【发布时间】:2015-02-14 08:09:17
【问题描述】:
我有一个名为 data 的 Map
data = {Salad=[Salad Bar Station, Green Pepper & Tomato Salad, Lo Mein Noodle Salad],
Pizza=[Cheese Pizza - Slice, Pepperoni Pizza, Tomato Bruschetta Flatbread Pizza, Herb
Seasoned Breadsticks, Spaghetti Sauce with Tomato Bits],
Dessert=[Carnival Cookies, Sweet Cinnamon Brownies, Oreo Crumble Pudding Cup,
Cherry Jell-O Parfait],
Cold Cereal=[Miscellaneous/Peripherals]}
现在我想重新排序,让它看起来像这样
data = Pizza=[Cheese Pizza - Slice, Pepperoni Pizza, Tomato Bruschetta Flatbread Pizza, Herb
Seasoned Breadsticks, Spaghetti Sauce with Tomato Bits],
Salad=[Salad Bar Station, Green Pepper & Tomato Salad, Lo Mein Noodle Salad],
Cold Cereal=[Miscellaneous/Peripherals]
Dessert=[Carnival Cookies, Sweet Cinnamon Brownies, Oreo Crumble Pudding Cup,
Cherry Jell-O Parfait]}
现在我有一些关于如何执行此操作的伪代码,只是不太确定如何在代码中执行此操作。
所以我想要一个按我想要的键顺序排列的数组。所以它会像
String[] desiredOrder = {"Pizza","Salad","Cold Cereal"};
int destination = 0;
然后有类似的东西
for (int i=0; i < desiredOrder.length; i++) {
//then find if the first desiredOrder object is in the data array
//then check if it is in the right spot
if (it is in right spot) {
//do nothing
}
else {
//move it to the right spot
}
}
非常感谢您能帮我填写其余部分,以便将键及其值移动到正确的位置。
提前感谢您的帮助。
编辑
为了做到这一点,我正在从网络上解析 xml,这就是我正在做的事情,请注意我刚刚将其更改为 LinkedHashMap,所以现在它按 xml 的顺序获取它,但我仍然想重新排列它我自己的订单。
String currentDay = "";
String currentMeal = "";
String counter = "";
LinkedHashMap<String, List<String>> itemsByCounter = new LinkedHashMap<String , List<String>>();
List<String> items = new ArrayList<String>();
while (eventType != XmlResourceParser.END_DOCUMENT) {
String tagName = xmlData.getName();
blogPost = new HashMap<String, String>();
switch (eventType) {
case XmlResourceParser.START_TAG:
if (tagName.equalsIgnoreCase("day")) {
currentDay = xmlData.getAttributeValue(null, "name");
}
if (tagName.equalsIgnoreCase("meal")) {
currentMeal = xmlData.getAttributeValue(null, "name");
}
if (tagName.equalsIgnoreCase("counter") && currentDay.equalsIgnoreCase(day) && currentMeal.equalsIgnoreCase(meal)) {
counter = xmlData.getAttributeValue(null, "name");
}
break;
case XmlResourceParser.TEXT:
if (currentDay.equalsIgnoreCase(day) && currentMeal.equalsIgnoreCase(meal)) {
if (xmlData.getText().trim().length() > 0) {
items.add(xmlData.getText());
}
}
break;
case XmlPullParser.END_TAG:
if (tagName.equalsIgnoreCase("counter")) {
if (items.size() > 0) {
itemsByCounter.put(counter, items);
items = new ArrayList<String>();
}
}
break;
}
eventType = xmlData.next();
}
对不起,它很长,但基本上它只是抓取项目,添加一个作为键,然后将其下的作为值。
这是我用数据填充 ListView 的方法。
MyCustomAdapter mAdapter = new MyCustomAdapter();
for (int i = 0; i < data.size(); i++) {
Object[] array = data.keySet().toArray();
String string = array[i].toString();
mAdapter.addSeparatorItem(string);
for (int i2 = 0; i2 < data.get(string).size(); i2++) {
mAdapter.addItem(data.get(string).get(i2));
}
}
public class MyCustomAdapter extends BaseAdapter{
ArrayList mData = new ArrayList();
LayoutInflater mLayoutInflater;
private static final int TYPE_ITEM = 0;
private static final int TYPE_SEPARATOR = 1;
private static final int TYPE_MAX_COUNT = TYPE_SEPARATOR + 1;
private TreeSet mSeparatorsSet = new TreeSet();
public MyCustomAdapter() {
mLayoutInflater = (LayoutInflater)getSystemService(Context.LAYOUT_INFLATER_SERVICE);
}
public void addItem(final String item) {
mData.add(item);
notifyDataSetChanged();
}
public void addSeparatorItem(final String item) {
mData.add(item);
// save separator position
mSeparatorsSet.add(mData.size() - 1);
notifyDataSetChanged();
}
@Override
public int getItemViewType(int position) {
return mSeparatorsSet.contains(position) ? TYPE_SEPARATOR : TYPE_ITEM;
}
@Override
public int getViewTypeCount() {
return TYPE_MAX_COUNT;
}
@Override
public int getCount() {
return mData.size();
}
@Override
public String getItem(int position) {
return mData.get(position).toString();
}
@Override
public long getItemId(int position) {
return position;
}
@Override
public View getView(int position, View convertView, ViewGroup parent) {
ViewHolder holder = null;
int type = getItemViewType(position);
System.out.println("getView " + position + " " + convertView + " type = " + type);
if (convertView == null) {
holder = new ViewHolder();
switch (type) {
case TYPE_ITEM:
convertView = mLayoutInflater.inflate(R.layout.item_cell_view, parent, false);
holder.textView = (TextView)convertView.findViewById(R.id.text);
break;
case TYPE_SEPARATOR:
Context context = DiningItemsActivity.this;
convertView = mLayoutInflater.inflate(R.layout.header_cell_view, parent, false);
holder.textView = (TextView)convertView.findViewById(R.id.textSeparator);
convertView.setBackgroundColor(colorBar);
break;
}
convertView.setTag(holder);
} else {
holder = (ViewHolder)convertView.getTag();
}
holder.textView.setText(mData.get(position).toString());
return convertView;
}
public class ViewHolder {
public TextView textView;
}
【问题讨论】:
-
为什么要这样做?重点是什么? Maps 通常没有特定的顺序(例如 HashMap 是按键的哈希排序的......)
-
@MichelMichaelMeyer 我正在使用此数据并将其显示在列表视图中,并且我希望更重要的数据位于顶部,我将键用作部分标题以及其中的值和数据部分
-
@iqueqiorio 看看答案,对你有帮助吗,使用
linkedHashMap -
你不能只遍历desiredOrder,使用它的Strings作为Keys,然后将它们依次传递到列表视图中吗?
-
@MichelMichaelMeyer 你介意张贴你的意思吗,这很难分辨
标签: java android arrays sorting hashmap