Question

我有这个JSON数组：

"locations": {
        "0": {
            "locationNo": "10014",
            "inventoryLocationNo": "10014"
        },
        "1": {
            "locationNo": "10499",
            "inventoryLocationNo": "10499"
        },
        "2": {
            "locationNo": "110868",
            "inventoryLocationNo": "110868"
        },

我想从此数组中删除此索引部分("0", "1", "2")，其余全部应该相同。

有什么建议吗？

响应应该是这样的：

"locations": {
    {
        "locationNo": "13137",
        "inventoryLocationNo": "13137"
   },
    {
        "locationNo": "15857",
        "inventoryLocationNo": "15857"
    },
    {
        "locationNo": "15903",
        "inventoryLocationNo": "15903"
    },

Answer 1

使用Object.values：

＆＃13;

var obj = {
  "locations": {
        "0": {
            "locationNo": "10014",
            "inventoryLocationNo": "10014"
        },
        "1": {
            "locationNo": "10499",
            "inventoryLocationNo": "10499"
        },
        "2": {
            "locationNo": "110868",
            "inventoryLocationNo": "110868"
        }
     }
};

obj.locations = Object.values(obj.locations);

console.log(obj);

＆＃13;

或Object.keys如果您不能使用Object.values：

＆＃13;

var obj = {
  "locations": {
        "0": {
            "locationNo": "10014",
            "inventoryLocationNo": "10014"
        },
        "1": {
            "locationNo": "10499",
            "inventoryLocationNo": "10499"
        },
        "2": {
            "locationNo": "110868",
            "inventoryLocationNo": "110868"
        }
     }
};

obj.locations = Object.keys(obj.locations).map(key => obj.locations[key]);

console.log(obj);

＆＃13;

Answer 2

使用传播运算符的简单方法

import pandas as pd
time_range = pd.date_range("2018-01-01", "2018-02-01", freq='W')
print(time_range)
DatetimeIndex(['2018-01-07', '2018-01-14', '2018-01-21', '2018-01-28'], dtype='datetime64[ns]', freq='W-SUN')

JSON格式化删除索引

2 个答案: