我正在为如何在标签上实施长按而苦苦挣扎。我尝试了几种不同的方法,但似乎都没有效果。
我确信这可能是基本的东西,或者我只是不了解Tablayout和分页器的点击次数。
我在这里发现了一些问题,但长按并没有提及。
这是我尝试过的一些方法。
main_tab_pager.iterator().forEach {
view ->
view.setOnLongClickListener{
Toast.makeText(applicationContext, "....", Toast.LENGTH_LONG).show()
true
}
main_tabs.setOnLongClickListener {
when(main_tabs.selectedTabPosition) {
0-> Toast.makeText(applicationContext, "tab 0", Toast.LENGTH_LONG).show()
1-> Toast.makeText(applicationContext, "tab 1", Toast.LENGTH_LONG).show()
2-> Toast.makeText(applicationContext, "tab 2", Toast.LENGTH_LONG).show()
}
true
}
main_tab_pager.setOnLongClickListener {
when(main_tab_pager.currentItem) {
0-> Toast.makeText(applicationContext, "tab 0", Toast.LENGTH_LONG).show()
1-> Toast.makeText(applicationContext, "tab 1", Toast.LENGTH_LONG).show()
2-> Toast.makeText(applicationContext, "tab 2", Toast.LENGTH_LONG).show()
}
true
}
标签页/寻呼机设置代码。
main_tab_pager.adapter = TabAdapter(supportFragmentManager, sections)
main_tabs.setupWithViewPager(main_tab_pager)
标签适配器
class TabAdapter(fragmentManager: FragmentManager, private val sections:
Array<BaseFragment>) : FragmentPagerAdapter(fragmentManager)
{
override fun getItem(position: Int): Fragment {
return sections[position]
}
override fun getCount(): Int {
return sections.size
}
override fun getPageTitle(position: Int): CharSequence? {
return sections[position].title
}
}
答案 0 :(得分:0)
解决方案,感谢Mike M。
val tabs = main_tabs.getChildAt(0) as LinearLayout
for (i in 0 until tabs.childCount) {
when(i){
0 -> tabs.getChildAt(0).setOnLongClickListener {
Toast.makeText(baseContext, "tab 0 ", Toast.LENGTH_LONG).show()
true
}
1 -> tabs.getChildAt(1).setOnLongClickListener {
Toast.makeText(baseContext, "tab 1", Toast.LENGTH_LONG).show()
true
}
2 -> tabs.getChildAt(2).setOnLongClickListener {
Toast.makeText(baseContext, "tab 2", Toast.LENGTH_LONG).show()
true
}
}
}