Question

我在Python工作，我有以JSON格式存储的数据。我想编译与某个键对应的所有数据。例如

{"menu": {
    "header": "SVG Viewer",
    "items": [
        {"id": "Open"},
        {"id": "OpenNew", "label": "Open New"},
        null,
        {"id": "ZoomIn", "label": "Zoom In"},
        {"id": "ZoomOut", "label": "Zoom Out"},
        {"id": "OriginalView", "label": "Original View"},
        null,
        {"id": "Quality"},
        {"id": "Pause"},
        {"id": "Mute"},
        null,
        {"id": "Find", "label": "Find..."},
        {"id": "FindAgain", "label": "Find Again"},
        {"id": "Copy"},
        {"id": "CopyAgain", "label": "Copy Again"},
        {"id": "CopySVG", "label": "Copy SVG"},
        {"id": "ViewSVG", "label": "View SVG"},
        {"id": "ViewSource", "label": "View Source"},
        {"id": "SaveAs", "label": "Save As"},
        null,
        {"id": "Help"},
        {"id": "About", "label": "About Adobe CVG Viewer..."}
    ]
}}

我想将所有“标签”键的内容编译成列表。

有没有办法在没有“走下树”的情况下做到这一点，即无需迭代包含的键（“menu”＆gt;＆gt;“item”＆gt;＆gt;“label”）？

最好是，我正在寻找一个基于JSON的解决方案，而不是使用REGEX来定位内容的技巧。

Answer 1

您可以使用jsonpath_rw：

import json
from jsonpath_rw import parse

d = json.loads("""{"menu": {
        "header": "SVG Viewer",
        "items": [
            {"id": "Open"},
            {"id": "OpenNew", "label": "Open New"},
            null,
            {"id": "ZoomIn", "label": "Zoom In"},
            {"id": "ZoomOut", "label": "Zoom Out"},
            {"id": "OriginalView", "label": "Original View"},
            null,
            {"id": "Quality"},
            {"id": "Pause"},
            {"id": "Mute"},
            null,
            {"id": "Find", "label": "Find..."},
            {"id": "FindAgain", "label": "Find Again"},
            {"id": "Copy"},
            {"id": "CopyAgain", "label": "Copy Again"},
            {"id": "CopySVG", "label": "Copy SVG"},
            {"id": "ViewSVG", "label": "View SVG"},
            {"id": "ViewSource", "label": "View Source"},
            {"id": "SaveAs", "label": "Save As"},
            null,
            {"id": "Help"},
            {"id": "About", "label": "About Adobe CVG Viewer..."}
        ]
    }}""")

    get_labels = parse("menu.items[*].label")

    for match in get_labels.find(d):
        print match.value   # value of label

请注意，这基本上可以为您“走下树”，但它的更高级别界面使其更易于使用。

在Python中编译来自JSON数据的多个关键内容

1 个答案: