表格：
categories = Table("categories", metadata,
Column("id", Integer, primary_key=True),
Column("name", String),
Column("parent_id", Integer, ForeignKey("categories.id"),
CheckConstraint('id!=parent_id'), nullable=True),
)
一个类别可以有很多孩子，但只有一个孩子。我使用CTE得到了如下字典值列表：例如。对于id：14，父亲是13并且从父8-> 10-> 12-> 13-> 14遍历，其中父8没有父id。
[
{
"id": 14,
"name": "cat14",
"parent_id": 13,
"path_info": [
8,
10,
12,
13,
14
]
},
{
"id": 15,
"name": "cat15",
"parent_id": 13,
"path_info": [
8,
10,
12,
13,
15
]
}
]
我想将父级的属性也作为子类别嵌入到列表中：
{
"id": 14,
"name": "cat14",
"parent_id": 13,
"subcats": [
{
"id: 8",
"name": "cat8",
"parent_id":null
},
{
"id: 10",
"name": "cat10",
"parent_id":8
},
{
"id: 12",
"name": "cat12",
"parent_id":10
},
and similarly for ids 13 and 14.....
]
},
{
"id": 15,
"name": "cat15",
"parent_id": 13,
"subcats": [
{
"id: 8",
"name": "cat8",
"parent_id":null
},
{
"id: 10",
"name": "cat10",
"parent_id":8
},
{
"id: 12",
"name": "cat12",
"parent_id":10
},
and similarly for ids 13, 14, 15.....
]
}
] 请注意，'path_info'已从字典中删除，并且每个id都已显示其详细信息。我想要以上面缩进格式的json转储。怎么去？使用烧瓶0.10，python 2.7
答案 0 :(得分：2)
使用一些list / dict理解可以容忍这种方法。
lst = [{"id": 14, "name": "cat14", "parent_id": 13, "path_info": [8, 10, 12, 13, 14]}, {"id": 15, "name": "cat15", "parent_id": 13, "path_info": [8, 10, 12, 13, 15]}]
master_dct = { d['id'] : d for d in lst}
for d in lst:
d['subcats'] = [{field : master_dct[i][field] for field in ['id', 'name', 'parent_id']} \
for i in d['path_info'] if i in master_dct]
import json
with open('out.json', 'w') as f:
json.dump(lst, f)
答案 1 :(得分：1)
您可以在python代码中执行它：
鉴于我们有一个json对象。我稍微修改了它 - 添加了缺少的节点并根据规范要求包装到对象中：
{
"array": [
{
"id": 14,
"name": "cat14",
"parent_id": 13,
"path_info": [
8,
10,
12,
13,
14
]
},
{
"id": 15,
"name": "cat15",
"parent_id": 13,
"path_info": [
8,
10,
12,
13,
15
]
},
{
"id": 13,
"name": "cat13",
"parent_id": 12,
"path_info": [
8,
10,
12,
13
]
},
{
"id": 12,
"name": "cat12",
"parent_id": 10,
"path_info": [
8,
10,
12
]
},
{
"id": 10,
"name": "cat10",
"parent_id": 8,
"path_info": [
8,
10
]
},
{
"id": 8,
"name": "cat8",
"parent_id": null,
"path_info": [
8
]
}
]
}
然后您可以使用以下代码：
# load data above from file
j=json.load(open('json_file_above.json')) #
# the array with real data we need
a=j['array']
# auxiliary dict which have node identificators as keys and nodes as values
d={x['id']:x for x in a}
# here the magic begins :)
for x in a:
# add new key with list to each element
x['subcats'] = [
# compose dict element for subcats
dict(id=i, name=d[i]['name'], parent_id=d[i]['parent_id'])
for
i
in [
# we take path_info id list and
# cut off the first element - itself
y for y in x['path_info'][1:]
]
]
del x['path_info']
确保你得到了你需要的东西：
>>> print(json.dumps(a, indent=True))
[
{
"name": "cat14",
"subcats": [
{
"name": "cat10",
"id": 10,
"parent_id": 8
},
{
"name": "cat12",
"id": 12,
"parent_id": 10
},
{
"name": "cat13",
"id": 13,
"parent_id": 12
},
{
"name": "cat14",
"id": 14,
"parent_id": 13
}
],
"id": 14,
"parent_id": 13
},
{
"name": "cat15",
"subcats": [
{
"name": "cat10",
"id": 10,
"parent_id": 8
},
{
"name": "cat12",
"id": 12,
"parent_id": 10
},
{
"name": "cat13",
"id": 13,
"parent_id": 12
},
{
"name": "cat15",
"id": 15,
"parent_id": 13
}
],
"id": 15,
"parent_id": 13
},
{
"name": "cat13",
"subcats": [
{
"name": "cat10",
"id": 10,
"parent_id": 8
},
{
"name": "cat12",
"id": 12,
"parent_id": 10
},
{
"name": "cat13",
"id": 13,
"parent_id": 12
}
],
"id": 13,
"parent_id": 12
},
{
"name": "cat12",
"subcats": [
{
"name": "cat10",
"id": 10,
"parent_id": 8
},
{
"name": "cat12",
"id": 12,
"parent_id": 10
}
],
"id": 12,
"parent_id": 10
},
{
"name": "cat10",
"subcats": [
{
"name": "cat10",
"id": 10,
"parent_id": 8
}
],
"id": 10,
"parent_id": 8
},
{
"name": "cat8",
"subcats": [],
"id": 8,
"parent_id": null
}
]
>>>
答案 2 :(得分：1)
这个pythonic代码：简单明了
RewriteEngine On
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^(.*)$ index.php/$1 [L]
<Files "index.php">
AcceptPathInfo On
</Files>