Question

我在下面有JSON响应，我想计算“ref_id”对象中字符串匹配的出现次数。但我不知道在循环数组和计算匹配时从哪里开始。我应该感谢任何帮助或建议

{
"tag": "checkthisServer",
"success": 1,
"error": 0,
"svd_ids": [
    {
        "ref_id": "f91e2651"
    },
    {
        "ref_id": "f91e2651"
    },
    {
        "ref_id": "ue87sr5d"
    },
    {
        "ref_id": "f91e2651"
    }
    {
        "ref_id": "ue87sr5d"
    }
]
}

输出应为：

f91e2651: 3
ue87sr5d: 2

Answer 1

您可以在ArrayList上使用Collections.frequency执行此操作。首先从JSONObeject获取JSONArray，然后将其转换为ArrayList：

JSONObject mainobj=new JSONObject("your json string");
// get svd_ids JSONArray
JSONArray json_array=mainobj.getJSONArray("svd_ids");

// get ArrayList from JSONArray

   ArrayList<String> listsvd_ids = new ArrayList<String>();  
   for (int i=0;i<json_array.length();i++){ 
    listsvd_ids.add(json_array.get(i).toString());
   } 

//  occurrence of f91e2651
int occur_f91e2651 = Collections.frequency(listsvd_ids, "f91e2651");
//  occurrence of ue87sr5d
int occur_ue87sr5d = Collections.frequency(listsvd_ids, "ue87sr5d");
//.....

Answer 2

尝试一下：这里的filterValues是您的JSON

filterValues = filterValues.reduce((newObj, obj) => {
let value = 1;
if(newObj[obj[ref_id]]) {
   value = value + 1;
   newObj[obj[ref_id]] = obj[ref_id] + ' (' + value + ')';
} else {
    newObj[obj[ref_id]] = obj[ref_id] + ' (' + value + ')';
}            
return newObj;
}, {});

Answer 3

试试这个，

JSONArray responseArray = jsonObject.getJSONArray("svd_ids");
int countFirst = 0, countSecond = 0;

for (int i = 0; i < responseArray.length(); i++) {
    JSONObject obj = responseArray.getJSONObject(i);
    if (obj.optString("ref_id").equals("f91e2651") {
        countFirst = countFirst + 1;
    }
    if (obj.optString("ref_id").equals("ue87sr5d") {
        countSecond = countSecond + 1;
    }
}

计算JSON数组中字符串的出现次数

3 个答案: