # #where! is identical to #where, except that instead of returning a new relation, it adds
     # the condition to the existing relation.
-    def where!(opts = nil, *rest) # :nodoc:
-      if opts.nil?
+    def where!(opts = :chain, *rest) # :nodoc:
+      if opts == :chain
         WhereChain.new(self)
       else
         references!(PredicateBuilder.references(opts)) if Hash === opts

activerecord/test/cases/associations/has_many_associations_test.rb

@@ -299,7 +299,7 @@ def test_finding_array_compatibility
   end
 
   def test_find_with_blank_conditions
-    [[], {}, ""].each do |blank|
+    [[], {}, nil, ""].each do |blank|
       assert_equal 2, Firm.all.merge!(:order => "id").first.clients.where(blank).to_a.size
     end
   end