[英]creating custom array from existing array in javascript
我有一个如下所示的数组:
var search = [
{ code: "t1", name1: "n1", name2: "n2" },
{ code: "t1", name1: "n5", name2: "n6" },
{ code: "t2", name1: "n10", name2: "n11" },
{ code: "t2", name1: "n18", name2: "n20" },
{ code: "t3", name1: "n18", name2: "n20" },
];
我想将此数组转换为以下格式:
var finald = [
{ code: "t1", name1: "n1,n5", name2: "n2,n6" },
{ code: "t2", name1: "n10,n18", name2: "11,n20" },
{ code: "t3", name1: "n18", name2: "n20" },
];
我已经尝试过如下代码。 但没有成功。 首先,我从数组中收集了所有唯一的code
:
var flags = [];
var codes = [];
for(var z=0; z<search.length; z++){
if( flags[data[z].code]) continue;
flags[data[z].code] = true;
codes.push(data[z].code);
}
var finald = [];
for(var i=0; i<search.length; i++){
var name1 = [];
var name2 = [];
for(var y=0; y<codes.length; y++){
if(codes[y] == search[i].code ){
var row = {
code: codes[y],
name1:search[i].name1,
name2:search[i].name2,
};
finald.push(row);
}
}
}
请帮忙。
这不是你所要求的,但仍然尝试这种方法。
您基本上想通过键“代码”对数组执行“加入”操作,因此可以尝试将数组转换为map
对象 - key: code , value : {name1, name2}
var search = [
{code:"t1", name1:"n1", name2:"n2"},
{code:"t1", name1:"n5", name2:"n6"},
{code:"t2", name1:"n10", name2:"n11"},
{code:"t2", name1:"n18", name2:"n20"},
{code:"t3", name1:"n18", name2:"n20"}];
const myMap = new Map();
search.forEach((obj) => {
const key = obj.code;
if(!myMap.has(key)) {
myMap.set(key, {
'name1' : obj.name1,
'name2' : obj.name2
});
} else {
var ele = myMap.get(key);
ele.name1 = `${ele.name1},${obj.name1}`;
ele.name2 = `${ele.name2},${obj.name2}`;
}
});
console.log(myMap);
/*
key: "t1" => value: {name1:"n1,n5", name2:"n2,n6"},
key: "t2" => value: {name1:"n10,n18", name2:"n11,n20"},
key: "t3" => value: {name1:"n18", name2:"n20"}
*/
如果你必须以数组形式将它添加到上面, jsfiddle
var resArray = [];
myMap.forEach((value, key) => {
resArray.push(Object.assign({'code': key}, value));
});
console.log(resArray);
/*
[{ code: "t1", name1: "n1,n5", name2: "n2,n6" },
{ code: "t2", name1: "n10,n18", name2: "11,n20" },
{ code: "t3", name1: "n18", name2: "n20" }]
*/
像这样的东西?
您可以使用 reduce 或 map ,但在我看来,这对于一个简单的任务来说更容易理解
我假设搜索是按代码排序的。
var search = [ {code:"t1", name1:"n1", name2:"n2"}, {code:"t1", name1:"n5", name2:"n6"}, {code:"t2", name1:"n10", name2:"n11"}, {code:"t2", name1:"n18", name2:"n20"}, {code:"t3", name1:"n18", name2:"n20"} ], finald = []; finald.push(search[0]); for (var i = 1; i < search.length; i++) { var f = finald[finald.length - 1]; if (f.code == search[i].code) { f.name1 += ","+search[i].name1; f.name2 += ","+search[i].name2; } else { finald.push(search[i]); } } console.log(finald);
这应该做你想做的:
var search = [ { code: "t1", name1: "n1", name2: "n2" }, { code: "t1", name1: "n5", name2: "n6" }, { code: "t2", name1: "n10", name2: "n11" }, { code: "t2", name1: "n18", name2: "n20" }, { code: "t3", name1: "n18", name2: "n20" } ]; const items = search.reduce((acc, item) => { const code = item.code; if (!acc[code]) acc[code] = {name1: [], name2: []}; acc[code].name1.push(item.name1); acc[code].name2.push(item.name2); return acc; }, {}); const finald = Object.keys(items).reduce((acc, key) => { const item = items[key]; acc.push({ code: key, name1: item.name1.join(','), name2: item.name2.join(',') }); return acc; }, []);
您可以为此创建自己的自定义逻辑:
var search = [ { code: "t1", name1: "n1", name2: "n2" }, { code: "t1", name1: "n5", name2: "n6" }, { code: "t2", name1: "n10", name2: "n11" }, { code: "t2", name1: "n18", name2: "n20" }, { code: "t3", name1: "n18", name2: "n20" }, ]; //this is the result array var result = []; //this is a flag var itemExist = false; search.forEach((item) => { //reset flag to false itemExist = false; //check if the object already exist in the result array or not for(var i=0; i<result.length; i++){ //if the object exist then merge the values of name1 and name2 if(result[i].code === item.code){ result[i].name1 = result[i].name1 +','+item.name1; result[i].name2 = result[i].name2 +','+item.name2; itemExist = true; //break the for loop break; } } //if object do not exist in the result array then push it if(!itemExist){ result.push(item); } }); console.log(result);
我写了这个函数:
function normalize(arr) {
var output = [];
for(var i = 0; i < arr.length; i++) {
var codeIndex = _findIndex(output, arr[i].code);
if(codeIndex !== -1) {
// Code already in output array, merge names.
