如何在同一条消息中多次循环使用不同类型的嵌套JSON对象 [英] How to loop different types of nested JSON objects multiple times in the same message
问题描述
Python noob再次出现.我正在尝试创建一个python脚本来自动生成具有多个项目的JSON，但使用for循环记录多次以生成它们，JSON消息的结构和基数如下:
Python noob here, again. I'm trying to create a python script to auto-generate a JSON with multiple item but records multiple times using a for loop to generate them, the JSON message is structured and cardinality are as follows:
messageHeader[1]
-item [1-*]
--itemAttributesA [0-1]
--itemAttributesB [0-1]
--itemAttributesC [0-1]
--itemLocaton [1]
--itemRelationships [0-1]
在遍历同一对象之前，我已经获得了一些非常好的帮助，但是对于一个记录，例如，仅是itemRelationships记录.但是，一旦我尝试创建包含多个项目(即5个)以及itemAttribute，itemLocation和itemRelationships的单个实例的一条消息，由于我不断收到关键错误，因此它不起作用.我试图定义与我想做的事情有关的keyError，但是不能将我做错的事情与其他地方的例子联系起来.
I've had some really good help before for looping through the same object but for one record for example just the itemRelationships record. However as soon as I try to create one message with many items (i.e. 5) and a single instance of an itemAttribute, itemLocation and itemRelationships it does not work as I keep getting a key error. I've tried to define what a keyError is in relation to what I am trying to do but cannot link what I am doing wrong to the examples else where.
这是我目前的代码:
import json
import random
data = {'messageID': random.randint(0, 2147483647), 'messageType': 'messageType'}
data['item'] = list()
itemAttributeType = input("Please selct what type of Attribute item has, either 'A', 'B' or 'C' :")
for x in range(0, 5):
data['item'].append({
'itemId': "I",
'itemType': "T"})
if itemAttributeType == "A":
data['item'][0]['itemAttributesA']
data['item'][0]['itemAttributesA'].append({
'attributeA': "ITA"})
elif itemAttributeType == "B":
data['item'][0]['itemAttributesB']
data['item'][0]['itemAttributesB'].append({
'attributeC': "ITB"})
else:
data['item'][0]['itemAttributesC']
data['item'][0]['itemAttributesC'].append({
'attributeC': "ITC"})
pass
data['item'][0]['itemLocation'] = {
'itemDetail': "ITC"}
itemRelation = input("Does the item have a relation: ")
if itemRelation > '':
data['item'][0]['itemRelations'] = {
'itemDetail': "relation"}
else:
pass
print(json.dumps(data, indent=4))
我尝试过也尝试了这段代码，这给了我更好的结果:
I have tried also tried this code which gives me better results:
import json
import random
data = {'messageID': random.randint(0, 2147483647), 'messageType': 'messageType'}
data['item'] = list()
itemAttributeType = input("Please selct what type of Attribute item has, either 'A', 'B' or 'C' :")
for x in range(0, 5):
data['item'].append({
'itemId': "I",
'itemType': "T"})
if itemAttributeType == "A":
data['item'][0]['itemAttributesA'] = {
'attributeA': "ITA"}
elif itemAttributeType == "B":
data['item'][0]['itemAttributesB'] = {
'attributeB': "ITB"}
else:
data['item'][0]['itemAttributesC'] = {
'attributeC': "ITC"}
pass
data['item'][0]['itemLocation'] = {
'itemDetail': "ITC"}
itemRelation = input("Does the item have a relation: ")
if itemRelation > '':
data['item'][0]['itemRelations'] = {
'itemDetail': "relation"}
else:
pass
print(json.dumps(data, indent=4))
这实际上给了我一个结果，但是给了我messageHeader，item，itemAttributeA，itemLocation，itemRelations，最后是四个item记录，如下所示:
This actually gives me a result but gives me messageHeader, item, itemAttributeA, itemLocation, itemRelations, and then four items records at the end as follows:
{
"messageID": 1926708779,
"messageType": "messageType",
"item": [
{
"itemId": "I",
"itemType": "T",
"itemAttributesA": {
"itemLocationType": "ITA"
},
"itemLocation": {
"itemDetail": "location"
},
"itemRelations": {
"itemDetail": "relation"
}
},
{
"itemId": "I",
