Javascript 使用下划线.js筛选数组

为了更好地理解JS，我正在尝试过滤一些对象，我正在使用下划线.JS

我来自C#背景，习惯LINQ，但下划线不太一样

你能帮我根据定义的测试筛选出这个数组吗？我遇到的问题是数组上的数组属性。

Where

运算符与我通常用于筛选项目的C不同

products = [
       { name: "Sonoma", ingredients: ["artichoke", "sundried tomatoes", "mushrooms"], containsNuts: false },
       { name: "Pizza Primavera", ingredients: ["roma", "sundried tomatoes", "goats cheese", "rosemary"], containsNuts: false },
       { name: "South Of The Border", ingredients: ["black beans", "jalapenos", "mushrooms"], containsNuts: false },
       { name: "Blue Moon", ingredients: ["blue cheese", "garlic", "walnuts"], containsNuts: true },
       { name: "Taste Of Athens", ingredients: ["spinach", "kalamata olives", "sesame seeds"], containsNuts: true }
    ];

it("given I'm allergic to nuts and hate mushrooms, it should find a pizza I can eat (functional)", function () {

      var productsICanEat = [];

      //This works but was hoping I could do the mushroom check as well in the same line
      var noNuts = _(products).filter(function (x) { return !x.containsNuts;});

      var noMushrooms = _(noNuts).reject(function(x){ return !_(x.ingredients).any(function(y){return y === "mushrooms";});});


      console.log(noMushrooms);

      var count = productsICanEat.length;
      expect(productsICanEat.length).toBe(count);
  });

---

您只需删除

拒绝

回调中选择code>，使其如下所示：

var noMushrooms = _(noNuts).reject(function(x){ 
    return _(x.ingredients).any(function(y){return y === "mushrooms";});
});

---

否则，您将拒绝不包含蘑菇的，而不是包含蘑菇的。

我设法将我的解决方案全部打包到一个筛选器调用中，因此我想发布它：

products = [
       { name: "Sonoma", ingredients: ["artichoke", "sundried tomatoes", "mushrooms"], containsNuts: false },
       { name: "Pizza Primavera", ingredients: ["roma", "sundried tomatoes", "goats cheese", "rosemary"], containsNuts: false },
       { name: "South Of The Border", ingredients: ["black beans", "jalapenos", "mushrooms"], containsNuts: false },
       { name: "Blue Moon", ingredients: ["blue cheese", "garlic", "walnuts"], containsNuts: true },
       { name: "Taste Of Athens", ingredients: ["spinach", "kalamata olives", "sesame seeds"], containsNuts: true }
    ];

 it("given I'm allergic to nuts and hate mushrooms, it should find a pizza I can eat (functional)", function () {

      var productsICanEat = [];

      productsICanEat = _(products).filter(function (x) { return !x.containsNuts && !_(x.ingredients).any(function(y){return y === "mushrooms";});});


      expect(productsICanEat.length).toBe(1);
  });

---

实现这一点的更简洁的方法是使用下划线的chain（）函数：

var noMushrooms = _(products).chain()
    .filter(function (x) { 
        return !x.containsNuts;})
    .reject(function(x){ 
        return _(x.ingredients).any(function(y){
            return y === "mushrooms";
        });
    })
    .value();

---

这将得到期望的结果

var no_nuts = _.filter(products,function(item) {
         return !item.containsNuts;
       });

var no_mushroom = _.reject(no_nuts,function(item) {
        return _.any(item.ingredients,function(item1) {
            return item1 === "mushrooms"
        }); 
     });

console.log(no_mushroom);

---

reject（）

与

filter（）

相反，而

any（）

相当于某种数组方法，当数组中的任何元素通过回调返回true时，该方法返回true。

DOH！谢谢你，我做到了。如果可能的话，你有什么想法来连锁吗？我已经试过了，但是幸运的是，你不需要先做

\uu.chain（list）

，然后你就可以整天把你的函数链起来了，就像这样

\u.chain（myArray）.filter（function（）{}）.reject（function（）{}）

看起来像是。any（）是去润滑的。下划线现在有一个u.some（）方法，根据是否有“某些”匹配项返回true或false。@skrile它不是不推荐的，

any

是的别名。谢谢@johnyhk-我今天学到了一些新东西：）您不需要链接，但通过这种链接方式可以非常清楚地知道您对列表执行的操作。