如何使用在Javascript中具有共同值的另一個數組元素的索引獲取數組？

Question

我有一個對象的數組arr ，每個對象的形式如下：

obj={id: /*some string*/,  //id is unique
     msgDetails: { content: /*some string*/,time : /*number*/ }
     }

為了通過其id值獲取特定元素的索引，我使用以下命令：

var idIndex=Babble.messages.findIndex(function(element){
   return element.id===num;
});

有沒有一種方法可以獲取arr中具有id>=num的元素的所有索引，其中num是給定的數字，而沒有for循環？

Answer 1

您可以使用filter代替for ：

data.filter(d => Number(d.id) > id);

 var data = [{ id: "1", msgDetails: { content: "abc1", time: 1 } },{ id: "2", msgDetails: { content: "abc2", time: 1 } },{ id: "3", msgDetails: { content: "abc3", time: 1 } },{ id: "4", msgDetails: { content: "abc4", time: 1 } }]; var filterData = function(id) { return data.filter(d => Number(d.id) > id); }; console.log(filterData(2)); // Another way var filterId = function(cond) { return data.filter(d => cond(Number(d.id))); }; console.log(filterId(id => id > 2)); 

Answer 2

您可以通過.map()和.filter()集合來獲取所需的索引。

var ids = Babble.messages.map((e, i) => [+e.id, i])
                         .filter(a => a[0] >= num)
                         .map(a => a[1]);

Answer 3

您將首先使用map獲取索引，然后對其進行鏈式filter ：

 var Babble = { messages: [{ id: "1", msgDetails: { content: "abc", time: 10 }}, { id: "3", msgDetails: { content: "word", time: 15 }}, { id: "5", msgDetails: { content: "phrase", time: 12 }}, { id: "7", msgDetails: { content: "test", time: 21 }}] }; var num = 4; var idIndexes = Babble.messages.map( (el, i) => el.id >= num ? i : null ) .filter(i => i !== null); console.log('indexes with id-values greater or equal than ' + num + ':'); console.log(idIndexes); 

Answer 4

這將記錄ID等於或大於指定ID的項目的索引。

 var messages = [ { id: 10 }, { id: 12 }, { id: 2 }, { id: 20 }, { id: 30 } ]; function getIndexesForId(id) { // Create a result array var indexes = []; // Loop over all messages. messages.forEach((item, index) => { // Check if message ID is equal to or larger than requested ID. if (item.id >= id) { // Push the index of the current item into the result array. indexes.push(index); } }); // Return the array. return indexes; } console.log(getIndexesForId(10)); console.log(getIndexesForId(20));