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Following information can help us to resolve the issue faster.
In addition, we do not accept issues unrelated to Fragmentation.
我们接了fragmentation,使用isupportfragment来做的插入。
但是发现只有pop()是回退的操作,因此在basefragment里面的onbackpressed()中执行了pop()做回退。
结果目前发现在monkey过程中,快速点击造成onbackpressed执行了2次,同时执行了2次pop,导致多回退了一个fragment。
请问有没有办法单独pop一个指定的当前的fragment?
The text was updated successfully, but these errors were encountered:
TransactionDelegate 这个类的的pop()方法中
void pop(final FragmentManager fm) { enqueue(fm, new Action(Action.ACTION_POP, fm) { @Override public void run() { handleAfterSaveInStateTransactionException(fm, "pop()"); FragmentationMagician.popBackStackAllowingStateLoss(fm); removeTopFragment(fm); } }); }
发现主要问题是这里没有做判断topfragment是否是当前framgent,导致当前fragment变动,快速点击触发的多余的action还是会执行。
另外大大真的不维护了吗??
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@GWYloved 这个应该是防抖动的操作,你的项目有使用SupportActivity(继承或者实现的方式)吗?
另外这个库,我在继续维护,已经发布了新版本#1237
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Following information can help us to resolve the issue faster.
In addition, we do not accept issues unrelated to Fragmentation.
我们接了fragmentation,使用isupportfragment来做的插入。
但是发现只有pop()是回退的操作,因此在basefragment里面的onbackpressed()中执行了pop()做回退。
结果目前发现在monkey过程中,快速点击造成onbackpressed执行了2次,同时执行了2次pop,导致多回退了一个fragment。
请问有没有办法单独pop一个指定的当前的fragment?
The text was updated successfully, but these errors were encountered: