Java 根据真/假条件从URL绘制位图

使用下面的代码，我只想在

chatMessageObj.left

为真时显示图像，但图像总是显示出来

public View getView(int position, View convertView, ViewGroup parent) {

...

String personPhotoUrl = "Image url";
personPhotoUrl = personPhotoUrl.substring(0, personPhotoUrl.length() - 2) + PROFILE_PIC_SIZE;
    if (chatMessageObj.left) {
        new LoadProfileImage(chatImage).execute(personPhotoUrl);
    }

...

}

private class LoadProfileImage extends AsyncTask<String, Void, Bitmap> {
    ImageView bmImage;

    public LoadProfileImage(ImageView bmImage) {
        this.bmImage = bmImage;
    }

    protected Bitmap doInBackground(String... urls) {
        String urldisplay = urls[0];
        Bitmap mIcon11 = null;
        try {
            InputStream in = new java.net.URL(urldisplay).openStream();
            mIcon11 = BitmapFactory.decodeStream(in);
        } catch (Exception e) {
            Log.e("Error", e.getMessage());
            e.printStackTrace();
        }
        return mIcon11;
    }
    protected void onPostExecute(Bitmap result) {
        bmImage.setImageBitmap(result);
    }
}

---

谢谢。

这是因为listview总是保持刷新，所以您需要做的只是添加else块

像这样的,

if (chatMessageObj.left) {
    new LoadProfileImage(chatImage).execute(personPhotoUrl);
} else {
    chatImage.setVisibility(View.GONE);
}

---

希望这能帮助您：）

我假设如果chatMessageObj.left不是真的，那么您不想执行异步任务？如果是这样，那么您在这里提供的内容看起来是正确的。您需要检查值或chatMessageObj.left的分配方式

if (chatMessageObj.left) {
    new LoadProfileImage(chatImage).execute(personPhotoUrl);
} else {
    chatImage.setVisibility(View.GONE);
}