[英]Unable to remove tableitem from table in swt
當我從left_group_table(List)中選擇List5時,應從middle_group_table(Contact)中刪除所有屬於List5的項。 如果列表中包含多個項目,則應刪除聯系表中的所有項目。 請在下面找到該應用程序的屏幕截圖和代碼段。 提前致謝!
public static ArrayList<String> allEmailsFortheSelectedList = new ArrayList<String>();
HashMap<Integer, ArrayList<String>> allEmailsForALLSelectedList;
tableCursor.addMouseListener(new MouseListener() {
@Override
public void mouseUp(MouseEvent arg0) {
final int selectionIndex = left_group_table.getSelectionIndex();
if(left_group_table.getItem(selectionIndex).getChecked()) {
int tempCount = 0;
left_group_table.getItem(selectionIndex).setChecked(false);
TableItem[] items = middle_group_table.getItems();
if(allEmailsForALLSelectedList.containsKey(selectionIndex)) {
allEmailsForALLSelectedList.remove(selectionIndex);
}
Set<Entry<Integer, ArrayList<String>>> set = allEmailsForALLSelectedList.entrySet();
Iterator<Entry<Integer, ArrayList<String>>> itr = set.iterator();
while(itr.hasNext())
{
HashMap.Entry<Integer, ArrayList<String>> entry = itr.next();
for(int i=0; i< entry.getValue().size(); i++) {
new TableItem(middle_group_table, SWT.NONE);
items[tempCount].setText(1, entry.getValue().get(i));
tempCount++;
}
}
tempCount = items.length;
middle_group_table.setRedraw(true);
}else {
int middleGroupTableItemCount = 0;
left_group_table.getItem(selectionIndex).setChecked(true);
sendEmailslistName = left_group_table.getItem(selectionIndex).getText(1);
int listId = SelectionDb.getUserContactListId(sendEmailslistName);
allEmailsForALLSelectedList.put(selectionIndex, SelectionDb.getAllContactEmail(listId));
for (int i = 0; i < SelectionDb.getAllContactEmail(listId).size(); i++) {
new TableItem(middle_group_table, SWT.NONE);
}
middle_group_table.setRedraw(true);
TableItem[] items = middle_group_table.getItems();
Set<Entry<Integer, ArrayList<String>>> set = allEmailsForALLSelectedList.entrySet();
Iterator<Entry<Integer, ArrayList<String>>> itr = set.iterator();
while(itr.hasNext())
{
HashMap.Entry<Integer, ArrayList<String>> entry = itr.next();
for(int i=0; i< entry.getValue().size(); i++) {
items[middleGroupTableItemCount].setText(1, entry.getValue().get(i));
middleGroupTableItemCount++;
}
}
middleGroupTableItemCount = items.length;
}
}
您的表應該有removeAll()方法可供使用。
middle_group_table.removeAll();
編輯:要刪除一行,您需要獲取要刪除的元素的正確索引,而沒有任何LINQ的最簡單方法是:
First get the right list you need:
List<String> itemsToRemove = ...getting the list5, from your code I don't understand how the list is holding.
Then you can just iterate in reverse way and remove.
for (int i = middle_group_table.getItemCount() - 1; i <= 0; i--)
{
if (itemsToRemove.contains(items[i].getText()))
middle_group_table.remove(i);
}
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