JavaScript 如何在不提供与 object 值匹配的确切搜索值的情况下查找 object 数组的数据

Question

How to make a function that finds data of array of object without giving exact search value that matches with value of object.

For example my array is:

 const array = [{ "name": "Max Messi", "age": 21, "gender": "male" }, { "name": "tina baidya", "age": 10, "gender": "female" }, { "name": "tina shrestha", "age": 100, "gender": "female" } ]

Now i want a function to return all the data that has name "tina" in it.

I tried using array.filter() method but it requires exact search name. Like i need to type tina shrestha instead of just tina

this is what I had tried:

 const array = [{ "name": "Max Messi", "age": 21, "gender": "male" }, { "name": "Lina baidya", "age": 10, "gender": "female" }, { "name": "tina shrestha", "age": 100, "gender": "female" } ] function findData(data, id){ const found = data.filter(element => element.name === id) return found } console.log(findData(array, "tina"))//logs empty array as i need to type full search value

So how can i make the function that searches json data without putting exact value.

Answer 1

You're almost there, you just need to check if the name includes your string, rather than being equal to it:

const found = data.filter(element => element.name.includes(id))

Answer 2

You can try using the includes() method like so:

const array = [
    {
        "name": "Max Messi",
        "age": 21,
        "gender": "male"
    },
    {
        "name": "Lina baidya",
        "age": 10,
        "gender": "female"
    },
    {
        "name": "tina shrestha",
        "age": 100,
        "gender": "female"
    }
]

const findData = (data, searchParam) => {
    return data.filter(element => element.name.includes(searchParam.toLowerCase()));
}

const results = findData(array, "tin");

console.log(results);

However, it won't work if you search for example tina2 or tinathy so it doesn't cover all edge-cases!