Question

我需要在很多活动中从扩展文件中访问很多图像。

执行：

expansionFile = APKExpansionSupport.getAPKExpansionZipFile(context, 8, -1);
fileStream = expansionFile.getInputStream("drawables/drawable-hdpi/" + image);
Drawable drawable = Drawable.createFromStream(fileStream, null);

每个活动都很慢，在某些活动中我需要同时加载4个以上的图像。

那么，如果我创建一个在每个活动中抽象出这个代码的新类，你会怎么想？

Answer 1

所以我创建了一个ExpansionFileRetriever类，它检索扩展文件：

公共类ExpansionFileRetriever {

private static ExpansionFileRetriever instance;

public static ExpansionFileRetriever get() {
   if(instance == null) instance = getSync();
   return instance;
}

private static synchronized ExpansionFileRetriever getSync() {
   if(instance == null) instance = new ExpansionFileRetriever();
   return instance;
}

public ExpansionFileRetriever(){
   // here you can directly access the Application context calling
   Context c = App.get();
   try {
        expansionFile = APKExpansionSupport.getAPKExpansionZipFile(c, 20, -1);
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
}

ZipResourceFile expansionFile;

public Drawable getDrawable(String path, String resource) {
    InputStream fileStream = null;
    String file = path + resource;
    try {
        fileStream = expansionFile.getInputStream(file);
        Drawable drawable = Drawable.createFromStream(fileStream, null);
        return drawable;
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
        return null;
    }   
}

}

每当我需要访问扩展文件的文件时

我只是用

ExpansionFileRetriever expFile = ExpansionFileRetriever.get();

并且得到，说Drawable我所做的就是：

expFile.getDrawable(path, name)

全局访问扩展文件

1 个答案: