Java onActivityResult中的requestCode和resultCode指的是什么?
大家好。我是android新手,现在在我的程序中使用startActivityForResult。在我的应用程序中,我有两个按钮和两个文本视图。用于打开对话框的两个按钮。如何检查在ActivityResult上按下的按钮,以便根据按钮设置文本视图Java onActivityResult中的requestCode和resultCode指的是什么?,java,android,android-studio,radio-button,onactivityresult,Java,Android,Android Studio,Radio Button,Onactivityresult,大家好。我是android新手,现在在我的程序中使用startActivityForResult。在我的应用程序中,我有两个按钮和两个文本视图。用于打开对话框的两个按钮。如何检查在ActivityResult上按下的按钮,以便根据按钮设置文本视图 int a1 = 1; int a2 = 2; button1.setOnClickListener(new View.OnClickListener() { @Override publ
int a1 = 1;
int a2 = 2;
button1.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AlertDialogRadio(a1);
}
});
button2.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
AlertDialogRadio(a2);
}
});
public void AlertDialogRadio(final int k) { //parameter k is never used
final CharSequence[] ClaimsModel = {"Project", "Petrol", "Car Maintenance"
, "Medical", "Other"};
AlertDialog.Builder alt_bld = new AlertDialog.Builder(getActivity());
alt_bld.setTitle("Select a Claims");
alt_bld.setSingleChoiceItems(ClaimsModel, -1, new DialogInterface
.OnClickListener() {
public void onClick(DialogInterface dialog, int item) {
if (item == 0) {
Intent intent = new Intent(getActivity().getApplicationContext(), Project1.class);
startActivityForResult(intent, 0);
} else if (item == 1) {
Intent intent = new Intent(getActivity().getApplicationContext(), Petrol.class);
startActivityForResult(intent, 1);
} else if (item == 2) {
Intent intent = new Intent(getActivity().getApplicationContext(), CarMainten.class);
startActivityForResult(intent, 2);
} else if (item == 3) {
Intent intent = new Intent(getActivity().getApplicationContext(), Medical.class);
startActivityForResult(intent, 3);
} else if (item == 4) {
Intent intent = new Intent(getActivity().getApplicationContext(), Other.class);
startActivityForResult(intent, 4);
}
dialog.dismiss();
}
});
AlertDialog alert = alt_bld.create();
alert.show();
}
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
if (resultCode == 1) { // if button1 was clicked
switch (requestCode) {
case 0:
String result = data.getStringExtra("text");
String b = data.getStringExtra("a");
c.setText(" " + b + "------" + "RM " + result);
Toast.makeText(getActivity(), "Not completed ", Toast.LENGTH_LONG).show();
break;
case 1:
String result1 = data.getStringExtra("text");
String b1 = data.getStringExtra("a");
c.setText(" " + b1 + "------" + "RM " + result1);
break;
case 2:
String result2 = data.getStringExtra("text");
String b2 = data.getStringExtra("a");
c.setText(" " + b2 + "------" + "RM " + result2);
break;
case 3:
String result3 = data.getStringExtra("text");
String b3 = data.getStringExtra("a");
c.setText(" " + b3 + "------" + "RM " + result3);
break;
case 4:
String result4 = data.getStringExtra("text");
String b4 = data.getStringExtra("a");
c.setText(" " + b4 + "------" + "RM " + result4);
break;
}
}
else if (resultCode == 2) { // if button2 was clicked
switch (requestCode) {
case 0:
String result = data.getStringExtra("text");
String b = data.getStringExtra("a");
d.setText(" " + b + "------" + "RM " + result);
break;
case 1:
String result1 = data.getStringExtra("text");
String b1 = data.getStringExtra("a");
d.setText(" " + b1 + "------" + "RM " + result1);
break;
case 2:
String result2 = data.getStringExtra("text");
String b2 = data.getStringExtra("a");
d.setText(" " + b2 + "------" + "RM " + result2);
break;
case 3:
String result3 = data.getStringExtra("text");
String b3 = data.getStringExtra("a");
d.setText(" " + b3 + "------" + "RM " + result3);
break;
case 4:
String result4 = data.getStringExtra("text");
String b4 = data.getStringExtra("a");
d.setText(" " + b4 + "------" + "RM " + result4);
break;
}
}
}
所以我的程序应该是这样工作的:
如果单击按钮1,…c.setText;
如果单击按钮2,…d.setText
但该程序现在在TextView上没有显示任何内容。错误是否来自于如果resultCode==1,或者如果resultCode==2??非常感谢
假设使用select Project1.class
项目1.2类
这里应该避免使用文字代码
if (resultCode == 1)
最好使用命名常量-
if (resultCode == RESULT_OK) {
结果确定为-1,所以这可能是您的问题。一旦我将所有请求代码设置为相同的代码。没用。 看起来你的申请代码不同,我习惯于这样看:
private static final int REQUEST_CODE_THIS = 0;
private static final int REQUEST_CODE_THAT = 1;
private static final int REQUEST_CODE_THE_OTHER = 1003;
public void onClick(DialogInterface dialog, int item) {
if (item == 0) {
Intent intent = new Intent(getActivity().getApplicationContext(), Project1.class);
startActivityForResult(intent, REQUEST_CODE_THIS);
检查这样更改onActivityResult是否有帮助
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == REQUEST_CODE_THIS) { // if button1 was clicked
switch (resultCode == RESULT_OK) {
case 0:
...
}
request\u code是调用函数标识,从哪里请求,result\u code是被调用函数标识符,它还将被调用消息的状态指定为Intent.ACTIVITY\u OK等等。有人可以帮忙吗?ThanksrequestCode代表谁,resultCode代表结果,你是否在其他活动中使用setResult?是的,我在5个活动中使用了setResult。将getActivity.getApplicationContext替换为getActivity;无法解析getActivity。如何知道按下了哪个按钮?哦,我明白了。您将返回一个int而不是标准结果代码。标准模式是发回RESULT_OK或RESULT_CANCELLED,然后通过此双参数setResult方法将结果数据发送回Intent。看起来您已经在发回Intent了。为什么不将按下的按钮放在其中?那么它是否有另一种方法来检查单击了哪个按钮?如何将参数k传递给new Intent,然后将参数k返回到onActivityResult以检查单击了哪个按钮?谢谢,但是如果你也能帮助解决我的问题,tt会更好:请先使用ACTIVITY_OK而不是1,然后在setResult之后尝试删除finish,它会解决你的问题
@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
if (requestCode == REQUEST_CODE_THIS) { // if button1 was clicked
switch (resultCode == RESULT_OK) {
case 0:
...
}