如何停止将项目添加到数组列表中？

Question

有两个标签，费用和收入。“费用”选项卡和“收入”选项卡包含不同的列表视图。当使用“费用”选项卡启动活动时，list显示数据库字符串数组项。如果列表中有3个项目，则在将选项卡切换到“收入”选项卡并返回到“费用”选项卡后，它将显示6个项目。再次将相同的值添加到列表中。怎么才能推翻这个？这是我的密码。

String[] theCategory;
String[] theMoneyType;
String[] theDescp;
String[] theDate;
String[] theAmount;

AllDataClass allData;

ListView exList;
private ArrayList<String> kId = new ArrayList<String>();           
private ArrayList<String> t_category = new ArrayList<String>();
private ArrayList<String> t_date = new ArrayList<String>();
private ArrayList<String> t_amount=new ArrayList<String>();
private ArrayList<String> t_moneytype=new ArrayList<String>();
private ArrayList<String> t_description=new ArrayList<String>();     <<< these array list values getting from SQlite database 

ArrayList<AllExpense> list=new ArrayList<AllExpense>();

theCategory=t_category.toArray(new String[t_category.size()]);

    theDate=t_date.toArray(new String[t_date.size()]);

    theAmount=t_amount.toArray(new String[t_amount.size()]);
    theMoneyType=t_moneytype.toArray(new String[t_moneytype.size()]);

    theDescp=t_description.toArray(new String[t_description.size()]);

    for(int i=0;i<theDate.length;i++) <<<<<< //this loop running again after switching the tab. I need this loop only one time.>>>>>>>>>>>
    {

        AllExpense ae=new AllExpense(theCategory[i],theDate[i],theAmount[i],theMoneyType[i],theDescp[i]);

        list.add(ae);

    }

Answer 1

在添加项目之前清除列表。

list.clear();

for(int i=0;i<theDate.length;i++) <<<<<< //this loop running again after switching the tab. I need this loop only one time.>>>>>>>>>>>
    {

        AllExpense ae=new AllExpense(theCategory[i],theDate[i],theAmount[i],theMoneyType[i],theDescp[i]);

        list.add(ae);

    }

Answer 2

用于验证天气字符串是否可用的写入方法。

您还可以使用

list.clear();

前循环

Answer 3

在添加之前检查：

private void test() {
    ArrayList<String> names=new ArrayList<String>();
    ArrayList<String> totalNames=db.getAllNames();
    for (int i = 0; i < totalNames.size(); i++) {
        if(!names.contains(totalNames.get(i))){
            names.add(totalNames.get(i));
        }
    }
}