6

编辑：由我项目的相关部分组成的 Pastebin：

这是更新的代码

ColouredItem 也是一个包装器：

     public class ColouredItem
     {//Only a wrapper class,no behaviour has been defined here
        String name,colour;
     }

尝试使用以下代码解析字符串中的颜色时，我得到一个 NumberFormatException：

     row.setBackgroundColor(Color.parseColor(item.colour));

我使用以下内容从资源创建项目列表：

    for(int i=0;i<list.length;i++)
    {
        item=new ColouredMenuItem();
        String[] cmenu =list[i].split("#");
        item.name=cmenu[0];
        item.colour="#"+cmenu[1];
        Log.d(TAG, item.colour);
        menuList.add(item);
    }

这是我得到的例外......我发现 view.setBackgroundColor 只需要一个整数值：

         #ffffff 
         #ffffBB 
         #fff45f 
         #ffff00 
         Shutting down VM
         threadid=1: thread exiting with uncaught exception (group=0x4001d800)
         FATAL EXCEPTION: main
             java.lang.NumberFormatException: ffffff 
         at java.lang.Long.parse(Long.java:364)
         at java.lang.Long.parseLong(Long.java:354)
         at android.graphics.Color.parseColor(Color.java:207)
         at com.example.samplelistproject.MadAdapter.getView(MadAdapter.java:60)
         at android.widget.AbsListView.obtainView(AbsListView.java:1315)
         at android.widget.ListView.makeAndAddView(ListView.java:1727)
         at android.widget.ListView.fillDown(ListView.java:652)
         at android.widget.ListView.fillFromTop(ListView.java:709)
         at android.widget.ListView.layoutChildren(ListView.java:1580)
         at android.widget.AbsListView.onLayout(AbsListView.java:1147)
         at android.view.View.layout(View.java:7035)
         at android.widget.FrameLayout.onLayout(FrameLayout.java:333)
         at android.view.View.layout(View.java:7035)
         at android.widget.LinearLayout.setChildFrame(LinearLayout.java:1249)
         at android.widget.LinearLayout.layoutVertical(LinearLayout.java:1125)
         at android.widget.LinearLayout.onLayout(LinearLayout.java:1042)
         at android.view.View.layout(View.java:7035)
         at android.widget.FrameLayout.onLayout(FrameLayout.java:333)
         at android.view.View.layout(View.java:7035)
         at android.view.ViewRoot.performTraversals(ViewRoot.java:1045)
         at android.view.ViewRoot.handleMessage(ViewRoot.java:1727)
         at android.os.Handler.dispatchMessage(Handler.java:99)
         at android.os.Looper.loop(Looper.java:123)
         at android.app.ActivityThread.main(ActivityThread.java:4627)
         at java.lang.reflect.Method.invokeNative(Native Method)
         at java.lang.reflect.Method.invoke(Method.java:521)
         at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:868)
         at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:626)
         at dalvik.system.NativeStart.main(Native Method)

添加 # 作为一些答案的建议并没有解决问题：

          java.lang.NumberFormatException: Invalid long: "#ffffff"
      at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2211)
      at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2261)
      at android.app.ActivityThread.access$600(ActivityThread.java:141)
      at android.app.ActivityThread$H.handleMessage(ActivityThread.java:1256)
      at android.os.Handler.dispatchMessage(Handler.java:99)
      at android.os.Looper.loop(Looper.java:137)
      at android.app.ActivityThread.main(ActivityThread.java:5103)
      at java.lang.reflect.Method.invokeNative(Native Method)

与此实现也没有区别：

          String cmenu=list[i];
          item.name=cmenu.substring(0, cmenu.indexOf("#"));
          item.colour=cmenu.substring(cmenu.indexOf("#"));
4

4 回答 4

12

使用此代码

row.setBackgroundColor(Color.parseColor("#424242"));

它也帮助了我，不要删除“#”。

我用了这段代码

private List<String> item;

item = new ArrayList<String>();
item.add("#424242");
row.setBackgroundColor(Color.parseColor(item.get(0)));

