小编Naz*_*141的帖子

我使用ionicframework来构建具有角度的混合应用程序.我有一个带有两个输入框的表单:日期和时间.日期输入的值保存在数据库中: Tue Mar 24 2015 00:00:00 GMT 0300 (AST)我在Android设备上运行表单时得到这种类型的数据,但我想保存日期简单如:24-04-2015

我尝试了什么

form.html

<form>
<input ng-model="app.date" class="positive" type="date">
<input ng-model="app.time" class="positive" type="time">
<button ng-click="makeApp(app)">send</button>
</form>

controller.js

$scope.makeApp = function (app) {
$http.post("http://www.foo.com/senddata.php?date="+app.date+"&time="+app.time)
.success(function(data){   
});

我有一个ArrayList<String> tokens;在ListViewClick上存储令牌值.在我从ArrayList<String> tokens其中检索数据时添加值后,仅显示最后添加的值

ArrayList<String> tokens;
...
....
public void onItemClick(AdapterView<?> listView, View view,
                                int position, long id) {
            tokens = new ArrayList<String>();

            Cursor cursor = (Cursor) listView.getItemAtPosition(position);
            String selectedtoken = cursor.getString(cursor.getColumnIndexOrThrow("ContactToken"));

            tokens.add(selectedtoken);
        }
...
...
Log.i("array: ", tokens.toString()); // shows the last added value

我试图从中取出值ArrayList并放入JSONObject.我编写了下面的代码,但它只将arraylist中的最后一个值放到jsonobject中

我正在努力实现这一目标.

{"lstContacts":"array_value"},{"lstContacts":"array_value"},{"lstContacts":"array_value"}

这是我的代码

ArrayList<String> tokens;
JSONObject contactsObj;
..
...
test.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            try {
                for (int i = 0; i < tokens.size(); i++) {
                    contactsObj.put("ContactToken", tokens.get(i));
                }

            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            String jsonStr = contactsObj.toString();
            Log.e("CONTACTS", jsonStr); // adds only last array to json object
        }
    });

我尝试在Android SQLite中搜索记录是否该记录存在于任何其他列"ITEM_NAME"或列"ITEM_CODE"中.

它永远不会奏效.

mCursor = mDb.query(true, ITEM_TABLE_NAME, new String[] {ITEM_ID, ITEM_CODE,
                ITEM_NAME, ITEM_CPRICE, ITEM_UPRICE}, 
                ITEM_NAME + " like '%" + inputText + "%'"
                + "OR" + ITEM_NAME + " like '%" + inputText + "%'" , null,
                null, null, null, null);

但在一栏中搜索有效

mCursor = mDb.query(true, ITEM_TABLE_NAME, new String[] {ITEM_ID, ITEM_CODE,
                ITEM_NAME, ITEM_CPRICE, ITEM_UPRICE}, 
                ITEM_NAME + " like '%" + inputText + "%'"
                null, null, null, null);

我试图将图像从url加载到我的android ImageView.但它没有为我的网址提供图片.但当我调用另一个示例网址时,它会加载到ImageView上

我的网址为空

https://192.168.100.15/HeyVoteWeb/Home/GetImage/d9cbd32c-47fc-4644-ab97-1f525c96e9ed/100000102

此示例网址适用于我

https://upload.wikimedia.org/wikipedia/commons/4/47/PNG_transparency_demonstration_1.png

这是我正在处理的代码

public class GetImage extends Activity{
ImageView postpic1;
Bitmap b;
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.all_posts);
    postpic1 = (ImageView) findViewById(R.id.postpic1);
    information info = new information();
    info.execute("");
}
public class information extends AsyncTask<String, String, String>
{
    @Override
    protected String doInBackground(String... arg0) {
        try
        {
            URL url = new URL("https://localhost/HeyVoteWeb/Home/GetImage/d9cbd32c-47fc-4644-ab97-1f525c96e9ed/100000102");
            InputStream is = new BufferedInputStream(url.openStream());
            b = BitmapFactory.decodeStream(is);
        } catch(Exception e){}
        return null;
    }
    @Override
    protected void onPostExecute(String result) {
        postpic1.setImageBitmap(b);

    }
}
}