小编LA_*_*LA_的帖子

我有一个表单位于几个Twitter的Bootstrap选项卡上:

<form id="personal-data" class="form-horizontal">
  <div class="tabbable">
    <ul id="tab" class="nav nav-tabs">
      <li class="active"><a href="#home" data-toggle="tab">Home</a></li>
      <li class=""><a href="#profile" data-toggle="tab">Profile</a></li>
      <!-- ... -->
    </ul>

    <div id="myTabContent" class="tab-content">
      <div class="tab-pane fade active in" id="home">
        <fieldset>
          <div class="control-group">
            <div class="controls">
              <input type="text" id="field1">
            </div>
          </div>
        </fieldset>
      </div>
      <div class="tab-pane fade" id="profile">
        <fieldset>
          <div class="control-group">
            <div class="controls">
              <input type="text" id="field2">
            </div>
          </div>
        </fieldset>          
      </div>

当我valid在活动选项卡上使用jQuery验证表单时,在同一选项卡上使用无效值的字段,则验证失败(这是正确的).但是当我在一个选项卡上并且另一个选项卡上的值无效时,验证将返回true,这是不正确的.我该如何解决？如何在另一个标签上突出显示该字段？

请参阅本演示(只需按下按钮,旁边 -它会显示错误信息,然后转到最后一个标签,然后按完成那里).

我有以下代码:

private Toast movieRecordToast;

    private void displayNextMovie() {
        if (movieRecordToast != null) movieRecordToast.cancel(); // cancel previous Toast (if user changes movies too often)
        movieRecordToast = Toast.makeText(getApplicationContext(), "Next", Toast.LENGTH_SHORT);
        movieRecordToast.show();

    private void displayPrevMovie() {
        if (movieRecordToast != null) movieRecordToast.cancel();
        movieRecordToast = Toast.makeText(getApplicationContext(), "Prev", Toast.LENGTH_SHORT);
        movieRecordToast.show();        

但如果displayNextMovie被快速调用几次然后displayPrevMovie被调用,"Next"Toast仍会显示,并且只有在显示"Prev"之后.看起来取消无法正常工作.

一旦第一次启动应用程序,我需要向SQLite数据库添加初始值.我该怎么办？

我怎么能抓到这样的事件？onCreateContextMenu很相似,但我不需要菜单.

我已经下载了python-Levenshtein档案并提取了Levenshtein目录.所以,结果我有以下文件结构:

Levenshtein
  - __init__.py
  - _levenshtein.c
  - _levenshtein.h
  - StringMatcher.py
myscript.py

以下myscript.py内容:

from Levenshtein import *
from warnings import warn

print Levenshtein.distance(string1, string2)

但是我收到以下错误 -

Traceback (most recent call last):
  File "myscript.py", line 1, in <module>
    from Levenshtein import *
  File "/path/to/myscript/Levenshtein/__init__.py", line 1, in <module>
    from Levenshtein import _levenshtein
ImportError: cannot import name _levenshtein

这有什么不对？

根据文件:

请注意,从Android 4.4开始 - 您的应用程序尝试中止SMS_RECEIVED_ACTION广播的任何尝试都将被忽略,因此所有感兴趣的应用都有机会接收它.

所以,只有默认的短信应用程序可以做到这一点,看起来没有任何改变与Android 5.0(由于未知的原因,根据问题被标记为过时).

有没有办法阻止传入短信的非默认短信应用程序如果手机是root(这是垃圾邮件拦截应用程序需要)？怎么做？

我有以下设置 -

文件夹结构:

myapp
  - conf
    - locale
      - ru
        - LC_MESSAGES
          - django.mo # contains "This is the title." translation
          - django.po
  - templates
    - index.html
  setting.py
  main.py

app.yaml:

...
env_variables:
 DJANGO_SETTINGS_MODULE: 'settings'

handlers:
...
- url: /locale/ # do I need this?
  static_dir: templates/locale

libraries:
- name: django
  version: "1.5"

settings.py:

USE_I18N = True

LANGUAGES = (
    ('en', 'EN'),
    ('ru', 'RU'),
)

LANGUAGE_CODE = 'ru'
LANGUAGE_COOKIE_NAME = 'django_language'

SECRET_KEY = 'some-dummy-value'

MIDDLEWARE_CLASSES = (
  'django.middleware.locale.LocaleMiddleware' …

这是代码:

public class BillingService extends Service implements ServiceConnection {
...
    @Override
    public void onStart(Intent intent, int startId) {
        handleCommand(intent, startId); // line 361
    }

    /**
     * The {@link BillingReceiver} sends messages to this service using intents.
     * Each intent has an action and some extra arguments specific to that action.
     * @param intent the intent containing one of the supported actions
     * @param startId an identifier for the invocation instance of this service
     */
    public void handleCommand(Intent intent, int startId) …

我TextSwitcher的每个记录ListView应显示第一个值(text1),然后显示另一个值(text2),然后再显示第一个值,依此类推.只有当text2不为空时才会发生.否则,应始终显示text1(不进行任何更改和动画).

我创建了Runnable(),它改变了布尔变量(time2)然后调用items.notifyDataSetChanged().它按预期工作,结果setViewValue()我ListView被称为.

这是代码:

items.setViewBinder(new SimpleCursorAdapter.ViewBinder() {

    @Override
    public boolean setViewValue(View view, Cursor cursor, int columnIndex) {
      int viewId = view.getId();
      switch(viewId) {
      case R.id.timetext:
          TextSwitcher itemTime = (TextSwitcher) view;
          if (itemTime.getChildCount() != 2) {
              itemTime.removeAllViews();
              itemTime.setFactory(new ViewSwitcher.ViewFactory() {
                  @Override
                  public View makeView() {
                      TextView t = new TextView(MyActivity.this);
                      t.setTextSize(18);
                      t.setTypeface(null, Typeface.BOLD);
                      t.setTextColor(Color.WHITE);
                      return t;
                    }
                  });
              itemTime.setAnimateFirstView(true);
              itemTime.setInAnimation(AnimationUtils.loadAnimation(MyActivity.this, …

我在TextView中有链接:

<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    ...

    <TextView
        android:layout_width="wrap_content"
        android:layout_height="wrap_content"
        android:autoLink="all"
        android:linksClickable="true"
        android:text="@string/app_description" />
</RelativeLayout>

该链接是资源的一部分:

<?xml version="1.0" encoding="utf-8"?>
<resources>
    ...
    <string name="app_description">Some text with the <a href="http://www.example.com">link</a>.</string>
</resources>

不知何故,链接不起作用.可能是什么原因？它显示为链接.但是点击不会打开浏览器.

班级代码:

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