如何基于属性值合并两个javascript数组？

我有两个数组

array1 = [{"id":1,"name":"Michale Sharma","gender":"Male","age":25,"salary":10000},{"id":2,"name":"Sunil Das","gender":"Male","age":24,"salary":5000},{"id":3,"name":"Robin Pandey","gender":"Male","age":35,"salary":45000},{"id":4,"name":"Mona Singh","gender":"Female","age":27,"salary":12000}] 

array2 = [{"Deptid":1,"Deptname":"IT"},{"Deptid":12,"Deptname":"HR"},{"Deptid":3,"Deptname":"HW"}] 

输出：

{ "0": { "id": 1, "name": "Michale Sharma", "gender": "Male", "age": 25, "salary": 10000, "Deptid": 1, "Deptname": "IT" }, "1": { "id": 2, "name": "Sunil Das", "gender": "Male", "age": 24, "salary": 5000}, "2": { "id": 3, "name": "Robin Pandey", "gender": "Male", "age": 35, "salary": 45000, "Deptid": 3, "Deptname": "HW" }, "3": { "id": 4, "name": "Mona Singh", "gender": "Female", "age": 27, "salary": 12000 }, "4" :{ "Deptid": 12, "Deptname": "HR" } } 

我想根据属性值将它们合并，例如array1的id和array2的Deptid

这意味着如果id值=1和Deptid值=1，则合并记录，如果不合并，则保留一个数组中的值，另一个数组将为null。换句话说，完全外部联接的功能。因为数组数据不能是连续的，也不能具有相同的长度

我已尝试使用Jquery.extend，如下所示

$.extend(true, array1,array2) 

它不接受任何属性，但合并数组

---

我也看到了，但没有帮助。

看起来您需要自定义逻辑。以下是使用lodash库的示例：

var array1 = [{"id":1,"name":"Michale Sharma","gender":"Male","age":25,"salary":10000},{"id":2,"name":"Sunil Das","gender":"Male","age":24,"salary":5000},{"id":3,"name":"Robin Pandey","gender":"Male","age":35,"salary":45000},{"id":4,"name":"Mona Singh","gender":"Female","age":27,"salary":12000}] 
var array2 = [{"Deptid":1,"Deptname":"IT"},{"Deptid":12,"Deptname":"HR"},{"Deptid":3,"Deptname":"HW"}] 

var defaults = {
    Deptid: null,
    DeptName: null
};

_.each(array1, function (obj) {
    var dept = _.find(array2, { Deptid: obj.id });
    _.defaults(obj, dept, defaults);
});

---

使用LoDash，大致如下：

_.reduce(array1, function(res, item1) {
  var found = _.find(array2, function(item2) {
    return item1.id === item2.Deptid;
  });
  if (found) res.push(_.merge({}, item1, found));
  return res;
}, []);

编辑：对不起，我忘了合并空道具

可能是这样的（未优化）：

---

任何通用方法都无法知道一个数组对象的id属性将与另一个数组对象的Deptid属性相匹配。你需要编写一些代码来循环每一个。这不会让你只与人力资源部门打交道。没错，还需要一个循环，但同样的概念也适用。

function join(array1, array2) {
  var keys = _.keys(array2[0]);
  var vals = _.map(keys, function() { return '' });

  return _.map(array1, function(item1) {
    var found = _.find(array2, function(item2) {
      return item1.id === item2.Deptid;
    });

    if (found)
      return _.merge({}, item1, found);

    return _.merge({}, item1, _.zipObject(keys, vals));
  }, []);  
}