Javascript：在数组中查找对象的索引，不知道索引，只有对象

Question

我有一个对象数组，我想删除其中一个对象。

[
  {"field":"ingredients","gte":"egg","lte":"egg"},
  {"field":"ingredients","gte":"bakepulver","lte":"bakepulver"},
  {"field":"ingredients","gte":"hvetemel","lte":"hvetemel"}
]

我不知道要删除的对象的索引，但是我知道整个对象。 即：

{"field":"ingredients","gte":"bakepulver","lte":"bakepulver"}

我需要通过对象的全部内容（所有属性）而不是通过field ， gte或lte来查找对象的索引。 如何仅使用普通JavaScript在数组中找到对象的索引？

Answer 1

您可以遍历数据，然后检查键的长度以及每个键的内容是否相同。

 var data = [{ "field": "ingredients", "gte": "egg", "lte": "egg" }, { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, { "field": "ingredients", "gte": "hvetemel", "lte": "hvetemel" }], search = { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, keys = Object.keys(search), index = -1; data.some(function (a, i) { if (Object.keys(a).length === keys.length && keys.every(function (k) { return a[k] === search[k]; })) { index = i; return true; } }); console.log(index); 

ES6

 var data = [{ "field": "ingredients", "gte": "egg", "lte": "egg" }, { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, { "field": "ingredients", "gte": "hvetemel", "lte": "hvetemel" }], search = { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, keys = Object.keys(search), index = data.findIndex(a => Object.keys(a).length === keys.length && keys.every(k => a[k] === search[k])); console.log(index); 

Answer 2

您可以使用Object.keys() ， Array.prototype.findIndex() ， Array.prototype.every()来检查每个属性名称，值和对象属性名称.length是否相等。

 let data = [ {"field":"ingredients","gte":"egg","lte":"egg"}, {"field":"ingredients","gte":"bakepulver","lte":"bakepulver"}, {"field":"ingredients","gte":"hvetemel","lte":"hvetemel"} ]; let props = {"field":"ingredients","gte":"bakepulver","lte":"bakepulver"}; let keys = Object.keys(props); let index = data.findIndex(o => keys.every(key => o[key] === props[key]) && Object.keys(o).length === keys.length); console.log(index); 

Answer 3

从您的进一步评论中可以理解，您仅需要针对此特定情况的解决方案，其中对象由3个给定的属性组成。 因此，我建议使用此ES6解决方案：

 var data = [ { "field": "ingredients", "gte": "egg", "lte": "egg" }, { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, { "field": "ingredients", "gte": "hvetemel", "lte": "hvetemel" }]; var search = { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }; var index = data.findIndex( a => a.field == search.field && a.gte == search.gte && a.lte == search.lte); console.log(index); 

如果您没有完整的ES6支持，则：

 var data = [ { "field": "ingredients", "gte": "egg", "lte": "egg" }, { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, { "field": "ingredients", "gte": "hvetemel", "lte": "hvetemel" }]; var search = { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }; var index = -1; data.some(function (a, i) { if (a.field == search.field && a.gte == search.gte && a.lte == search.lte) return index = i, true; }); console.log(index); 

Answer 4

您可以使用此代码解决问题

var array = [
    {"field":"ingredients","gte":"egg","lte":"egg"},
    {"field":"ingredients","gte":"bakepulver","lte":"bakepulver"},
    {"field":"ingredients","gte":"hvetemel","lte":"hvetemel"}
];
var searchObject =      
    {"field":"ingredients","gte":"bakepulver","lte":"bakepulver"};

var indexOfSearchResult;

for (var i = 0; i < array.length; i++) {
    var checkEqualBool = true;
    for (var key in array[i]) {
        if(array[i][key]!=searchObject[key]){
            checkEqualBool = false;
            break;
        }
    }
    if(checkEqualBool){
        indexOfSearchResult = i;
        break;
    }
}

Answer 5

我使用比较对象的旧方法，将其字符串化 。 尝试审查这一

 var x = [{ "field": "ingredients", "gte": "egg", "lte": "egg" }, { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }, { "field": "ingredients", "gte": "hvetemel", "lte": "hvetemel" }]; var control = { "field": "ingredients", "gte": "bakepulver", "lte": "bakepulver" }; function getIndex(arr, key) { var got = false, result = -1; arr.every(function(e, i) { if (JSON.stringify(e) === JSON.stringify(key)) { console.log('match found'); result = i; return false; } return true; }) return result; } console.log(getIndex(x, control));