java - 如何将 sqlite 中的数组与 json 数组的内容进行比较

Question

您好，我在比较来自 sqlite 数据库的数组和来自 Web 服务响应的 json 数组的最佳策略时遇到了问题。请告诉我比较这两个数组的最佳方法。这是我的代码:

public void getAllElements() {
     Map<String, String> map = new HashMap<String, String>();
     try {
        //select all query
        db.open();
        Cursor cc= db.getAllEntries();
        try {
            // looping through all rows and adding to list
            if (cc.moveToFirst()) {
                do {
                   map.put("ID",cc.getString(cc.getColumnIndex(DbHelper.ENTRY_ID)));
                } while (cc.moveToNext());
            }
        } finally {
            try { cc.close(); } catch (Exception ignore) {}
        }
    } finally {
         try { db.close(); } catch (Exception ignore) {}
    }
    // return  list 
}

protected String doInBackground(String... params) {
    // Creating service handler class instance

    JSONParser jParser = new JSONParser();
    DefaultHttpClient client = new DefaultHttpClient(); // Thread
    String uri = "url";

    HttpGet httpget = new HttpGet(uri);
    httpget.setHeader("Cookie", cookie);

    try {
        response = client.execute(httpget);
        // response
        String res = EntityUtils.toString(response.getEntity());

        // jarray with the json responde
        jArr = new JSONArray(res);
        System.out.println("response" + jArr);

        for (int i = 0; i < jArr.length(); i++) {
            // create json object
            JSONObject jsonObject = jArr.getJSONObject(i);
            // easy parse fields
            String vid = jsonObject.getString("vid");
            String nid = jsonObject.getString("nid");
            // uid value
            String field_text = jsonObject.getString("field_text");
            String field_edit_ts = jsonObject.getString("field_editts");
            String field_deleted_flag = jsonObject.getString("field_deleted");
            String field_device_name = jsonObject.getString("field_devicename");
            String field_creation_ts = jsonObject.getString("field_creationts");
            String field_device_key = jsonObject.getString("field_devicekey");
         }
    } catch(Exception e){
    }
}

Answer 1

我建议将数据库和 JSON 数据包装在 class 中，并实现 equals(Object) 方法。

对于您的 JSON 数据，一旦您有了一个类(如上所述)，我将使用 GSON ( https://code.google.com/p/google-gson/ ) 直接转换为对象

然后，为了进行比较，将数据库或 JSON 数据转换为 List(来自 Collections 的内容，或 HashTable/HashMap(搜索速度更快))并使用Collections.contains(Object)(在List上)方法和一个包含计数，您可以将其与列表的大小。举例如下，填空:

List<YourObject> dbDataList = new ArrayList<YourObject>(Arrays.asList(yourDatabaseDataArray));
int containsCount = 0;
for (YourObject o : yourJsonDataArray){
    if (dbDataList.contains(o))
        containsCount ++;
}
// Success condition is up to you, I'm just comparing the sizes
if (containsCount == dbDataList.size()){
    // Success!
} else {
    // Failure
}