output[_findIndex(output, arr[i].code)].name1 += ', ' + arr[i].name1;
output[_findIndex(arr[i].code)].name2 += ', ' + arr[i].name2;
}
else {
// Code new to output array, pus into.
output.push(arr[i]);
}
}
return output;
//
// Privates
//
function _findIndex(arr, code){
var index = -1;
for(var i = 0; i < arr.length; i++){
if(arr[i].code == code) {
index = i;
break;
}
}
return index;
}
}
输出:
0 : {code: "t1", name1: "n1, n5", name2: "n2, n6, n20"}
1 : {code: "t2", name1: "n10, n18", name2: "n11"}
2 : {code: "t3", name1: "n18", name2: "n20"}
您可以使用reduce
和map
let search = [{code:"t1", name1:"n1", name2:"n2"},{code:"t1", name1:"n5", name2:"n6"},{code:"t2", name1:"n10", name2:"n11"},{code:"t2", name1:"n18", name2:"n20"},{code:"t3", name1:"n18", name2:"n20"},]; let result = Object.values( search.reduce( (c,v) => { c[ v.code ] ? c[ v.code ].push( v ) : c[ v.code ] =[ v ]; return c; },{}) ).map( v => { let n1 = [], n2 = []; v.forEach( e => { n1.push( e.name1 ); n2.push( e.name2 ); }); return { code : v[0].code, name1 : n1.join(","), name2 : n2.join(",") } }); console.log( result );
另一个解决问题的方法 - 这段代码强调数据的不变性,从所有迭代中返回新对象。 这可能会防止在执行此类操作时难以发现错误:
var finald = transform([ { code: "t1", name1: "n1", name2: "n2" }, { code: "t1", name1: "n5", name2: "n6" }, { code: "t2", name1: "n10", name2: "n11" }, { code: "t2", name1: "n18", name2: "n20" }, { code: "t3", name1: "n18", name2: "n20" }, ]); console.log(finald); function transform(source) { const keyedByCode = source.reduce((acc, item) => { const { name1 = [], name2 = [] } = acc[item.code] || {}; return Object.assign({}, acc, { [item.code]: { name1: [item.name1, ...name1], name2: [item.name2, ...name2], }, }); }, {}); return Object.keys(keyedByCode).map((code) => ({ code, name1: keyedByCode[code].name1.sort().join(','), name2: keyedByCode[code].name2.sort().join(','), })); }
.as-console-wrapper { max-height: 100% !important; }
代码的顺序在这里无关紧要:
const search = [{code:"t1", name1:"n1", name2:"n2"}, {code:"t1", name1:"n5", name2:"n6"}, {code:"t2", name1:"n10", name2:"n11"}, {code:"t2", name1:"n18", name2:"n20"}, {code:"t3", name1:"n18", name2:"n20"}]; function combineObjectsByCode(objects) { let codes = new Set(); objects.forEach(object => codes.add(object.code)); let combinedObjects = []; codes.forEach(code => { let objectsWithSameCode = objects.filter(object => object.code === code); let combinedObject; objectsWithSameCode.forEach( obj => { if (!combinedObject) { combinedObject = obj; } else { combinedObject.name1 = combinedObject.name1 + ', ' + obj.name1; combinedObject.name2 = combinedObject.name2 + ', ' + obj.name2; } }) combinedObjects.push(combinedObject); }); return combinedObjects; } console.log(combineObjectsByCode(search));
您可以使用array#reduce
根据code
对数组进行分组,并将相同的值推送到数组中。 使用Object.values()
获取所有值,然后使用array#map
使用array#join()
将数组转换为字符串。
var search = [{ code: "t1", name1: "n1", name2: "n2" },{ code: "t1", name1: "n5", name2: "n6" },{ code: "t2", name1: "n10", name2: "n11" },{ code: "t2", name1: "n18", name2: "n20" },{ code: "t3", name1: "n18", name2: "n20" }], result = Object.values(search.reduce((r,{code,name1, name2}) => { r[code] = r[code] || {code, name1: [], name2: []}; r[code].name1.push(name1); r[code].name2.push(name2); return r; },{})) .map(({code,name1,name2}) => ({code, name1: name1.join(','), name2: name2.join(',')})); console.log(result);
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