Question

我会开始从我的分裂开始新的活动。

startActivityForResult(intent, 1);

并且希望处理导致父母分裂活动的结果:

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    Log.d(TAG, "onActivityResult, requestCode: " + requestCode + ", resultCode: " + resultCode);
    if (requestCode == 1) {
        // bla bla bla
    }
}

问题在于,我从来不使用<代码>requestCode。页: 1

象0x40001,0x20001。等等,有一套随机较高的借方。 docs没有说什么。任何想法?

Answer 1

页: 1 当你这样做时,<条码>requestCode通过<条码>加以改动,该代码有<条码>。

如果您希望获得正确的<代码>resultCode。在你的活动中:

变化:

startActivityForResult(intent, 1);

:

getActivity().startActivityForResult(intent, 1);

Answer 2

要求书并非错误。在使用v4支持图书馆的碎块时,碎块指数在请求书的16个顶点上编码,你的请求书码在16个底层。碎片指数后来用于寻找正确的碎片以达到结果。

因此,活动开始成为碎片物体,处理主动反应要求守则如下:

originalRequestCode = changedRequestCode - (indexOfFragment << 16)
      6             =      196614        -       (3 << 16)

Answer 3

<<>Easier>:

Java: int unmaskedRequestCode = requestCode & 0x0000ffff

Kotlin: val unmaskedRequestCode = requestCode and 0x0000ffff

对16个下限进行核对,仅使其做符合逻辑的工作,16个上限为零

protected void onActivityResult(int requestCode, int resultCode, Intent data) {

    final int unmaskedRequestCode = requestCode & 0x0000ffff

    if(unmaskedRequestCode == ORIGINAL_REQUEST_CODE){
      //Do stuff

    }
}

Answer 4

如果你不断公布,然后在<条码>上使用

例如:

public static final int REQUEST_CODE =1;
getActivity().startActivityForresult(intent, REQUEST_CODE);

Answer 5

You can also define
super.onActivityResult(requestCode, resultCode, data)
in Activity (if you overrideonActivityResult) at this

@Override
protected void onActivityResult(int requestCode, int resultCode, Intent data) {
    switch (requestCode) {

        ...

        default:
            super.onActivityResult(requestCode, resultCode, data);
    }
}

电话:startActative forResult(intent, requestCode) within Fragment

Answer 6

缩略语

  getActivity().startActivityForResult(builder.build(getActivity()), PLACE_PICKER_REQUEST);

主要活动:

if (requestCode == PLACE_PICKER_REQUEST) {
            if (resultCode == RESULT_OK) {    
     //what ever you want to do
            }

友情链接