使用python/bash的

Question

这是我写的，这将改变/缩短JSON对象上的所有按键的功能在一个JSON对象，缩短按键的名称

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 
    } 

我所寻找的是一个方法来做到这在终端使用bash或python或其他，但我想在终端做到这一点。所以我会在file.json上运行该脚本，结果将是file2.json与file2.json具有较短的键名。 我该怎么做？

对于bash我想使用sed，但我不认为这会取代我不想取代的值。 我知道的python的小可能是要走的路。

这里是一个控制台打印：

>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] 

这里是样本数据来测试我的功能上：

var json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"[email protected]","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"[email protected]","image":"some_image_url","name":"Name 2"}]'; 

Answer 1

假设你知道如何JSON的加载到一个Python数据结构。您提供的样本数据将是dict s的Python list。我认为最直接的方法是构建一个将旧密钥映射到新密钥的字典，然后使用字典解析遍历列表，以便使用您创建的密钥名称映射中的新密钥名称重建dict 。使用您的样本：

In [8]: jobj = [{"_id":"5078c3a803ff4197dc81fbfb","email":"[email protected]","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"[email protected]","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': '[email protected]', 
    'id': '5078c3a803ff4197dc81fbfb', 
    'img': 'some_image_url', 
    'n': 'Name 1'}, 
{'e': '[email protected]', 
    'id': '5078c3a803ff4197dc81fbfc', 
    'img': 'some_image_url', 
    'n': 'Name 2'}] 

编辑添加的

如果你想确保所有的钥匙留在顺序，您需要从使用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) 

输出：

[OrderedDict([('id', '5078c3a803ff4197dc81fbfb'), 
       ('e', '[email protected]'), 
       ('img', 'some_image_url'), 
       ('n', 'Name 1')]), 
OrderedDict([('id', '5078c3a803ff4197dc81fbfc'), 
       ('e', '[email protected]'), 
       ('img', 'some_image_url'), 
       ('n', 'Name 2')])] 

Answer 2

Just使用sed替换密钥。

#!/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

如果你将要操纵在命令行JSON，我建议安装jq。

你可以把你的钥匙地图关联数组在bash 4+，这样的事情：

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

环通过它来建立一个jq脚本来代替钥匙。

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 

你像这样运行：

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

在我的样本JSON：

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

不都在地图上的按键确实有一键不在地图上，它产生了这样的结果：

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