如何根據字串陣列過濾javascript物件陣列

鑒于以下物件，我想搜索多個字串的所有鍵。我一直在嘗試和搜索，但在這個問題上卻一無所獲。任何人都可以在這里為我提供一些幫助嗎？

搜索這個：

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];

對于這個字串陣列：

const query = ['chris', 'com'];

將回傳這樣的結果陣列：

[
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
]

如果我對此不清楚，請告訴我。謝謝！

// 更新

請注意，我正在尋找的關鍵是每個物件都必須能夠通過任何鍵（不僅僅是創建者）進行搜索，并且在物件中具有所有搜索字串，而不僅僅是一個或另一個。

uj5u.com熱心網友回復：

const array = [{
    name: "Blue Iron Chow Chow",
    status: "Complete",
    creator: "John"
  },
  {
    name: "Purple Steel Husky",
    status: "Error",
    creator: "Chris"
  },
  {
    name: "Purple Composite Husky",
    status: "Ready",
    creator: "Chris"
  },
  {
    name: "Aqua Zinc Spaniel",
    status: "Complete",
    creator: "Chris"
  },
  {
    name: "Fuschia Silver Corgi",
    status: "Complete",
    creator: "John"
  },
];

const query = ['chris', 'com'];

const filtered = array.filter(obj => {
  return query.every(val =>
  Object.values(obj).join(" ").toLowerCase().includes(val.toLowerCase()))
})

console.log(filtered)

uj5u.com熱心網友回復：

你可以試試這個：

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];

const query = ['chris', 'com'];

const result = array.filter(({creator}) => query.includes(creator.toLowerCase()))

console.log(result)

uj5u.com熱心網友回復：

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];

let filtered = []


array.map((x)=>{
//x contains individual objects of array,
for (let key in x) {
//key in x will get all the keys inside the x , now all we have to check is does the values in x are included in query or not to find that we use the below code
  const contains = ['Chris', 'com'].includes(x[key])
   console.log(contains)
  if(contains) filtered.push(x)
}

})

uj5u.com熱心網友回復：

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];


const query = ['chris', 'com'];

var result = array.filter(({creator, name, status}) => {
    return query.every(item => {
            item = item.toLowerCase();
            return creator.toLowerCase().includes(item) || name.toLowerCase().includes(item) || status.toLowerCase().includes(item)
        })
});

輸出：-

[    
  { name: 'Purple Composite Husky', status: 'Ready', creator: 'Chris' },
  { name: 'Aqua Zinc Spaniel', status: 'Complete', creator: 'Chris' }   
]

uj5u.com熱心網友回復：

您可以使用.filter()with Object.values()、RegExp()和.every()方法，如下面的演示所示：

const array = [
    { name: "Blue Iron Chow Chow", status: "Complete", creator: "John" },
    { name: "Purple Steel Husky", status: "Error", creator: "Chris" },
    { name: "Purple Composite Husky", status: "Ready", creator: "Chris" },
    { name: "Aqua Zinc Spaniel", status: "Complete", creator: "Chris" },
    { name: "Fuschia Silver Corgi", status: "Complete", creator: "John" }, 
];
const query = ['chris', 'com'];
const filtered = array.filter(
    obj => query.every(
        term => new RegExp(term, 'i').test( 
            Object.values(obj).join(" ")
        )            
    )
);
console.log( filtered );

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