Question

我有一个＆＃34; N＆＃34; Objets，我需要做的是仅恢复或保留同一数组中的Objets或其他newField与所有Objects的价值相同的数据Array Array的。{注意：新的Array或最终的export class BucketPipe implements PipeTransform { public arr = [] transform(value: any, args?: any): any { return this.seleccion(value); } seleccion(value) { console.log("val: " +JSON.stringify(value)) var newValue = new Array() for(var i = 0; i < value.length; i++) { for(var y = 0; y < value[i].length; y++) { if(value[i][y]) { var val = value[i][y] do { newValue.push(value[i][y]) val = value[i][y] } while (value[i][y].newField !== val.newField) } } } return this.filter(newValue) } filter(values) { var field: string[] = new Array() for(var i = 0; i < values.length; i++) { Object.keys(values[i]).map((obj,key) => { if(values[i].newField == values[i][obj]) field.push(values[i]) }) } this.arr = field var result = []; this.arr.forEach((subArr, index) => { if(this.validate(subArr, index)) result.push(subArr); }); console.log("Result: ",result) } validate(subArr, index) { var test = true; for(var i = 0; i < subArr.length && test; i++){ var obj = subArr[i]; var subtest = false; for(var j = 0; j < this.arr.length && !subtest; j++){ if(index == j) continue; var found = this.arr[j].find(function(e){ return e["newField"] == obj["newField"]; }); if(found) subtest = true; } test = test && subtest; } return test; } }每个重复元素只能有一个Object。注意2：我必须编辑它，因为返回值必须在所有对象中都是通用的。例如，如果具有相同newField的3/3对象将返回一个，如果具有相同newField的2/3对象则不返回任何内容

这就是我所拥有的，但它真的没什么......：/

（这是来自Angular 2的过滤器）我会感激任何帮助。

谢谢

 [[{"id":42,
    "idBucket":"patrimonial",
    "originalField":"c",
    "newField":"nCliente",
    "name":"Número Cliente",
    "description":"Número del Cliente",
    "filter":true,
    "visible":true,
    "idDataType":1},{"id":43,
    "idBucket":"patrimonial",
    "originalField":"ct",
    "newField":"nContrato",
    "name":"Número Contrato",
    "description":"Número de Contrato",
    "filter":true,
    "visible":true,
    "idDataType":1},{"id":45,
    "idBucket":"patrimonial",
    "originalField":"s",
    "newField":"sucursal",
    "name":"Sucursal",
    "description":"Sucursal",
    "filter":true,
    "visible":true,
    "idDataType":1
    },{"id":47,
    "idBucket":"patrimonial",
    "originalField":"sp",
    "newField":"sProducto",
    "name":"Subproducto",
    "description":"Subproducto",
    "filter":true,
    "visible":true,
    "idDataType":1
    }],
    [{
    "id":11,
    "idBucket":"expunic",
    "originalField":"nc",
    "newField":"nCliente",
    "name":"Número Cliente",
    "description":"Número del Cliente",
    "filter":true,
    "visible":true,
    "idDataType":1
    },{
    "id":12,
    "idBucket":"expunic",
    "originalField":"t",
    "newField":"titulo",
    "name":"Título Aplicación",
    "description":"Título de la Aplicación o Gaveta",
    "filter":true,
    "visible":true,
    "idDataType":1
    }],
    [{
    "id":16,
    "idBucket":"bastanteo",
    "originalField":"t",
    "newField":"titulo",
    "name":"Título Aplicación",
    "description":"Nombre de la Gaveta",
    "filter":true,
    "visible":true,
    "idDataType":1
    }]]

这是数组的一个例子：

newField

在此示例中，不会返回任何内容，因为常见的[[{ "id":11, "idBucket":"expunic", "originalField":"nc", "newField":"nCliente", "name":"Número Cliente", "description":"Número del Cliente", "filter":true, "visible":true, "idDataType":1 },{ "id":12, "idBucket":"expunic", "originalField":"t", "newField":"titulo", "name":"Título Aplicación", "description":"Título de la Aplicación o Gaveta", "filter":true, "visible":true, "idDataType":1 }], [{ "id":16, "idBucket":"bastanteo", "originalField":"t", "newField":"titulo", "name":"Título Aplicación", "description":"Nombre de la Gaveta", "filter":true, "visible":true, "idDataType":1 }]]值为＆＃34; titulo＆＃34;但它只在阵列位置1和2中但不在0中，如果是newField值＆＃34; titulo＆＃34;在所有3中它将返回一个只有一个Object的新数组，并不关心其中一个3.

