Question

我正在使用Yummly API中的数据来获取食谱详细信息。我正在创建一个Android应用程序，它将根据用户输入的匹配成分显示食谱。我试图进入“recipeName”标签，但它抱怨它没有价值，我尝试了各种方法。我正在尝试将所有配方名称添加到HashMap中，稍后将在ListView中显示。

下面是JSON对象形式的一个配方（39个）的示例。

Click here for full response

{
    "attribution": {
        "html": "<a href='http://www.yummly.com/recipes/soup'>soup recipes</a> search powered by <img src=''/>",
        "url": "http://www.yummly.com/recipes/soup",
        "text": "soup recipes: search powered by Yummly",
        "logo": ""
    },
    "totalMatchCount": 39,
    "facetCounts": {},
    "matches": [
        {
            "attributes": {
                "course": [
                    "Soups"
                ],
                "cuisine": [
                    "Italian"
                ]
            },
            "flavors": {
                "salty": 0.6666666666666666,
                "sour": 0.8333333333333334,
                "sweet": 0.6666666666666666,
                "bitter": 0.5,
                "meaty": 0.16666666666666666,
                "piquant": 0.5
            },
            "rating": 4.6,
            "id": "Vegetarian-Cabbage-Soup-Recipezaar",
            "smallImageUrls": [],
            "sourceDisplayName": "Food.com",
            "totalTimeInSeconds": 4500,
            "ingredients": [
                "garlic cloves",
                "ground pepper",
                "diced tomatoes",
                "celery",
                "tomato juice",
                "salt",
                "cabbage",
                "bell peppers",
                "oregano",
                "carrots",
                "basil",
                "vegetable broth",
                "chili pepper flakes",
                "green beans",
                "onions",
                "onion soup mix"
            ],
            "recipeName": "Vegetarian Cabbage Soup"
        }
}

我的搜索类的片段

//imports above  

    private static final String TAG_TOTAL_MATCH = "totalMatchCount";
    private static final String TAG_MATCHES = "matches";
    private static final String TAG_NAME = "recipeName";
    private static final String TAG_RATING = "rating";

        JSONParser jsonParser = new JSONParser();

        protected String doInBackground(String... args) {    
            JSONObject json = jsonParser.makeHttpRequest(full_url, "GET", params);

            // check log cat for JSON response
            Log.d("Response", json.toString());

            // check for success tag
            try {
                int count = json.getInt(TAG_TOTAL_MATCH);
                String name = json.getString(TAG_NAME);
                String matches = json.getString(TAG_MATCHES);
                JSONObject namelist = json.getJSONObject(matches).getJSONObject(name);

                HashMap<String, String> pairs = new HashMap<String, String>();

                if (count > 0) {
                    int i;

                    //populate Hashmap with recipe names
                    for (i=1; i<=count; i++){
                        Iterator it = namelist.keys();
                        while (it.hasNext()) {
                            String n = (String) it.next();
                            pairs.put(n,name);
                        }
                    }
                    //display Hashmap in terminal
                    for(int j =0; j<pairs.size(); j++) {
                        System.out.println(pairs.get(j));

                }

            }
        }
    }

我没有包含所有内容，因为它不相关。 Log.d（）确实给了我一个JSON对象，就像这个问题顶部显示的那样，所以JSON被成功解析，只是Android工作室说没有“recipeName”的值。我做错了什么？

Answer 1

recipeName和rating位于matches内，但您尝试访问它们，就像它们与totalMatchCount处于同一级别一样。

首先，请考虑删除此行：

String matches = json.getString(TAG_MATCHES);

并更改下一行以从根对象中获取matches数组：

JSONArray matches = json.getJSONArray(TAG_MATCHES);

我确定你可以通过迭代匹配数组并独立处理每个食谱来从中获取它。

从JSON中解析嵌套值并添加到HashMap中

1 个答案: