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社区首页 >问答首页 >如何动态获取json路径,并使用路径修改json字段的值

如何动态获取json路径,并使用路径修改json字段的值
EN

Stack Overflow用户
提问于 2019-06-14 07:16:48
回答 2查看 317关注 0票数 2

我有很多json文件,它们的结构是不同的。我每次都需要更改json中一个字段的值,其他字段的值保持不变。

现在,我已经能够动态地获得json的路径代码,如下所示

代码语言:javascript
复制
def get_paths(source):
    paths = []
    if isinstance(source, collections.MutableMapping):  # found a dict-like structure...
        for k, v in source.items():  # iterate over it; Python 2.x: source.iteritems()
            paths.append([k])  # add the current child path
            paths += [[k] + x for x in get_paths(v)]  # get sub-paths, extend with the current
    # else, check if a list-like structure, remove if you don't want list paths included
    elif isinstance(source, collections.Sequence) and not isinstance(source, str):
        for i, v in enumerate(source):
            paths.append([i])
            paths += [[i] + x for x in get_paths(v)]  # get sub-paths, extend with the current
    return paths

其中一个json示例如下,它只是众多json中的一个:

代码语言:javascript
复制
{
    "glossary": {
        "title": "example glossary",
        "GlossDiv": {
            "title": "S",
            "GlossList": {
                "GlossEntry": {
                    "ID": "SGML",
                    "SortAs": "SGML",
                    "GlossTerm": "Standard Generalized Markup Language",
                    "Acronym": "SGML",
                    "Abbrev": "ISO 8879:1986",
                    "GlossDef": {
                        "para": "A meta-markup language, used to create markup languages...",
                        "GlossSeeAlso": [
                            "GML",
                            "XML"
                        ]
                    },
                    "GlossSee": "markup"
                }
            }
        }
    }
}

使用以下代码获取路径

代码语言:javascript
复制
def loadJson():
    try:
        with open('../json/test1.json', 'r') as loadf:
            load_dict = json.load(loadf)
            return load_dict
    except Exception as e:
        raise Exception("load json fail")

t_json = loadJson()
paths = get_paths(loadJson())

此示例中的路径如下:

代码语言:javascript
复制
[['glossary'],
 ['glossary', 'title'],
 ['glossary', 'GlossDiv'],
 ['glossary', 'GlossDiv', 'title'],
 ['glossary', 'GlossDiv', 'GlossList'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'ID'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'SortAs'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossTerm'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Acronym'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'Abbrev'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'para'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso'],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso', 0],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossDef', 'GlossSeeAlso', 1],
 ['glossary', 'GlossDiv', 'GlossList', 'GlossEntry', 'GlossSee']]

现在的问题是如何通过获取的路径动态修改其中一个字段的值?

示例:我想将"title": "S"修改为"title": "M"

如何使用['glossary', 'GlossDiv', 'title']做到这一点呢?

代码语言:javascript
复制
for path in paths:
   # How to get t_json[path[0]][path[1]][path[2]]?
EN

回答 2

Stack Overflow用户

发布于 2019-06-14 08:41:31

如何获取t_json[path[0]][path[1]][path[2]]

简单:我们只需要遍历path,一次应用一个索引操作。这要求我们在每一步之后记住我们的进度,最简单的方法是重用一个在整个路径中跟踪其路径的变量。因此,例如:

代码语言:javascript
复制
element = t_json
for path_item in path:
    element = element[path_item]
票数 1
EN

Stack Overflow用户

发布于 2019-06-14 09:36:02

下面是如何使用path的基本思想,它是一个键序列:

代码语言:javascript
复制
from functools import reduce

# From https://stackoverflow.com/a/28225747/355230
def recursive_get(d, *keys):
    return reduce(lambda c, k: c.get(k, {}), keys, d)

t_json = loadJson()
path = ['glossary', 'GlossDiv', 'title']
sub_dict = recursive_get(t_json, *path[:-1])
sub_dict['title'] = 'M'
print(t_json)
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/56589613

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