另一个例子

Object

在此示例中，它将返回一个newField，因为在所有2个对象中，Objects具有相同的值，并不关心其中之一。虽然如果阵列中有10个Objets，则所有10个newField必须具有相同的Objects值。请注意，只有在newFields

中共有2 Objects个值时，才能返回2

<div ng-init="addEditRuleSetCtrl.initSelectors()">

            <query-builder group="addEditRuleSetCtrl.filter.group"
                           rule-condition-operators="addEditRuleSetCtrl.ruleConditionOperators"
                           rule-condition-set-operators="addEditRuleSetCtrl.ruleConditionSetOperators"
                           fields="addEditRuleSetCtrl.fieldCodes"
                           ></query-builder>
</div>

以上

self.initSelectors  = function()
        {
            self.getRuleConditionOperators();
            self.getRuleConditionSetOperators();
            self.fieldCodes = self.getCorrespondingFields();
            console.log("initSelectors : fieldCodes: " + JSON.stringify(self.fieldCodes ));
        }


self.getCorrespondingFields = function()
        {
            return self.origin == 'DDE' ? self.fieldsDDE : self.fieldsATT;
        };

Answer 1

这是一种可行的方法：

function getFirstObjectForEachRepeatedKey(values, keyName)
{
  if (!keyName)
    keyName = "newField";

  var haveSeenKey = {};
  var alreadySelectedValue = {};
  var result = [];

  for (var i = 0; i < values.length; i++)
  {
    var key = values[i][keyName];

    if (!(key in haveSeenKey))
      haveSeenKey[key] = values[i];
    else
    {
      var selectedValue = haveSeenKey[key];

      if (selectedValue !== alreadySelectedValue)
      {
        result.push(selectedValue);
        haveSeenKey[key] = alreadySelectedValue;
      }
    }
  }

  return result;
}

Answer 2

这就是我的理解。（我用你的阵列测试了它并给出了正确的结果）：

// define your array here (must be named arr or you can modify the code accordingly
var arr = [[{}, {}, {}], [{}], [{}, {}]]; // ...

function validate(subArr, index){
    var test = true;
    for(var i = 0; i < subArr.length && test; i++){
        var obj = subArr[i];
        var subtest = false;
        for(var j = 0; j < arr.length && !subtest; j++){
            if(index == j) continue;
            var found = arr[j].find(function(e){
                return e["newField"] == obj["newField"];
            });
            if(found) subtest = true;
        }
        test = test && subtest;
    }
    return test;
}

var result = [];
arr.forEach(function(subArr, index){
    if(validate(subArr, index))
        result.push(subArr);
});

代码段：

var arr = [[{
        "id": 42,
        "idBucket": "patrimonial",
        "originalField": "c",
        "newField": "nCliente",
        "name": "Número Cliente",
        "description": "Número del Cliente",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }, {
        "id": 43,
        "idBucket": "patrimonial",
        "originalField": "ct",
        "newField": "nContrato",
        "name": "Número Contrato",
        "description": "Número de Contrato",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }, {
        "id": 45,
        "idBucket": "patrimonial",
        "originalField": "s",
        "newField": "sucursal",
        "name": "Sucursal",
        "description": "Sucursal",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }, {
        "id": 47,
        "idBucket": "patrimonial",
        "originalField": "sp",
        "newField": "sProducto",
        "name": "Subproducto",
        "description": "Subproducto",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }],
    [{
        "id": 11,
        "idBucket": "expunic",
        "originalField": "nc",
        "newField": "nCliente",
        "name": "Número Cliente",
        "description": "Número del Cliente",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }, {
        "id": 12,
        "idBucket": "expunic",
        "originalField": "t",
        "newField": "titulo",
        "name": "Título Aplicación",
        "description": "Título de la Aplicación o Gaveta",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }],
    [{
        "id": 16,
        "idBucket": "bastanteo",
        "originalField": "t",
        "newField": "titulo",
        "name": "Título Aplicación",
        "description": "Nombre de la Gaveta",
        "filter": true,
        "visible": true,
        "idDataType": 1
    }]];

function validate(subArr, index){
    var test = true;
    for(var i = 0; i < subArr.length && test; i++){
        var obj = subArr[i];
        var subtest = false;
        for(var j = 0; j < arr.length && !subtest; j++){
            if(index == j) continue;
            var found = arr[j].find(function(e){
                return e["newField"] == obj["newField"];
            });
            if(found) subtest = true;
        }
        test = test && subtest;
    }
    return test;
}

var result = [];
arr.forEach(function(subArr, index){
    if(validate(subArr, index))
        result.push(subArr);
});

console.log(result);

仅返回对象

2 个答案: