Question

这是我编写的一个函数，它将改变/缩短json对象上的所有键

function replaceKeyWithNewKey(jsonObj, new_keys, old_keys){
        console.log("test")
        for(i=0;i<jsonObj.length;i++){

            for(el in jsonObj){
                //console.log(new_keys[el])
                jsonObj[i][new_keys[el]]=jsonObj[i][old_keys[el]] //add new key
                delete jsonObj[i][old_keys[el]] // delete old key
            }
        }

        return jsonObj
    }

我正在寻找的是使用bash或python或其他方式在终端中执行此操作的方法，但我想在终端中执行此操作。所以我会在file.json上运行脚本，结果是file2.json，file2.json有更短的键名。 我该怎么做？

对于bash，我正在考虑使用sed，但我认为这不会像我可能会取代我不想替换的价值一样好。我知道的小蟒蛇可能是要走的路。

这是一个控制台打印输出：

    >data[0]
    Object {Rec_Open_Date: "2016-07-07", MSISDN: 123, IMEI: 223, Data_Volume_Bytes: "673", Device_Manufacturer: "Samsung Korea"…}
    >old_keys=Object.keys(data[0])
    ["Rec_Open_Date", "MSISDN", "IMEI", "Data_Volume_Bytes", "Device_Manufacturer", "Device_Model", "Product_Description", "Data_Volume_MB"]
    >new_keys=["r", "m", "i", "d", "f", "l", "s", "d2"]
    ["r", "m", "i", "d", "f", "l", "s", "d2"]
    >function replaceKeyWithNewKey(jsonObj, new_keys, old_keys){
        console.log("test")
        for(i=0;i<jsonObj.length;i++){

            for(el in jsonObj){
                //console.log(new_keys[el])
                jsonObj[i][new_keys[el]]=jsonObj[i][old_keys[el]] //add new key
                delete jsonObj[i][old_keys[el]] // delete old key
            }
        }

        return jsonObj
    }
    undefined
    > replaceKeyWithNewKey(data, new_keys, old_keys)
    VM129:2 test
    [Objectd: "673"d2: "0.000641823"f: "Samsung Korea"i: 223l: "Samsung GT-I9505"m: 123r: "2016-07-07"s: "PREPAY PLUS - $1 - #33"__proto__: Object, Object, Object, Object, Object, Object, Object, Object, Object]

这是测试我的函数的示例数据：

var json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}]';

Answer 1

假设您知道如何将json加载到python数据结构中。您提供的示例数据将是list个dict的python dict。我认为最直接的方法是构建一个字典，将旧密钥映射到新密钥，然后使用字典理解迭代列表，以使用密钥名称映射中的新密钥名重建In [8]: jobj = [{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}] In [9]: keymap = {'email':'e', 'image':'img', 'name':'n', '_id':'id'} In [10]: for i in range(len(jobj)): ...: jobj[i] = {keymap[k]:jobj[i][k] for k in jobj[i]} ...: In [11]: jobj Out[11]: [{'e': 'user1@gmail.com', 'id': '5078c3a803ff4197dc81fbfb', 'img': 'some_image_url', 'n': 'Name 1'}, {'e': 'user2@gmail.com', 'id': '5078c3a803ff4197dc81fbfc', 'img': 'some_image_url', 'n': 'Name 2'}]你已经建成了。使用您的样本：

OrderedDict

编辑添加

如果您想确保所有按键都按顺序排列，则需要使用collections中的import json from collections import OrderedDict keymap = {'email':'e', 'image':'img', 'name':'n', '_id':'id'} with open('ordered_example.json') as f: jobj = json.load(f, object_pairs_hook=OrderedDict) for i in range(len(jobj)): jobj[i] = OrderedDict((keymap[k],jobj[i][k]) for k in jobj[i]) print(jobj)

[OrderedDict([('id', '5078c3a803ff4197dc81fbfb'),
              ('e', 'user1@gmail.com'),
              ('img', 'some_image_url'),
              ('n', 'Name 1')]),
 OrderedDict([('id', '5078c3a803ff4197dc81fbfc'),
              ('e', 'user2@gmail.com'),
              ('img', 'some_image_url'),
              ('n', 'Name 2')])]

输出：

unused: false

Answer 2

如果您要从命令行操作JSON，我建议您安装jq。

您可以将键映射放在bash 4+中的关联数组中，如下所示：

declare -A map=([foobar]=foo [poohbah]=pooh [zoowicky]=zoo)

循环遍历它以构建jq脚本来替换密钥。

jq_script=
for old in "${!map[@]}"; do
  new="${map[$old]}"
  jq_script+="${jq_script:+|}if has(\"$old\") then { \"$new\": .[\"$old\"] } + del(.[\"$old\"]) else . end"
done

你运行的

：

jq "$jq_script" <old.json >new.json

在我的示例JSON上：

{
  "foobar": 1,
  "zoowicky": 3,
  "different": 4
}

它没有地图中的所有键，并且确实有一个不在地图中的键，它会产生这样的结果：

{
  "zoo": 3,
  "foo": 1,
  "different": 4
}

Answer 3

只需使用sed代替键。

#!/bin/bash

function replaceKeyWithNewKey() {

    json_file="$1"
    old_key="$2"
    new_key="$3"

    # you'd better to backup first.
    sed -i "" "s/${old_key}/${new_key}/g" ${json_file}
}

old_keys=("Rec_Open_Date" "MSISDN" "IMEI" "Data_Volume_Bytes" "Device_Manufacturer" "Device_Model" "Product_Description" "Data_Volume_MB")

new_keys=("r" "m" "i" "d" "f" "l" "s" "d2")

# input json file
json_file="input"

for ((i = 0; i < ${#old_keys[@]}; ++i)) do
    replaceKeyWithNewKey ${json_file} "${old_keys[$i]}" "${new_keys[$i]}"
done

使用python / bash缩短json对象中的键名

3 个答案: