Question

我尝试从Facebook sdk获取数据：

public void onCompleted(JSONObject object, GraphResponse response) {
    String name = null;
    try {
        name = object.getString("name");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    String profile = null;
    try {
        profile = object.getJSONObject("picture").getJSONObject("data").getString("url");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    String cover = null;
    try {
        cover = object.getJSONObject("cover").getString("source");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    String email = null;
    try {
        email = object.getString("email");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    String gender = null;
    try {
        gender = object.getString("gender");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    String birthday = null;
    try {
        birthday = object.getString("birthday");
    } catch (JSONException e) {
        e.printStackTrace();
    }
    new SessionManager(activity).storeUserInfo(name, profile, cover, email, gender, birthday);
}

这是非常不利的，因为有很多字段，每个字段可以为null，每个字段都需要一个try-cach博客。

我需要一个解决方案来更清晰地呈现代码。

Answer 1

如果您只想删除try catch块，请使用“opt”替换“get” （例如。optInt(Key)，optString(Key) ..）

如果密钥不存在，请注意“opt”将返回null，这可能会导致许多空指针异常而不进行检查。

Answer 2

好的，您可以通过调用object.has("key")来检查密钥是否存在，如下所示：

public void onCompleted(JSONObject object, GraphResponse response) {
    String name = "", profile = "", cover = "", email = "", gender = "", birthday = "";
    try {
        if (object.has("name")){
            name = object.getString("name");
        }
        if (object.has("picture")){
            profile = object.getJSONObject("picture").getJSONObject("data").getString("url");
        }
        if (object.has("cover")){
            cover = object.getJSONObject("cover").getString("source");
        }
        if (object.has("email")){
            email = object.getString("email");
        }
        if (object.has("gender")){
            gender = object.getString("gender");
        }
        if (object.has("birthday")){
            birthday = object.getString("birthday");
        }

    } catch (JSONException e) {
        e.printStackTrace();
    }
    new SessionManager(activity).storeUserInfo(name, profile, cover, email, gender, birthday);
}

Answer 3

我宁愿这样做：

String value = object.has(key) && !object.isNull(key) ? object.getString(key) : null;

一条线，更清洁，更容易〜：）

Answer 4

您的所有字段都是强制性的吗？
你可以这样做：

try{
   name = object.getString("name");
   ... // more methods here
   new SessionManager(activity).storeUserInfo(name, profile, cover, email, gender, birthday);
}catch(JSONException e){
  e.printStackTrace( );
}

这样，只有在检索完所有字段后才会到达SessionManager。即使缺少单个字段，您也可以直接进入catch块。

其他：

public String getString(JSONObject obj,String key,String defVal){
    if( obj.has(key) )
        return obj.getString( key );
    return defVal; 
}

现在你可以这样做：

name = getString(json,"name","UNDEFINED");

Answer 5

在一个try块中添加所有语法如果任何一个语法有异常而不是go in catch ....

public void onCompleted(JSONObject object, GraphResponse response) { String name = null; try { name = object.getString("name"); String profile = null; profile = object.getJSONObject("picture").getJSONObject("data").getString("url"); String cover = null; cover = object.getJSONObject("cover").getString("source"); String email = null; email = object.getString("email"); String gender = null; gender = object.getString("gender"); String birthday = null; birthday = object.getString("birthday"); } catch (JSONException e) { e.printStackTrace(); } new SessionManager(activity).storeUserInfo(name, profile, cover, email, gender, birthday); }

在解析json时提示许多try-catch博客

5 个答案: