本文介绍了getSupportFragmentManager().getFragments()已弃用?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在使用viewpager和编译API23.我在以下语句的代码中显示了编译错误,但项目确实可以编译.

I am working with a viewpager and compiling API 23. I'm showing a compile error in my code for the following statement, but the project does compile.

List<Fragment> fragments = getSupportFragmentManager().getFragments();

此外,我在Android文档中找不到支持片段管理器和非支持片段管理器的此方法.有人知道这是怎么回事吗?

Furthermore, I cannot find this method in the android documentation for either the support fragment manager and the non-support one. Does anyone know what's going on here?

推荐答案

我实际上是这样做的,以获取对所有片段的引用:

I actually did this to get a reference to all the fragments:

private List<WeakReference<Fragment>> mFragList = new ArrayList<WeakReference<Fragment>>();

@Override
public void onAttachFragment (Fragment fragment) {
    mFragList.add(new WeakReference(fragment));
}

public List<Fragment> getActiveFragments() {
    ArrayList<Fragment> ret = new ArrayList<Fragment>();
    for(WeakReference<Fragment> ref : mFragList) {
        Fragment f = ref.get();
        if(f != null) {
            if(f.isVisible()) {
                ret.add(f);
            }
        }
    }
    return ret;
}

public Fragment findFragement(String tClass) {

    List<Fragment> fragments = getActiveFragments();
    for (Fragment fragment : fragments) {
        if (tClass.equalsIgnoreCase("Home")) {
            if (fragment instanceof ToggleFragment) {
                return fragment;

            }
        } else if (tClass.equalsIgnoreCase("Contacts")) {
            if (fragment instanceof ContactFragment) {
                return fragment;
            }
        }
    }
    return null;
}

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09-20 23:21