在数组内查找数组中的元素并返回带有父对象的元素

Question

Given...

    var a = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }];

What is a good way to find the object, for example, with i = 5 plus the object that contains the array that it's in?

So, the result would be two references...

var r0 = { "i": 5 };
var r1 = { "b": [{ "i": 4 }, { "i": 5 } ]};

Answer 1

You can do that with forEach :

 var a = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }]; var r0; var r1; a.forEach(function(item){ item.b.forEach(function(inner){ if(inner.i==5){ r0 = inner; r1 = item; } }); }); console.log(r0); console.log(r1);

Answer 2

Using arrow functions :

 var a = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }]; var result = a.filter(x => xbfind(b => bi == 5)); console.log(result)

Or using normal functions:

 var a = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }]; var result = a.filter(function (x) { return xbfind(function(b) { return bi == 5; }); }); console.log(result);

Answer 3

I think the solution may be different based on your actual background, like which factors you are considering are the most important such as time, space, read/write efficiency etc.

Below are two solutions for basic usage:

 var a = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }] //The commom way: it has to loop whole array for each query queryItem1 = function(searchTarget, searchVal) { return searchTarget.filter( function(val){ return val.b.filter(function(item){return item.i===searchVal;}).length>0; } ) } console.log('---common way----') console.log(queryItem1(a, 5)) console.log('---end----') //Considering `query` efficiency; //build index once, then query by one step instead of loop buildIndexes = function(target) { let indexes = new Object(); for(index in target) { for(itemIndex in target[index].b) { let queryVal = target[index].b[itemIndex].i; let queryIndex = index; if(!indexes[queryVal]) { indexes[queryVal] = []; } indexes[queryVal].push(queryIndex); } } return indexes; } console.log('---consider `query` efficiency way----') indexes = buildIndexes(a) //console.log(indexes) //display index tree console.log('---test case 1----') console.log(indexes[5]) //if query vale is 5, get the index, then call like a[indexes[5][0]] console.log(a[indexes[5][0]]) console.log('---test case 2----') console.log(indexes[6]) //if query vale is 6, get the index, then call like a[indexes[6][0]] console.log(a[indexes[6][0]]) console.log('---end----')

Answer 4

You could take Array#some for both iterations, because if found, the iterations stop by returning true .

This proposal returns the first found elements on which the condition is true .

 var array = [{ b: [{ i: 2 }, { i: 3 }] }, { b: [{ i: 4 }, { i: 5 }] }, { b: [{ i: 6 }, { i: 7 }] }], i = 5, r0, r1; array.some(function (a) { return absome(function (b) { if (bi === i) { r0 = b; r1 = a; return true; } }); }); console.log(r0); console.log(r1);

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