我制作的应用程序需要互联网访问。我希望它会显示带有两个按钮("重试"和"退出")的警报对话框。所以,我试试这个:
void prepareConnection() {
if(!checkInternetConnection()) {
AlertDialog.Builder alert = new AlertDialog.Builder(this);
alert.setMessage(R.string.internet_not_available);
alert.setTitle(R.string.app_name);
alert.setPositiveButton(R.string.retry, new OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
prepareConnection();
}});
alert.setNegativeButton(R.string.quit, new OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
finish();
}});
alert.show();
}
}
boolean checkInternetConnection() {
ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
if ((cm.getActiveNetworkInfo() != null) && cm.getActiveNetworkInfo().isAvailable() && cm.getActiveNetworkInfo().isConnected()) {
return true;
}
return false;
}
但是 AlertDialog 与 OnClickListener 异步工作,并且 prepareConnection() 不会等待互联网连接并用户单击"重试"。我认为我在代码结构中的问题。如何使它正确?
我使用了这样的东西
boolean connection = checkNetworkConnection();
if(!connection){
createAlertDialog();
}
else{
whenConnectionActive();
}
和 createAlertDialog() 函数
public void createAlertDialog(){
final Dialog dialog = new Dialog(this);
dialog.setContentView(R.layout.custom_dialog);
dialog.setTitle("Message");
Button continueButton = (Button) dialog.findViewById(R.id.dialogContinueButton);
TextView tw = (TextView) dialog.findViewById(R.id.dialogText);
Button finishButton = (Button) dialog.findViewById(R.id.dialogFinishButton);
tw.setText("Message");
continueButton.setOnClickListener(new OnClickListener(){
public void onClick(View v) {
dialog.dismiss();
boolean connection = checkNetworkConnection();
if(!connection){
dialog.show();
}
else{
prepareConnection();
}
}
});