使用python/bash的

Question

這是我寫的，這將改變/縮短JSON對象上的所有按鍵的功能在一個JSON對象，縮短按鍵的名稱

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 
    } 

我所尋找的是一個方法來做到這在終端使用bash或python或其他，但我想在終端做到這一點。所以我會在file.json上運行該腳本，結果將是file2.json與file2.json具有較短的鍵名。 我該怎麼做？

對於bash我想使用sed，但我不認爲這會取代我不想取代的值。 我知道的python的小可能是要走的路。

這裏是一個控制檯打印：

>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] 

這裏是樣本數據來測試我的功能上：

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

假設你知道如何JSON的加載到一個Python數據結構。您提供的樣本數據將是dict s的Python list。我認爲最直接的方法是構建一個將舊密鑰映射到新密鑰的字典，然後使用字典解析遍歷列表，以便使用您創建的密鑰名稱映射中的新密鑰名稱重建dict 。使用您的樣本：

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'}] 

編輯添加的

如果你想確保所有的鑰匙留在順序，您需要從使用OrderedDictcollections

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) 

輸出：

[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')])] 

Answer 2

Just使用sed替換密鑰。

#!/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 

Answer 3

如果你將要操縱在命令行JSON，我建議安裝jq。

你可以把你的鑰匙地圖關聯數組在bash 4+，這樣的事情：

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

環通過它來建立一個jq腳本來代替鑰匙。

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 

你像這樣運行：

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

在我的樣本JSON：

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

不都在地圖上的按鍵確實有一鍵不在地圖上，它產生了這樣的結果：

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