我有一个从我希望在Java对象中分层表示的模式生成的XPath列表。基本上我想从XPath中分离每个“/”并将它们视为单个对象，没有重复。目前，我已将列表加载到具有包含子对象的HashMap的对象中。
我想做类似的事情但是使用ArrayList。这是因为我想生成一个没有HashMap键的JSON字符串。该消息将用于显示树视图（使用jstree）。
输入：
Root/Customers/Customer/CompanyName
Root/Customers/Customer/ContactName
Root/Customers/Customer/ContactTitle
Root/Customers/Customer/Phone
Root/Customers/Customer/Fax
Root/Customers/Customer/FullAddress/Address
Root/Customers/Customer/FullAddress/City
Root/Customers/Customer/FullAddress/Region
Root/Customers/Customer/FullAddress/PostalCode
Root/Customers/Customer/FullAddress/Country
Root/Orders/Order/CustomerID
Root/Orders/Order/EmployeeID
Root/Orders/Order/OrderDate
Root/Orders/Order/RequiredDate
Root/Orders/Order/ShipInfo/ShipVia
Root/Orders/Order/ShipInfo/Freight
Root/Orders/Order/ShipInfo/ShipName
Root/Orders/Order/ShipInfo/ShipAddress
Root/Orders/Order/ShipInfo/ShipCity
Root/Orders/Order/ShipInfo/ShipRegion
Root/Orders/Order/ShipInfo/ShipPostalCode
Root/Orders/Order/ShipInfo/ShipCountry
这是我目前的输出：
{
"text": "Root",
"children": {
"Root": {
"text": "Root",
"children": {
"Orders": {
"text": "Orders",
"children": {
"Order": {
"text": "Order",
"children": {
"RequiredDate": {
"text": "RequiredDate"
},
"ShipInfo": {
"text": "ShipInfo",
"children": {
"ShipName": {
"text": "ShipName"
},
"ShipCity": {
"text": "ShipCity"
},
"ShipAddress": {
"text": "ShipAddress"
},
"ShipVia": {
"text": "ShipVia"
},
"ShipPostalCode": {
"text": "ShipPostalCode"
},
"ShipCountry": {
"text": "ShipCountry"
},
"Freight": {
"text": "Freight"
},
"ShipRegion": {
"text": "ShipRegion"
}
}
},
"CustomerID": {
"text": "CustomerID"
},
"EmployeeID": {
"text": "EmployeeID"
},
"OrderDate": {
"text": "OrderDate"
}
}
}
}
},
"Customers": {
"text": "Customers",
"children": {
"Customer": {
"text": "Customer",
"children": {
"CompanyName": {
"text": "CompanyName"
},
"FullAddress": {
"text": "FullAddress",
"children": {
"Address": {
"text": "Address"
},
"Region": {
"text": "Region"
},
"PostalCode": {
"text": "PostalCode"
},
"Country": {
"text": "Country"
},
"City": {
"text": "City"
}
}
},
"Phone": {
"text": "Phone"
},
"Fax": {
"text": "Fax"
},
"ContactName": {
"text": "ContactName"
},
"ContactTitle": {
"text": "ContactTitle"
}
}
}
}
}
}
}
}
}
这是我想要的输出：
"data": [{
"text": "Root",
"children": [{
"text": "Orders",
"children": [{
"text": "Order",
"children": [{
"text": "RequiredDate"
}, {
"text": "ShipInfo",
"children": [{
"text": "ShipName"
}, {
"text": "ShipCity"
}, {
"text": "ShipAddress"
}, {
"text": "ShipCity"
}, {
"text": "ShipRegion"
}, {
"text": "ShipPostcode"
}, {
"text": "ShipCountry"
}]
}
}]
}]
}]
}]
有没有人对实现这一目标的最佳方法有任何想法？感谢任何答案！
编辑：这里要求的是代码..
