Android 如何解析这个JSON字符串？

我正在开发一个应用程序，我必须解析下面的JSON

有人能告诉我怎么做吗

{
    "navigation_entries": {
        "home": {
            "children": [
                "favorites",
                "calendar",
                "offers",
                "wineries",
                "dining",
                "lodging",
                "things_to_do",
                "weather",
                "settings"
            ],
            "banner_image": "home_page_banner",
            "icon": {
                "small": "icon_home_small",
                "large": "icon_home_large"
            },
            "type": "home_page",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "link_name": "Home"
        },
        "wineries": {
            "display_name": "Wineries",
            "icon": {
                "small": "icon_wineries_small",
                "large": "icon_wineries_large"
            },
            "line_template": "wineries_list_template",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "link_name": "Wineries",
            "type": "leaf_list",
            "leaf_template": "wineries_leaf_template",
            "section": "wineries"
        },
        "dining": {
            "display_name": "Dining",
            "icon": {
                "small": "icon_dining_small",
                "large": "icon_dining_large"
            },
            "line_template": "dining_list_template",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "link_name": "Dining",
            "type": "leaf_list",
            "leaf_template": "dining_leaf_template",
            "section": "dining"
        },
        "offers_dining": {
            "display_name": "Offers => Dining",
            "list_name": "Dining",
            "line_template": "offers_dining_list_template",
            "type": "leaf_list",
            "leaf_template": "offers_dining_leaf_template",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "section": "offers_dining"
        },
        "favorites": {
            "display_name": "Favorites",
            "icon": {
                "small": "icon_favorites_small",
                "large": "icon_favorites_large"
            },
            "type": "favorites",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "link_name": "Favorites"
        },
        "offers": {
            "display_name": "Offers",
            "children": [
                "offers_wineries",
                "offers_dining",
                "offers_lodging",
                "offers_things_to_do"
            ],
            "icon": {
                "small": "icon_offers_small",
                "large": "icon_offers_large"
            },
            "type": "navigation_list",
            "forward_icon": {
                "small": "icon_right_arrow_small",
                "large": "icon_right_arrow_large"
            },
            "link_name": "Offers"
        }
    },
    "type": "navigation"
}

您可以这样使用和：

String json = "{"
     + "  \"query\": \"Pizza\", "
     + "  \"locations\": [ 94043, 90210 ] "
     + "}";

JSONObject object = (JSONObject) new JSONTokener(json).nextValue();
String query = object.getString("query");
JSONArray locations = object.getJSONArray("locations");

您可以使用将JSON字符串格式化为一种更容易查看的格式。 Android提供了一个合适的库来做你想做的事情

是您需要了解如何使用Android JSON API的手册页面

您可以看到一个教程，了解如何解析JSON字符串

{
    "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 such as DocBook.",
                        "GlossSeeAlso": ["GML", "XML"]
                    },
                    "GlossSee": "markup"
                }
            }
        }
    }
}

如果上面的示例是您的字符串，您可以将其解析并传递给JSONObject构造函数：

String jsonString = "...."; // the above string
try {
    JSONObject obj = new JSONObject(jsonString);
} catch (JSONException e) {
    // This exception is thrown if jsonString is not correctly formatted
}

现在，如果您想在上面的字符串中获得标记为“GlossList”的JSONObject，可以执行以下操作：

JSONObject glossList = obj.getJSONObject("glossary").getJSONObject("GlossDiv").getJSONObject("GlossList");

还有另一种可能性：您还可以获得一些JSONArray。在我们的示例中，数组glossSee还包括：

JSONArray glossSee = glossList.getJSONObject("GlossEntry").getJSONObject("GlossDef").getJSONArray("GlossSeeAlso");

要直接获取此数组的值，可以使用方法

getString（inti）

；如果数组的内容是JSONObject，则可以使用方法

getJSONObject（inti）

。您还可以使用方法

getString（stringlable）

直接获取JSONObject中的字符串：

String title = glossDive.getString("title");
String firstElement = glossSee.getString(0);

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其他示例可以在中找到。

jsonlint.com有一个非常好的json格式化程序。这会使你的文章容易理解

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要解析此数据，可以使用JSONObject（org.json.JSONObject）

好的格式是您的朋友。上面提供的json字符串无效。请提供正确的Json字符串。您可以使用以下链接检查其有效性：上面的JSON字符串无效，因为开头有“show”标签可能与Vito Shadow重复，请告诉我如何在android中使用google图表和Fusion图表。如果你喜欢这个答案，你能接受我的答案吗？我从来没有用过谷歌和福森图表。。。对不起，我帮不了你。。。试着打开另一个问题，有人会帮助你；）