"itemType": "T"
},
{
"itemId": "I",
"itemType": "T"
},
{
"itemId": "I",
"itemType": "T"
},
{
"itemId": "I",
"itemType": "T"
}
]
}
我要实现的是此输出:
{
"messageID": 2018369867,
"messageType": "messageType",
"item": [{
"itemId": "I",
"itemType": "T",
"itemAttributesA": {
"attributeA": "ITA"
},
"itemLocation": {
"itemDetail": "Location"
},
"itemRelation": [{
"itemDetail": "D"
}]
}, {
"item": [{
"itemId": "I",
"itemType": "T",
"itemAttributesB": {
"attributeA": "ITB"
},
"itemLocation": {
"itemDetail": "Location"
},
"itemRelation": [{
"itemDetail": "D"
}]
}, {
"item": [{
"itemId": "I",
"itemType": "T",
"itemAttributesC": {
"attributeA": "ITC"
},
"itemLocation": {
"itemDetail": "Location"
},
"itemRelation": [{
"itemDetail": "D"
}]
}, {
"item": [{
"itemId": "I",
"itemType": "T",
"itemAttributesA": {
"attributeA": "ITA"
},
"itemLocation": {
"itemDetail": "Location"
},
"itemRelation": [{
"itemDetail": "D"
}]
},
{
"item": [{
"itemId": "I",
"itemType": "T",
"itemAttributesB": {
"attributeA": "ITB"
},
"itemLocation": {
"itemDetail": "Location"
},
"itemRelation": [{
"itemDetail": "D"
}]
}]
}
]
}]
}]
}]
}
我一整天都在努力工作，以使其正常工作，但是远离代码，我哪里出错了，任何帮助将不胜感激
I've been at this for the best part of a whole day trying to get it to work, butchering away at code, where am I going wrong, any help would be greatly appreciated
推荐答案
您的关闭信息.我认为您缺少的部分是将dict添加到当前的dict中，并使用for循环进行缩进.
Your close. I think the part your are missing is adding the dict to your current dict and indentation with your for loop.
import json
import random
data = {'messageID': random.randint(0, 2147483647), 'messageType': 'messageType'}
data['item'] = list()
itemAttributeType = input("Please selct what type of Attribute item has, either 'A', 'B' or 'C' :")
for x in range(0, 5):
data['item'].append({
'itemId': "I",
'itemType': "T"})
if itemAttributeType == "A":
# First you need to add `itemAttributesA` to your dict:
data['item'][x]['itemAttributesA'] = dict()
# You could also do data['item'][x] = {'itemAttributesA': = dict()}
data['item'][x]['itemAttributesA']['attributeA'] = "ITA"
elif itemAttributeType == "B":
data['item'][x]['itemAttributesB'] = dict()
data['item'][x]['itemAttributesB']['attributeC'] = "ITB"
else:
data['item'][x]['itemAttributesC'] = dict()
data['item'][x]['itemAttributesC']['attributeC'] = "ITC"
data['item'][x]['itemLocation'] = {'itemDetail': "ITC"}
itemRelation = input("Does the item have a relation: ")
if itemRelation > '':
data['item'][x]['itemRelations'] = {'itemDetail': "relation"}
else:
pass
print(json.dumps(data, indent=4))
如果您的示例接近您真正想要的代码，则也可以大大缩短此代码:
This code can also be shortened considerably if your example is close to what you truly desire:
import json
import random
data = {'messageID': random.randint(0, 2147483647), 'messageType': 'messageType'}
data['item'] = list()
itemAttributeType = input("Please selct what type of Attribute item has, either 'A', 'B' or 'C' :")
for x in range(0, 5):
new_item = {
'itemId': "I",
'itemType': "T",
'itemAttributes' + str(itemAttributeType): {
'attribute' + str(itemAttributeType): "IT" + str(itemAttributeType)
},
'itemLocation': {'itemDetail': "ITC"}
}
itemRelation = input("Does the item have a relation: ")
if itemRelation > '':
new_item['itemRelations'] = {'itemDetail': itemRelation}
data['item'].append(new_item)
print(json.dumps(data, indent=4))
另一个说明::如果您想让messageID真正独一无二，那么您可能应该考虑一下UUID；否则，您的邮件ID可能会匹配.
Another note: If you want messageID to be truly unique than you should probably look into a UUID; otherwise you may have message ids that match.
import uuid
unique_id = str(uuid.uuid4())
print(unique_id)
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