我已经用alertDialog.dismiss();
问题解决了我的问题,但现在我有一个用我的两行对话框。看到对话框弹出时它只显示这个alertDialog.setMessage("1st line" + System.getProperty("line.separator") + "2nd line");
,我也需要它来显示这个alertDialog.setMessage("1st line that doesn't display" + System.getProperty("line.separator") + "2nd line that doesn't display");
我真的不明白为什么会这样,所以任何人都可以帮我这个。
final AlertDialog.Builder alertDialog = new AlertDialog.Builder(this);
alertDialog.setTitle("ApplicationTitle");
alertDialog.setIcon(R.drawable.ic_launcher);
alertDialog.setMessage("1st line that doesn't display" + System.getProperty("line.separator") + "2nd line that doesn't display");
alertDialog.setMessage("1st line" + System.getProperty("line.separator") + "2nd line");
alertDialog.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
}
});
alertDialog.show();
答案 0 :(得分:2)
您必须在对方法setMessage
的一次调用中指定整个文本:
alertDialog.setMessage("1st line that doesn't display"
+ System.getProperty("line.separator")
+ "2nd line that doesn't display"
+ System.getProperty("line.separator") + "1st line"
+ System.getProperty("line.separator") + "2nd line");
如果您使用该方法两次或更多次,则仅显示最后一次通话中设置的文本。
答案 1 :(得分:0)
StringBuilder build = new StringBuilder();
build.append("1st line")
.append("\n")
.append("2nd line")
.append("\n")
.append("3rd line");
final AlertDialog.Builder alertDialog = new AlertDialog.Builder(this);
alertDialog.setTitle("ApplicationTitle");
alertDialog.setMessage(build.toString());
alertDialog.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
dialog.cancel();
}
});
alertDialog.show();