的TreeModel
public class TreeNode {
String id;
String text;
HashMap<String, TreeNode> children;
public TreeNode(String text)
{
this.text = text;
}
@Override
public String toString() {
return "TreeModel [id=" + id + ", text=" + text + ", children="
+ children + "]";
}
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
public HashMap<String, TreeNode> getChildren() {
return children;
}
public void setChildren(HashMap<String, TreeNode> children) {
this.children = children;
}
}
守则
File file = new File("xpaths.txt");
try {
BufferedReader br = new BufferedReader(new FileReader(file));
TreeNode root = new TreeNode("Root");
String currentLine;
try {
while((currentLine = br.readLine()) != null)
{
XpathUtils.processXPath(currentLine, root);
}
} catch (IOException e) {
e.printStackTrace();
}
System.out.println(new Gson().toJson(root));
XpathUtils
public static void processXPath(String xpath, TreeNode parent)
{
String[] elements = xpath.split("/");
for(int i=0; i < elements.length; i ++)
{
parent = processElement(elements, parent, i);
}
}
private static TreeNode processElement(
String[] xpath,
TreeNode parent,
int position)
{
if(parent.getChildren() == null)
{
parent.setChildren(new HashMap<String, TreeNode>());
}
if(!parent.getChildren().containsKey(xpath[position]))
{
TreeNode element = new TreeNode(xpath[position]);
parent.getChildren().put(xpath[position], element);
return element;
} else {
return parent.getChildren().get(xpath[position]);
}
}
编辑：短暂的休息后，我以新的视角回到了问题。看来问题很容易解决！基本上只是用ArrayList替换了HashMap，并添加了一些额外的方法来检查是否已经添加了XPath元素。可能不是最有效的方式，因为它每次循环数组，但它设法完成工作。
完成代码：
/**
* Processes an XPath by splitting each element and
* adding them into individual @TreeNode objects.
*
* @param xpath The XPath that is being processed
* @param parent The top level parent @TreeNode
*/
public static void processXPath(String xpath, TreeNode parent) {
String[] elements = xpath.split("/");
for (int i = 0; i < elements.length; i++) {
parent = processElement(elements, parent, i);
}
}
/**
* Add an element of an XPath array to a @TreeNode object
* firstly checking if the element already has a corresponding
* @TreeNode.
*
* @param xpath The Xpath that is being processed
* @param parent The parent TreeNode of the xpath element
* @param position The the element is in the xpath array
* @return
*/
private static TreeNode processElement(String[] xpath, TreeNode parent,
int position) {
if (parent.getChildren() == null) {
parent.setChildren(new ArrayList<TreeNode>());
}
if (doesNodeExist(xpath[position], parent.getChildren())) {
return getExistingNode(xpath[position], parent.getChildren());
} else {
TreeNode element = new TreeNode(xpath[position]);
parent.getChildren().add(element);
return element;
}
}
/**
* Loop through the parent nodes children and returns a @Boolean
* depicting if the node has already been added to the @ArrayList
*
* @param element The name of the element that is being processed
* @param children The list of children from the parent node
* @return
*/
private static boolean doesNodeExist(String element,
ArrayList<TreeNode> children) {
for (TreeNode node : children) {
if (node.getText().equals(element)) {
return true;
}
}
return false;
}
/**
* Loops through the parent nodes children and returns the
* @TreeNode object that was specified
*
* @param element
* @param children
* @return
*/
private static TreeNode getExistingNode(String element,
ArrayList<TreeNode> children) {
for (TreeNode node : children) {
if (node.getText().equals(element)) {
return node;
}
}
return null;
}
答案 0 :(得分：0)
我使用具有以下属性的Node对象创建一个简单的树：
String pathElement
boolean isComplete // true if this is a complete path for cases where you have a path a/b and and a path a/b/x a would have this set to false, but b and x will have it set to true
List<Node> children
答案 1 :(得分：0)
如果您使用的是Java 8，则应该查看我的开源项目：unXml。
unXml基本上从Xpath映射到Json属性。
可在Maven Central上找到。使用＆gt;版本0.8.1，以获取递归的东西。
它使用Jackson进行Json处理。如果你需要，杰克逊可以优雅地映射到Objects。
输入XML（简化，但Java代码也适用于您的xml）
<Root>
<Orders>
<Order>
<CustomerID></CustomerID>
<EmployeeID></EmployeeID>
</Order>
</Orders>
</Root>
Java代码
import com.fasterxml.jackson.databind.node.ObjectNode;
import com.nerdforge.unxml.Parsing;
import com.nerdforge.unxml.factory.ParsingFactory;
import com.nerdforge.unxml.parsers.Parser;
import org.w3c.dom.Document;
class TreeParser {
private Parsing parsing = ParsingFactory.getInstance().create();
/**
* This method will parse your XML, and return a Json ObjectNode
*/
public ObjectNode parseXml(String inputXML){
// get Xml as Document
Document document = parsing.xml().document(inputXML);
// create parser
Parser<ObjectNode> parser = parsing.obj()
.attribute("data", "Root", recursiveParser())
.build();
return parser.apply(document);
}
public Parser<ObjectNode> recursiveParser(){
return parsing.obj()
.attribute("text", parsing.simple().nodeNameParser())
.attribute("children",
parsing.arr("node()", parsing.with(this::recursiveParser))
).build();
}
}
输出JSON
{
"data":{
"children":[{
"children":[{
"children":[{
"children":[],
"text":"CustomerID"
},{
"children":[],
"text":"EmployeeID"
}],
"text":"Order"
}],
"text":"Orders"
}],
"text":"Root"
}
}