Android：动态设置ImageView？

发布于 2024-11-01 17:56:21 字数 466 浏览 3 评论 0原文

我正在尝试下载图像，并动态创建 ImageView 并将其添加到我的布局中，但它不会显示图像。这是我的代码：

ImageView image = new ImageView(this);
LinearLayout.LayoutParams vp = new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.FILL_PARENT);
image.setLayoutParams(vp);
image.setScaleType(ImageView.ScaleType.CENTER_CROP);
image.setMaxHeight(50);
image.setMaxWidth(50);
image.setImageDrawable(avatar);
theLayout.addView(image);

也许我需要在添加 ImageView 后刷新布局？如何“刷新”？

原文

I'm trying to download an image, and create an ImageView dynamically and add it to my layout, but it won't display the image. Here is my code:

ImageView image = new ImageView(this);
LinearLayout.LayoutParams vp = new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.FILL_PARENT);
image.setLayoutParams(vp);
image.setScaleType(ImageView.ScaleType.CENTER_CROP);
image.setMaxHeight(50);
image.setMaxWidth(50);
image.setImageDrawable(avatar);
theLayout.addView(image);

maybe I need to refresh the layout after I add the ImageView? How do you "refresh"?

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过去的过去 2024-11-08 17:56:21

尝试以下代码，您将不需要刷新。将图像 url 放入 inputurl 变量中。

InputStream is = null;
String inputurl = " Enter url of ur image ";
try {
        URL url = new URL(inputurl);
        Object content = url.getContent();
        is = (InputStream) content;
        avatar = Drawable.createFromStream(is,"src");
} catch (MalformedURLException e) {
        e.printStackTrace();
} catch (IOException e) {
        e.printStackTrace();
}

ImageView image = new ImageView(this);
LinearLayout.LayoutParams vp = new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.FILL_PARENT);
image.setLayoutParams(vp);
image.setScaleType(ImageView.ScaleType.CENTER_CROP);
image.setMaxHeight(50);
image.setMaxWidth(50);
image.setImageDrawable(avatar);
theLayout.addView(image);

Try the following code, and you will not need to refresh. Put your image url into the inputurl variable.

InputStream is = null;
String inputurl = " Enter url of ur image ";
try {
        URL url = new URL(inputurl);
        Object content = url.getContent();
        is = (InputStream) content;
        avatar = Drawable.createFromStream(is,"src");
} catch (MalformedURLException e) {
        e.printStackTrace();
} catch (IOException e) {
        e.printStackTrace();
}

ImageView image = new ImageView(this);
LinearLayout.LayoutParams vp = new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT, LayoutParams.FILL_PARENT);
image.setLayoutParams(vp);
image.setScaleType(ImageView.ScaleType.CENTER_CROP);
image.setMaxHeight(50);
image.setMaxWidth(50);
image.setImageDrawable(avatar);
theLayout.addView(image);

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Hello爱情风 2024-11-08 17:56:21

我认为你必须定义副总裁的参数，

如果你定义参数，它将显示如下图像-

LinearLayout.LayoutParams Params=new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT,LayoutParams.WRAP_CONTENT);
        Params.setMargins(0,0,0,0);

        Params=new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT,LayoutParams.WRAP_CONTENT);

I think u have to deifne the parameters for the vp

if u will define parameters it will display the image like-

LinearLayout.LayoutParams Params=new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT,LayoutParams.WRAP_CONTENT);
        Params.setMargins(0,0,0,0);

        Params=new LinearLayout.LayoutParams(LayoutParams.FILL_PARENT,LayoutParams.WRAP_CONTENT);

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