[英]Java Android - Dynamic load fragment (ViewPager)
In settngs existed checkbox, if it is checked, then a specific fragment must not be loaded. 在settngs existed复选框中,如果选中该复选框,则不得加载特定片段。 I have just 4 fragment and I use FragmentStatePagerAdapter to show them.
我只有4个片段,我使用FragmentStatePagerAdapter来显示它们。
public class TabPagerAdapter extends FragmentStatePagerAdapter {
public TabPagerAdapter(FragmentManager fm) {
super(fm);
}
@Override
public Fragment getItem(int i) {
switch (i) {
case 0:
return new Fragment_One();
case 1:
return new Fragment_Two();
case 2:
return new Fragment_Three();
case 3:
return new Fragment_Four();
}
return null;
}
@Override
public int getCount() {
return 4;
}
}
As show fragments, which are not only check in the settings? 作为显示片段,不仅要检查设置? I get value (true of false (Check or Uncheck) fragment, but how do not show this fragment, i dont know.
我得到了价值(假(检查或取消检查)为假的片段的真值,但我不知道如何不显示该片段。
You have to adapt your getItem()
method as well as the getCount()
method. 您必须调整
getItem()
方法以及getCount()
方法。
Let's assume you have a method shouldShowFragment(int fragmentNumber)
that tells me for a given fragment number from 0 to 3, whether it should be shown or not (depending on the settings). 假设您有一个方法
shouldShowFragment(int fragmentNumber)
,它告诉我给定的片段号(从0到3),是否应该显示(取决于设置)。
Now, implement getCount() like so to return the number of fragments that should be shown: 现在,像这样实现getCount()以返回应该显示的片段数:
public int getCount() {
int cnt = 0;
for (int i = 0; i < 4; i++) {
if (shouldShowFragment(i)) cnt++;
}
return cnt;
}
And implement getItem() like so to take the not showing fragments into account: 并像这样实现getItem(),以将未显示的片段考虑在内:
public Fragment getItem(int position) {
int cnt = -1;
for (int i = 0; i < 4; i++) {
if (shouldShowFragment(i)) cnt++;
if (cnt == position) {
switch(i) {
case 0 : return new Fragment_One();
case 1 : return new Fragment_Two();
case 2 : return new Fragment_Three();
case 3 : return new Fragment_Four();
}
}
}
return null;
}
First of all Save the all check buttons state globally(ie in shared preferences like btn1.setChecked == true/false whatever) and in above code do like below:- 首先,全局保存所有检查按钮的状态(例如,在共享首选项中,如btn1.setChecked == true / false),并且在上面的代码中,如下所示:
@Override
public Fragment getItem(int i) {
switch (i) {
case 0:
if(btn0.isChecked == true)
return new Fragment_One();
case 1:
if(btn1.isChecked == true)
return new Fragment_Two();
case 2:
if(btn2.isChecked == true)
return new Fragment_Three();
case 3:
if(btn3.isChecked == true)
return new Fragment_Four();
}
return null;
}
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