我是Java中使用json-simple库的新手,我已经完成了编码和解码示例.复制编码示例很好,但是我无法使用混合类型JSON来解码.
我的一个问题是库中有太多的类没有正确记录,而且我没有源(为了能够阅读并理解它们的用途).因此,我很难理解如何使用这些类.
看完这个例子后:
String jsonText = "{\"first\": 123, \"second\": [4, 5, 6], \"third\": 789}";
JSONParser parser = new JSONParser();
ContainerFactory containerFactory = new ContainerFactory(){
public List creatArrayContainer() {
return new LinkedList();
}
public Map createObjectContainer() {
return new LinkedHashMap();
}
};
try {
Map json = (Map)parser.parse(jsonText, containerFactory);
Iterator iter = json.entrySet().iterator();
System.out.println("==iterate result==");
while(iter.hasNext()) {
Map.Entry entry = (Map.Entry)iter.next();
System.out.println(entry.getKey() + "=>" + entry.getValue());
}
System.out.println("==toJSONString()==");
System.out.println(JSONValue.toJSONString(json));
} catch(ParseException pe) {
System.out.println(pe);
}
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我有.png图像,我res/drawable-hdpi/delete.png想将它作为图标添加到我的菜单中item.
我用了:
delete.xml
<bitmap
xmlns:android="http://schemas.android.com/apk/res/android"
android:src="@drawable/deletebutton">
</bitmap>
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菜单activity_main.xml
<menu xmlns:android="http://schemas.android.com/apk/res/android" >
<item
android:id="@+id/create"
android:showAsAction="always"
android:title="@string/createTask"/>
<item
android:id="@+id/delete"
android:showAsAction="always"
android:icon="@drawable/delete"
android:title="@string/deleteTask" />
</menu>
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但它不起作用.
我删除了xml文件并尝试直接加载图像,这里是日志文件.
04-02 17:46:50.887: E/AndroidRuntime(11957): FATAL EXCEPTION: main
04-02 17:46:50.887: E/AndroidRuntime(11957): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.example.devicecontrolpanel/com.example.devicecontrolpanel.Main}: android.content.res.Resources$NotFoundException: Resource ID #0x7f020008
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2205)
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2240)
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.app.ActivityThread.access$600(ActivityThread.java:139)
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1262)
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.os.Handler.dispatchMessage(Handler.java:99)
04-02 17:46:50.887: E/AndroidRuntime(11957): at android.os.Looper.loop(Looper.java:156)
04-02 17:46:50.887: …Run Code Online (Sandbox Code Playgroud) 我写的代码,Android这既需要Readable和Writable Database在同一时间.我收到以下错误:
05-21 13:27:47.079: E/SQLiteDatabase(8326): close() was never explicitly called on database '/data/data/com.example.devicecontrolpanel/databases/AlarmSystem'
05-21 13:27:47.079: E/SQLiteDatabase(8326): android.database.sqlite.DatabaseObjectNotClosedException: Application did not close the cursor or database object that was opened here
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.database.sqlite.SQLiteDatabase.<init>(SQLiteDatabase.java:2052)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.database.sqlite.SQLiteDatabase.openDatabase(SQLiteDatabase.java:1087)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.database.sqlite.SQLiteDatabase.openDatabase(SQLiteDatabase.java:1050)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.database.sqlite.SQLiteDatabase.openOrCreateDatabase(SQLiteDatabase.java:1136)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.app.ContextImpl.openOrCreateDatabase(ContextImpl.java:1041)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.content.ContextWrapper.openOrCreateDatabase(ContextWrapper.java:221)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at android.database.sqlite.SQLiteOpenHelper.getWritableDatabase(SQLiteOpenHelper.java:165)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at com.example.devicecontrolpanel.DataBaseAdapter.addDeviceOnSearch(DataBaseAdapter.java:215)
05-21 13:27:47.079: E/SQLiteDatabase(8326): at …Run Code Online (Sandbox Code Playgroud) 我遇到过很多java程序,我找到了一种新的编程风格:
Class_name<...>
任何人都可以解释这是什么.?
我已经做了四span这对mouseOver动画来top:20px和mouseOut动画回top:-20px.
我写了这段代码:
$(".pull_down").mouseover(function(){
$(this).animate({top:'20px'});
});
$(".pull_down").mouseout(function(){
$(this).animate({top:'-20px'});
});
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所有span有相同的类pull_down有这个CSS style:
.pull_down
{
-webkit-msie-transform: rotate(-90deg);
-webkit-transform: rotate(-90deg);
-moz-transform: rotate(-90deg);
-o-transform: rotate(-90deg);
background:#900;
color:#FFF;
font-size:20px;
text-transform:capitalize;
font-weight:bold;
padding:5px;
position:absolute;
border:#000 solid 2px;
border-bottom-left-radius:10px;
padding-right:100px;
z-index:500;
width:100px;
}
.pull_down:hover
{
cursor:pointer;
}
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基本上这是没用的.
这里的问题是,如果鼠标超过这些元素超过1次,比如7次,那么这些动画就会排队,这看起来很尴尬
所以我想要的是当我将鼠标悬停在跨度上它开始动画时,当我切换到另一个跨度时,最后一个跨度将恢复到它的位置.
一个例子就在这里
我还阅读.stop()的相关帖子,但无法弄清楚如何做到这一点.
在最近浏览各种程序时,我看到了以下代码:
protected Object doInBackground(Object... arg0)
{
....
....
....
}
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我不明白的意义 Object...
我从未见过...任何数据类型.