Question

我将URL从一个活动发送到另一个活动，如下所示：

startActivity(new Intent(MainActivity.this, SecondActivity.class).putExtra("key", fullurl));

得到这样的话：

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.test);
    url = getIntent().getStringExtra("key");
}

如何把它放在这里？

jsonobject = JSONfunctions.getJSONfromURL(url);

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.test);

    url = getIntent().getStringExtra("key");

    TextView txt = (TextView) findViewById(R.id.textView2);
    txt.setText(url);

    new DownloadJSON().execute();
}

private class DownloadJSON extends AsyncTask<Void, Void, Void> {
    @Override
    protected Void doInBackground(Void... params) {
        world = new ArrayList<>();

        jsonobject = JSONfunctions.getJSONfromURL(url);
        try {
            // Some code
        }
    }
}

Answer 1

你可以做这样的事情

//发出网址请求并获得回复 String jsonStr = sh.makeServiceCall（url，ServiceHandler.GET）;

        Log.d("Response: ", "> " + jsonStr);

        if (jsonStr != null) {
            try {
                JSONObject jsonObj = new JSONObject(jsonStr);

                // Getting JSON Array node
                contacts = jsonObj.getJSONArray(TAG_CONTACTS);

                // looping through All Contacts
                for (int i = 0; i < contacts.length(); i++) {
                    JSONObject c = contacts.getJSONObject(i);

                    String id = c.getString(TAG_ID);
                    String name = c.getString(TAG_NAME);
                    String email = c.getString(TAG_EMAIL);
                    String address = c.getString(TAG_ADDRESS);
                    String gender = c.getString(TAG_GENDER);

                    // Phone node is JSON Object
                    JSONObject phone = c.getJSONObject(TAG_PHONE);
                    String mobile = phone.getString(TAG_PHONE_MOBILE);
                    String home = phone.getString(TAG_PHONE_HOME);
                    String office = phone.getString(TAG_PHONE_OFFICE);

                    // tmp hashmap for single contact
                    HashMap<String, String> contact = new HashMap<String, String>();

                    // adding each child node to HashMap key => value
                    contact.put(TAG_ID, id);
                    contact.put(TAG_NAME, name);
                    contact.put(TAG_EMAIL, email);
                    contact.put(TAG_PHONE_MOBILE, mobile);

                    // adding contact to contact list
                    contactList.add(contact);
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        } else {
            Log.e("ServiceHandler", "Couldn't get any data from the url");
        }

Answer 2

假设您的网址是一个字符串，并且您想将其转换为网址对象。

    URL mUrl = new URL(url);
    URI uri = new URI(mUrl.getProtocol(), mUrl.getUserInfo(), mUrl.getHost(), mUrl.getPort(), mUrl.getPath(), mUrl.getQuery(), mUrl.getRef());
    mUrl = uri.toURL();

在此之后你可以使用JSONfunctions .getJSONfromURL（mUrl）;

Answer 3

第1步

更改

private class DownloadJSON extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... params) {

到

private class DownloadJSON extends AsyncTask<URL, Void, Void> {
@Override
protected Void doInBackground((URL... url) {

第2步

更改

new DownloadJSON().execute();

到

try{
    URL myUrl = new URL(url);
    new DownloadJSON().execute(myUrl);
}
catch(MalformedURLException mue){
    Log.e("Invalid url","Invalid url");
}

第3步

更改

jsonobject = JSONfunctions.getJSONfromURL(url);

到

jsonobject = JSONfunctions.getJSONfromURL(url[0]);

从其他活动获取URL时如何设置URL？

3 个答案: