比较2个对象以在javascript中创建对象的新数组

Question

我有两个要素：水果和板条箱

fruits是一个包含不同水果列表的数组，例如：

["apple","orange","mango","pear"]

crates是一系列包含水果的对象，例如：

[{id:1,fruit_name: "apple"},{id:2, fruit_name: "pear"}]

我想根据以下条件创建一个新的对象数组：-检查水果数组中的实体是否存在于任何包装箱数组中。

如果存在，则最终对象应该具有fruit_name属性以及另一个称为tested属性设置为true 。 如果不存在，那么tested应该是false ;

考虑以上情况，最终输出应如下：

[
    {fruit_name: "apple", tested: true},
    {fruit_name: "orange", tested: false},
    {fruit_name: "mango", tested: false},
    {fruit_name: "pear", tested: true},
]

我尝试过如下：

fruits.forEach(x => {
                        crates.filter((y) => {
                            let temp;
                            if (x == y.fruit_name) {
                                temp = {
                                    fruit: x,
                                    tested: true
                                }
                            }
                            else {
                                temp = {
                                    time: x,
                                    tested: false
                                }
                            }
                            testedCrates.push(temp);
                        })
                    })

这样做的问题是，它会将每个水果两次返回，并同时包含两个用于测试属性的值。

我得到的输出如下：

[
    {fruit: "apple", tested: true}, 
    {time: "apple", tested: false},
    {time: "orange", tested: false},
    {time: "orange", tested: false},
    {time: "mango", tested: false},
    {time: "mango", tested: false},
    {time: "pear", tested: false},
    {time: "pear", tested: false}
]

请为此提供一些解决方案。 此外，如果有更好的方法使用es6方法，将很有帮助。 提前致谢。

Answer 1

您可以使用array#map和array#some检查crates是否存在水果。

 var fruits = ["apple","orange","mango","pear"], crates = [{id:1,fruit_name: "apple"},{id:2, fruit_name: "pear"}]; var result = fruits.map(fruit => { var tested = crates.some(({fruit_name}) => fruit_name === fruit); return {'fruit_name' : fruit, tested}; }); console.log(result); 

Answer 2

创建一组存在于板条箱中的水果，然后将水果数组映射到所需的形式：

 const fruits = ["apple","orange","mango","pear"] const crates = [{id:1,fruit_name: "apple"},{id:2, fruit_name: "pear"}]; const cratesSet = new Set(crates.map(({ fruit_name }) => fruit_name)); const result = fruits.map((fruit_name) => ({ fruit_name, tests: cratesSet.has(fruit_name) })); console.log(result); 

Answer 3

您可以在检查集合的同时为crates Set ，并为新数组迭代fruits 。

 var fruits = ["apple", "orange", "mango", "pear"], crates = [{ id: 1, fruit_name: "apple" }, { id: 2, fruit_name: "pear" }], cSet = new Set(crates.map(({ fruit_name }) => fruit_name)), result = fruits.map(fruit_name => ({ fruit_name, tested: cSet.has(fruit_name) })); console.log(result); 

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Answer 4

fruits.map(fruit => ({
  fruit_name: fruit,
  tested: crates.some(crate => crate.fruit === fruit),
}));

Answer 5

fruits.map(fruit=>({
    fruit_name: fruit,
    tested: crates.some(x=>x.fruit_name === fruit)
}))

Answer 6

fruits.forEach(x => {
    let temp = { fruit: x, tested: false }

    crates.filter(y => {
        if (x === y.fruit_name)
            temp = { fruit: x, tested: true }
})
testedCrates.push(temp);