将一行数据从一个对象传输到另一个对象。

Question

我的目标是将一行数据从一个对象传输到另一个对象，其中selected_city_codes将说明要使用city_id从airport_data_2中删除哪些项以将其添加到airport_data_1中。不确定如何处理将数据从airport_data_2追加到airport_data_1的问题，但我演示了从airport_data_2删除行的结果。是否有更有效的方法在两个对象之间传输数据？

airport_data_1 = [{"departure_time":"12:00","arrival_time":"03:00","city_id":"BOS"},  
{"departure_time" :"12:00","arrival_time":"03:00","city_id":"BOS"},
{"departure_time" :"01:00","arrival_time":"04:00","city_id":"SFO"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"BOS"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"SFO"}]; 


airport_data_2 = [{"departure_time" :"03:00","arrival_time":"05:00","city_id":"DEN"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"SJC"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"JFK"},
{"departure_time" :"06:00","arrival_time":"09:00","city_id":"SJC"}]; 


selected_city_codes = ['SJC','JFK']; //items to remove from `airport_data_2` to `airport_data_1`

//my attempt to remove the listed values from selected_city_codes
for(i in selected_city_codes) { 
    airport_data_2 = airport_data_2.filter(item => item.city_id != selected_city_codes[i]);
}

之前的结果

airport_data_1 = [{"departure_time":"12:00","arrival_time":"03:00","city_id":"BOS"},  
{"departure_time" :"12:00","arrival_time":"03:00","city_id":"BOS"},
{"departure_time" :"01:00","arrival_time":"04:00","city_id":"SFO"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"BOS"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"SFO"}]; 


airport_data_2 = [{"departure_time" :"03:00","arrival_time":"05:00","city_id":"DEN"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"SJC"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"JFK"},
{"departure_time" :"06:00","arrival_time":"09:00","city_id":"SJC"}]; 

在 -- 我们期望的结果之后的结果

airport_data_1 = [{"departure_time":"12:00","arrival_time":"03:00","city_id":"BOS"},  
{"departure_time" :"12:00","arrival_time":"03:00","city_id":"BOS"},
{"departure_time" :"01:00","arrival_time":"04:00","city_id":"SFO"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"BOS"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"SFO"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"SJC"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"JFK"},
{"departure_time" :"06:00","arrival_time":"09:00","city_id":"SJC"}];

airport_data_2 = [{"departure_time" :"03:00","arrival_time":"05:00","city_id":"DEN"}]; 

- bombombs

---

你可以循环遍历 airport_data_2，并使用 if else 条件语句创建两个新数组。一个将成为新的 airport_data_2，第二个将与 airport_data_2 进行 concat() 操作。但我怀疑这对性能影响不大。 - Lluser

Answer 1

我对您的实现进行了轻微修改，并添加了相等的部分。

airport_data_1 = [{"departure_time":"12:00","arrival_time":"03:00","city_id":"BOS"},  
{"departure_time" :"12:00","arrival_time":"03:00","city_id":"BOS"},
{"departure_time" :"01:00","arrival_time":"04:00","city_id":"SFO"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"BOS"},
{"departure_time" :"03:00","arrival_time":"05:00","city_id":"SFO"}]; 


airport_data_2 = [{"departure_time" :"03:00","arrival_time":"05:00","city_id":"DEN"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"SJC"},
{"departure_time" :"04:00","arrival_time":"06:00","city_id":"JFK"},
{"departure_time" :"06:00","arrival_time":"09:00","city_id":"SJC"}]; 

selected_city_codes = ['SJC','JFK']; 

for(i in selected_city_codes) { 
    airport_data_2 = airport_data_2.filter((item) =>{ 
        if (item.city_id === selected_city_codes[i]){
        airport_data_1.push(item);
      }
      return item.city_id !== selected_city_codes[i]
    });
}

console.log(airport_data_1)
console.log(airport_data_2)

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- cmgchess