python3中字典与列表相互嵌套取值
如何从字典中取值?
如何从字典中的列表中取值?
如何从字典的列表的字典中取值?
实例代码:
dian = {
"name":"护脸霜",
"innerCode":"",
"clusterSn":{
"dataSourceType": "regular",
"bizSystem": "RDRSEW",
},
"description":"<p>1234567</p>",
"skuOptions":[
{
"id":1,
"name":"颜色",
"naed":"尺码",
"values":[
"黑色"
]
}
],
"categoryId":"343b5dbff0654d779fd81f4d8259ce60",
"MinOrderType":{
"keywords": "",
"keyposswr":[
{
"userName": "18110522",
"onedisty": [
{
"zhu_leg": "猪猪侠的腿",
"zhu_hand": "猪猪侠的手",
}
],
"pwegdt": "猪猪侠的手表"
}
],
"gbCode": "GBM852741",
"tableName": "test_table01",
}
}
上边是一个字典 ,字典中包含列表,包含字典,还有列表中包含字典。对于该如何取值:
1 print(type(dian))
2 print(dian["categoryId"])
3 print(type(dian["categoryId"])) # 取出的是字符串
4 print(dian["clusterSn"]["bizSystem"])
5 print(dian["skuOptions"]) # 取出的是list
6 print(type(dian["skuOptions"]))
7 print(dian["skuOptions"][0]) # 取出的是dict
8 print(type(dian["skuOptions"][0]))
9 print(dian["skuOptions"][0]["name"]) # 取dict中list中包含的dict指定的值
10 print(dian["MinOrderType"]["keyposswr"][0]["onedisty"][0]["zhu_leg"])
执行结果如下:
学如逆海行舟,不进则退。