如何从 Android 上的 ListView 中删除 ListView 项目?
但是,我想在删除每个项目之前将其点击次数设置为 3。
因此,如果第一个位置的项目被单击一次,第二个项目被单击两次,则在第一个项目点击达到 3 之前不要删除任何项目。然后仅删除该项目,对于 ListView 中的其他项目,每个项目都必须单击 3次。
listi.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view,final int position, long id) {
final PopupMenu pop = new PopupMenu(Danger.this, listi);
pop.getMenuInflater().inflate(R.menu.menu_location, pop.getMenu());
pop.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
switch (item.getItemId()) {
case R.id.Remove:
items.remove(position);
}//swithc
return false;
最佳答案
创建一个整数ArrayList,并使用与 ListView 相同的元素计数对其进行初始化,并将列表中所有元素的值设置为= 0
ArrayList<integers> counterList = new Arraylist();
for(int i = 0; i < listi.getAdapter.getChildrenCount(); i++){ // get total elements in adapter
counterList.add(0); // set each element of array list to 0
}
然后这里:
listi.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view,final int position, long id) {
final PopupMenu pop = new PopupMenu(Danger.this, listi);
pop.getMenuInflater().inflate(R.menu.menu_location, pop.getMenu());
pop.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
switch (item.getItemId()) {
case R.id.Remove:
if(counterList.get(position) >= 2){
items.remove(position); // remove current position item from arraylist adapter and notify data set changed
counterList.remove(position); // remove the current position element from counter list too
} else {
counterList.set(position, counterList.get(position) + 1); // if 3 clicks have not happened then increase the counter.
}
}//swithc
return false;
关于java - 单击 3 次后删除 ListView 项目,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45079382/