Java解析器Json作为多维数组
我在解析Json时遇到了一些麻烦。这是我拥有的Json:Java解析器Json作为多维数组,java,arrays,json,Java,Arrays,Json,我在解析Json时遇到了一些麻烦。这是我拥有的Json: { "responseHeader": { "status": 0, "QTime": 1 }, "response": { "numFound": 3447, "start": 0, "docs": [ { "sku": "5848-05855-0064", "skus": [ "5848-05855-0064",
{
"responseHeader": {
"status": 0,
"QTime": 1
},
"response": {
"numFound": 3447,
"start": 0,
"docs": [
{
"sku": "5848-05855-0064",
"skus": [
"5848-05855-0064",
"5848-05855-0064-5340"
]
},
{
"sku": "5848-05849-0059",
"skus": [
"5848-05849-0059",
"5848-05849-0059-5340"
]
}
]
},
"facet_counts": {
"facet_queries": {},
"facet_fields": {},
"facet_dates": {},
"facet_ranges": {}
}
}
]
},
"facet_counts": {
"facet_queries": {},
"facet_fields": {},
"facet_dates": {},
"facet_ranges": {}
}
}
我可以通过以下代码获得“sku”值:
String jsonData = responses.body().string();
JSONObject Jobject = new JSONObject(jsonData);
Jobject = (JSONObject) Jobject.get("response");
Object skuValues = Jobject.get("docs");
JSONArray jArray = (JSONArray) skuValues;
JSONObject skuConfig = jArray.getJSONObject(1);
String sku = skuConfig.getString("sku");
但我无法获得“SKU”值。有人能指导我如何从JSON中获取“SKU”值吗?要检索SKU,您可以执行以下操作:-
JSONArray skus = skuConfig.getJSONArray("skus");
skus1 = skus.getString(0);
要获得这些值,可以执行以下操作:-
JSONArray skus = skuConfig.getJSONArray("skus");
skus1 = skus.getString(0);
以下几点对我很有用:
Gson gson = new Gson();
Map fromJson = gson.fromJson("{ \"responseHeader\": { \"status\": 0, \"QTime\": 1 }, \"response\": { \"numFound\": 3447, \"start\": 0, \"docs\": [ { \"sku\": \"5848-05855-0064\", \"skus\": [ \"5848-05855-0064\", \"5848-05855-0064-5340\" ] }, { \"sku\": \"5848-05849-0059\", \"skus\": [ \"5848-05849-0059\", \"5848-05849-0059-5340\" ] } ] }, \"facet_counts\": { \"facet_queries\": {}, \"facet_fields\": {}, \"facet_dates\": {}, \"facet_ranges\": {} }}", Map.class);
JSONObject Jobject = new JSONObject(fromJson);
StringMap Jobject1 = (StringMap) Jobject.get("response");
Object skuValues = Jobject1.get("docs");
ArrayList jArray = (ArrayList) skuValues;
StringMap skuConfig = (StringMap) jArray.get(0);
System.out.println(skuConfig);
System.out.println(skuConfig.get("skus"));
第1点:您的示例JSON错误,请使用以下链接验证并格式化JSON 第2点:使用以下代码
JSONObject Jobject;
Jobject = new JSONObject(jsonData);
Jobject = (JSONObject) Jobject.get("response");
JSONArray docs = Jobject.getJSONArray("docs");
for (int i = 0; i < docs.length(); i++) {
JSONObject skuDoc = docs.getJSONObject(i);
String skuValue = skuDoc.optString("sku");
JSONArray skusArr = skuDoc.getJSONArray("skus");
System.out.println(skusArr);
}
JSONObject作业对象;
Jobject=新的JSONObject(jsonData);
Jobject=(JSONObject)Jobject.get(“响应”);
JSONArray docs=Jobject.getJSONArray(“docs”);
对于(int i=0;i
这个问题看起来很相似。这可能会有帮助,谢谢你!!为我工作!!