小编vis*_*nth的帖子

如何使用Javascript或jQuery过滤JSON数据？

这是我的JSON数据:

[{"name":"Lenovo Thinkpad 41A4298","website":"google"},
{"name":"Lenovo Thinkpad 41A2222","website":"google"},
{"name":"Lenovo Thinkpad 41Awww33","website":"yahoo"},
{"name":"Lenovo Thinkpad 41A424448","website":"google"},
{"name":"Lenovo Thinkpad 41A429rr8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ff8","website":"ebay"},
{"name":"Lenovo Thinkpad 41A429ss8","website":"rediff"},
{"name":"Lenovo Thinkpad 41A429sg8","website":"yahoo"}]

JavaScript的:

obj1 = JSON.parse(jsondata);

现在我只想要包含网站的名称和网站数据等于"雅虎"

我需要input一个div没有表单的框,当用户输入内容并返回时,它应该运行Javascript函数.

没有提交按钮.

我怎样才能做到这一点？

每次我尝试在Windows中生成Rails控制器时，都会在生成必要文件后启动服务器。

$ rails g controller supprefresh index process                                                                
io/console not supported; tty will not be manipulated                                                         
C:/Rubies/jruby-1.7.9/lib/ruby/shared/Win32API.rb:11 warning: warning: variable $KCODE is no longer effective 
      create  app/controllers/supprefresh_controller.rb                                                       
       route  get "supprefresh/process"                                                                       
       route  get "supprefresh/index"                                                                         
      invoke  erb                                                                                             
      create    app/views/supprefresh                                                                         
      create    app/views/supprefresh/index.html.erb                                                          
      create    app/views/supprefresh/process.html.erb                                                        
      invoke  rspec                                                                                           
      create    spec/controllers/supprefresh_controller_spec.rb                                               
      create    spec/views/supprefresh                                                                        
      create    spec/views/supprefresh/index.html.erb_spec.rb                                                 
      create    spec/views/supprefresh/process.html.erb_spec.rb                                               
      invoke  helper                                                                                          
      create    app/helpers/supprefresh_helper.rb                                                             
      invoke    rspec                                                                                         
      create      spec/helpers/supprefresh_helper_spec.rb                                                     
      invoke  assets                                                                                          
      invoke    coffee                                                                                        
      create      app/assets/javascripts/supprefresh.js.coffee                                                
      invoke    scss                                                                                          
      create      app/assets/stylesheets/supprefresh.css.scss                                                 
[2019-04-09 16:18:34] INFO  WEBrick 1.3.1                                                                     
[2019-04-09 16:18:34] INFO  ruby 2.0.0 (2013-12-06) …

这是红宝石中多级继承的示例，这里有3个类A，B和C。B继承自A，C继承自B，因此最后，类C具有A，B和C的所有方法。

class A
  def hello_by_a
    puts "A says hello"
  end
end

class B < A
  def hello_by_b
    puts "B says hello"
  end
end

class C < B
  def hello_by_c
    puts "C says hello"
  end
end

c = C.new
c.hello_by_a #=> A says hello
c.hello_by_b #=> B says hello
c.hello_by_c #=> C says hello

p c.methods-Object.methods #=> [:hello_by_c, :hello_by_b, :hello_by_a]

这和mixin一样，这里不是类A和B，而是包含了模块A和B，它们已包含在类C中。现在，类C具有所有3种方法

module A
  def hello_by_a
    puts "A says hello"
  end
end

module B
  def hello_by_b
    puts "B says hello"
  end
end …

请考虑下表:

SELECT id, Bill_Freq, Paid_From, Paid_To, Paid_Dt, rev_code FROM psr_20160708091408;

要求是获取已rev_code填充字符串**SUM**的行.我还注意到,对于rev_code填充为**SUM**的每一行,它Bill_Freq都不会为null或零.

所以我写了两个查询来获取id最低的行

基于where子句中的字符串检查查询:

select
        min(id) as head_id,
        bill_freq,
        Paid_From,
        Paid_To,
        Paid_Dt
from
    `psr_20160708091408` where rev_code = "**SUM**";

基于真实条件的查询:

select
    min(id) as head_id,
        bill_freq,
        Paid_From,
        Paid_To,
        Paid_Dt
from
    `psr_20160708091408` where bill_freq;

我没有看到有人使用第二种类型,想知道它的可靠性和失败的情况.

我需要解析2d数组的以下哈希，其中第一个数组具有键，其余数组具有值。

input = {
  "result": [
    [
      "id",
      "name",
      "address"
    ],
    [
      "1",
      "Vishnu",
      "abc"
    ],
    [
      "2",
      "Arun",
      "def"
    ],
    [
      "3",
      "Arjun",
      "ghi"
    ]
  ]
}

这是我想出的结果。

input[:result].drop(1).collect{|arr| Hash[input[:result].first.zip arr]}

在这里，我在遍历result数组时忽略了它的第一个子数组（包含键的那个），然后zip是键数组和值数组以进行哈希处理，然后将哈希收集到另一个数组中。

上面的解决方案给我我想要的是一个哈希

[{"id"=>"1", "name"=>"Vishnu", "address"=>"abc"}, {"id"=>"2", "name"=>"Arun", "address"=>"def"}, {"id"=>"3", "name"=>"Arjun", "address"=>"ghi"}]

有没有更好的方法来达到相同的结果？

在ruby中，我们可以Object.methods获取特定Object或类的所有方法。

例如：

irb(main):001:0> Object.methods
=> [:new, :allocate, :superclass, :<=>, :include, :<=, :>=, :==, :===, :included_modules, :include?, :name, :ancestors, :instance_methods, :public_instance_methods, :protected_instance_methods, :private_instance_methods, :constants, :const_get, :const_set, :const_defined?, :class_variables, :remove_class_variable, :class_variable_get, :class_variable_set, :class_variable_defined?, :public_constant, :private_constant, :deprecate_constant, :singleton_class?, :module_exec, :class_exec, :freeze, :inspect, :const_missing, :class_eval, :method_defined?, :public_method_defined?, :prepend, :<, :>, :private_method_defined?, :protected_method_defined?, :public_class_method, :module_eval, :to_s, :private_class_method, :autoload, :autoload?, :instance_method, :public_instance_method, :instance_of?, :kind_of?, :is_a?, :tap, :public_send, :remove_instance_variable, :instance_variable_set, :method, :public_method, :singleton_method, :extend, :define_singleton_method, :to_enum, :enum_for, :=~, :!~, :eql?, :respond_to?, :object_id, :send, :display, :nil?, …

irb(main):007:0> %w[1 2 3 4 5]&.each { |a| puts a }
1
2
3
4
5
=> ["1", "2", "3", "4", "5"]
irb(main):008:0> %w[1 2 3 4 5].each { |a| puts a }
1
2
3
4
5
=> ["1", "2", "3", "4", "5"]

双方&.each并.each似乎给了相同的结果

ruby-doc似乎没有关于此功能的任何内容

两者有什么区别？

我在 jruby irb 上玩，遇到了这种现象，其中参数方法[[:rest]]在字符串方法上调用时返回。这不仅是字符串的情况，而且我将举一个关于字符串的例子。

irb(main):042:0> String.new.methods-Object.methods
[:valid_encoding?, :casecmp, :to_java_bytes, :squeeze!,
 :is_utf8?, :slice, :hex, :[]=, :initialize_copy, :empty?,
 :oct, :rindex, :unseeded_hash, :%, :rjust, :chop, :index,
 :gsub!, :chomp!, :*, :+, :concat, :capitalize, :singularize,
 :titlecase, :each_line, :size, :deconstantize, :downcase!,
 :capitalize!, :to_sym, :humanize, :setbyte, :force_encoding,
 :sub, :reverse!, :swapcase, :scan, :gsub, :sum, :partition,
 :to_str, :codepoints, :swapcase!, :byteslice, :end_with?,
 :upto, :tr!, :[], :intern, :parameterize, :tableize, :chomp,
 :pluralize, :reverse, :mb_chars, :succ, :underscore, :titleize,
 :start_with?, :ljust, :tr, :chars, :chop!, :encode, :<<,
 :lstrip!, :dasherize, :prepend, :replace, :strip, :split, …

这是Author模型和Book模型

class Author < ApplicationRecord
  has_many :books, dependent: :destroy
end

class Book < ApplicationRecord
  belongs_to :author
end

我创建了2位作者（author1和author2），并为每位作者添加了2本书。

之后，我删除了author1并创建了一个新作者（author3）。现在，我想给author3两本author1的书。

是否有ActiveRecord方法用新作者来更新书籍author_id？