2,835
社区成员
发帖
与我相关
我的任务
分享android获取channel出错
从AndroidManifest.xml中获取channel出现错误提示：Key xx expected String but value was a java.lang.Integer. The default value <null> was reurned。
...全文
请发表友善的回复…
发表回复
ANDYMFC_No1 2016-09-23
- 打赏
- 举报
正确方法如下：
1 public String getChannel() {
2 String channel = null;
3 try {
4 ApplicationInfo ai = getPackageManager().getApplicationInfo(getPackageName(), PackageManager.GET_META_DATA);
5 channel = ai.metaData.getString("UMENG_CHANNEL");
6 if (channel == null) {
7 channel = String.valueOf(ai.metaData.getInt("UMENG_CHANNEL"));
8 }
9 } catch (NameNotFoundException e) {
10 e.printStackTrace();
11 }
12
13 return channel;
14 }
flutter 使用webview_flutter 运行在android设备上报错：Unhandled Exception: PlatformException(channel-error, Unab