8

我正在使用以下方法检查 JSON 字符串中的对象是否存在：

JSONObject json = null;

         try {
            json = new JSONObject(myJsonString);
        } catch (JSONException e) { e.printStackTrace(); } 


        if(json.has("myObject")) System.out.println("EXISTS");

        else System.out.println("DOESN'T EXIST");

当我尝试检查子对象是否存在时出现问题。例如：

...,"queue":{"building":{"q0":{"id":177779,...

队列总是存在并且也在构建，但q0并不总是存在。那么，如何检查q0的存在？而且，有没有办法使用 Gson 库来检查它？

先感谢您！

4

2 回答 2

11

如果尝试失败，您可以简单地尝试一下并返回 null 。或者你可以将你的尝试分解成小块来监控它失败的地方。

/**
 * This method will return the JSONObject q0, if it exists
 * If it doesn't exist it will return NULL
 *
 */
private JSONObject getQZero(JSONObject json)
{   
    try
    {
        return json.getJSONObject("queue").getJSONObject("building").getJSONObject("q0");
    }
    catch (JSONException e)
    {
        // This could be triggered either because there is no q0
        //   or because the JSON structure is different from what was expected.
        return null;
    }    
}

如果要打印每个级别的日志，也可以逐步进行；

/**
 * This method will show where your jsonparsing fails.
 * It will throw a JSONOException if the json is way different from what 
 *   was expected, and otherwise it will print a log of where the parsing
 *   failed.
 */
private JSONObject getQZero(JSONObject json) throws JSONException
{       
    // Stop if no queue
    if (! myObject.has("queue") 
    {
        Log.d(TAG, "no queue!");
        return null;
    }

    JSONObject queue = myObject.getJSONObject("queue");

    // Stop if no building
    if (! queue.has("building")
    {
        Log.d(TAG, "no building!");
        return null;
    }

    JSONObject building = queue.getJSONObject("building")

    // Stop if no q0
    if (! building.has("q0"))
    {
        Log.d(TAG, "no q0!");
        return null;
    }

    JSONObject q0 = building.getJSONObject("q0");
    // Q0 is returned here. If the method returned earlier, it returned NULL
    // You could also do nested ifs, but the indentation gets crazy
    return q0;
}
于 2012-04-08T15:51:46.877 回答
1

利用例外对您有利

   try {
        JSONObject i = json. getJSONObject("q0");
        // Is there do something
    } catch (JSONException e) { 
       // Isn't there
    }

http://www.json.org/javadoc/org/json/JSONObject.html#getJSONObject(java.lang.String)

JSONException - 如果未找到键或值不是 JSONObject。

于 2012-04-08T15:38:25.780 回答