我有这样的方法:
codes.each do |code|
company = Company.find_or_create_by(code: code)
company.foo = some_value
company.bar = some_value2
company.save
end
为了加快速度,我想用update_all
codes.each do |code|
Company.find_or_create_by(code: code)
Company.where(code: code).update_all(foo: some_value, bar: some_value2)
end
但每次执行SQL命令时find_or_create_by
都会运行。
有没有办法一次创建多个模型实例?
我想写成Company.create_all_if_not_exist(code: codes)
。
答案 0 :(得分:0)
我看到了更好的方法:
existing_companies = Company.where(code: codes)
existing_codes = existing_companies.pluck(:code) # +1 request
non_existing_codes = codes - existing_codes
attrs = {foo: some_value, bar: some_value2}
existing_companies.update_all(attrs) # +1 request
# + (codes - existing_codes) requests
non_existing_codes.each do |code|
Company.create({code: code}.merge(attrs))
end