它对我来说很有效，可能是你分开的东西不好用

或为您的代码

Button btn;
ColouredMenuItem item;
ArrayList<ColouredMenuItem> menuList = new ArrayList<ColouredMenuItem>();
String[] list = new String[] { "Page1 #ffffff", "Page2 #ffffBB" };

@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);

    setContentView(R.layout.sample);

    try {
        btn = (Button) findViewById(R.id.button1);
        for (int i = 0; i < list.length; i++) {
            item = new ColouredMenuItem();
            String[] cmenu = list[i].split("#");
            item.name = cmenu[0];
            item.color = "#" + cmenu[1];
            Log.d("colored", item.color);
            menuList.add(item);
        }



        btn.setBackgroundColor(Color.parseColor(menuList.get(1).color));
    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

这对我来说很好用

这是新代码

像这样使用 getter 和 setter 将您的彩色项目作为 bean 类

public class ColouredMenuItem {// Only a wrapper class,no behaviour has been defined
                        // here
String name, colour;

List<ColouredMenuItem> list=new ArrayList<ColouredMenuItem>();

public List<ColouredMenuItem> getList() {
    return list;
}

public void setList(List<ColouredMenuItem> menuList) {
    this.list = menuList;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getColour() {
    return colour;
}

public void setColour(String colour) {
    this.colour = colour;
}

}

然后在您的适配器中使用此代码

try {
        Log.d(TAG, menuList.get(position).colour);
        textView.setText(menuList.get(position).getName());

        {
            row.setBackgroundColor(Color.parseColor(menuList.get(position).getColour()));
        }
    } catch (Exception ex) {
        Log.e(TAG, "Still does not work");
    }

试一试，它在我身边有效

你的数组也只有这样 na

<string-array name="menu_array">
    <item>Page1 #ff7788</item>
    <item>Page1 #ff6688</item>
    <item>Page1 #424242</item>
</string-array>
于 2013-10-23T13:23:46.073 回答
0

尝试Color.parseColor("#ffffff");代替Color.parseColor("ffffff");

于 2013-10-23T13:14:16.163 回答
0

看看它会告诉你的堆栈跟踪：

 java.lang.NumberFormatException: ffffff  
     at java.lang.Long.parse(Long.java:364)
     at java.lang.Long.parseLong(Long.java:354)
     at android.graphics.Color.parseColor(Color.java:207)
     at com.example.samplelistproject.MadAdapter.getView(MadAdapter.java:60)

逐行：

you are trying to format Hexadecimal (base 16) value "0xffffff" to a decimal (base 10) value
you're trying to parse hexadecimal string "ffffff" to type Long
same as above.
error is thrown when calling `Color.parseColor()`
error is thrown from your MadAdapter.java Class on line 60.

因此，您需要找到一种从十六进制而不是十进制值解析它的方法。十六进制值通常以 0x[value] 或 #[value] 开头

于 2013-10-23T13:16:21.837 回答
0

假设：从字符串对象“item”解析颜色时，不是从列表数组中获取，而是从 ColureMenuItem 的实例变量中获取。 

    ColouredMenuItem item;
        ArrayList<ColouredMenuItem> menuList = new ArrayList<ColouredMenuItem>();
        String[] list = new String[]{"#ffffff","#00ffff"};





// parsing your string here, no change in this
for(int i=0;i<list.length;i++)
            {
                item=new ColouredMenuItem();
                String[] cmenu =list[i].split("#");
                item.name=cmenu[0];
                item.color="#"+cmenu[1];
                Log.d("colored", item.color);
                menuList.add(item);
            }

// 确认值是否正在解析。

            for(int i=0;i<menuList.size();i++)
            {
                int color = Color.parseColor(menuList.get(i).color);
                Log.d("color",""+menuList.get(i).color);
            }

和你的 ColouredMenuItem 类。

public class ColouredMenuItem {

    public String color;
    public String name;

}
于 2013-10-23T13:50:23.060 回答