[英]Best way to represent list of XPaths in java
我有一個列表,這些列表是根據我想在Java對象中按層次表示的模式生成的。 基本上,我想從XPath拆分每個“ /”,並將它們視為單獨的對象,沒有重復項。 目前,我已將列表加載到具有包含子對象的HashMap的對象中。
我想做類似的事情,但改用ArrayList。 這是因為我想生成一個沒有HashMap鍵的JSON字符串。 該消息將用於顯示樹視圖(使用jstree)。
輸入:
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
這是我當前的輸出:
{
"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"
}
}
}
}
}
}
}
}
}
這是我想要的輸出:
"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"
}]
}
}]
}]
}]
}]
是否有人對實現此目標的最佳方法有任何想法? 感謝任何答案!
編輯:按要求這里是代碼..
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;
}
}
編碼
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]);
}
}
編輯:短暫休息后,我以新的視角回到了問題。 看來問題很容易解決! 基本上只是將HashMap替換為ArrayList,並添加了一些額外的方法來檢查XPath元素是否已添加。 可能不是最有效的方法,因為它每次都會循環數組,但可以設法完成工作。
完成的代碼:
/**
* 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;
}
我將使用具有以下屬性的Node對象創建一個簡單的樹:
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
如果您使用的是Java 8,則應簽出我的開源項目: unXml 。 unXml
基本上從Xpaths映射到Json屬性。
它在Maven Central上可用。 使用>版本0.8.1
來獲取遞歸內容。
它使用Jackson進行Json處理。 如果需要,Jackson可以優雅地映射到Objects
。
輸入XML(已簡化,但是Java代碼也可以與您的xml一起使用)
<Root>
<Orders>
<Order>
<CustomerID></CustomerID>
<EmployeeID></EmployeeID>
</Order>
</Orders>
</Root>
Java代碼
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();
}
}
輸出JSON
{
"data":{
"children":[{
"children":[{
"children":[{
"children":[],
"text":"CustomerID"
},{
"children":[],
"text":"EmployeeID"
}],
"text":"Order"
}],
"text":"Orders"
}],
"text":"Root"
}
}
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