怎么把xml转换成json格式的文件

1. 在java中怎么把xml文件转换成json格式

在java中把xml文件转换成json格式

1、前提需要jar包：

json-lib-2.4-jdk15.jar 和 xom-1.2.5.jar ,maven 仓库：

.sf.json-lib

json-lib

2.4

jdk15

xom

xom

1.2.5

2、代码部分：

public static JSON xmlToJson(String xml) {

XMLSerializer xmlSerializer = new XMLSerializer()

if(xml!=null &&!”“.equals(xml)){

xml = xml.replaceAll(“\r|\n”, “”)

JSON json = xmlSerializer.read(xml)

return json

}else{

return null

}

}

public static void main(String[] args) {

String xmlStr = "01marry"

JSON json = xmlToJson(xmlStr)

System.out.println(json.toString())

}

3、结果：

{“id”:”01”,”name”:”marry”}

2. 把xml格式的字符串转换成json

To convert an XML node contained in string xml into a JSON string

XmlDocument doc = new XmlDocument()

doc.LoadXml(xml)

string jsonText = JsonConvert.SerializeXmlNode(doc)

To convert JSON text contained in string json into an XML node

XmlDocument doc = (XmlDocument)JsonConvert.DeserializeXmlNode(json)

XmlNote myXmlNode = JsonConvert.DeserializeXmlNode(myJsonString)

or .DeserilizeXmlNode(myJsonString, "root")

if myJsonString does not have a root

string jsonString = JsonConvert.SerializeXmlNode(myXmlNode)

XmlDictionaryReader reader = JsonReaderWriterFactory.CreateJsonReader(Encoding.UTF8.GetBytes(xml), XmlDictionaryReaderQuotas.Max)

XmlDocument xdoc = new XmlDocument()

xdoc.Load(reader)

3. xml文件怎么解析成json

org.jsonjson 20171018 import org.json.JSONObjectimport org.json.XMLpublic class Main {public static int PRETTY_PRINT_INDENT_FACTOR = 4public static String TEST_XML_STRING ="<?xml version=\"1.0\" ?>Turn this to JSON "public static void main(String[] args) {try {JSONObject xmlJSONObj = XML.toJSONObject(TEST_XML_STRING)String jsonPrettyPrintString = xmlJSONObj.toString(PRETTY_PRINT_INDENT_FACTOR)System.out.println(jsonPrettyPrintString)} catch (JSONException je) {System.out.println(je.toString())}}}。

引入using Newtonsoft.Json

数据查询后填充到DataTable ,再转　JsonConvert.SerializeObject

简单例宴卖毕子：晌芹

DataTable dt = new DataTable()

DataColumn dcName = new DataColumn("Name")

DataColumn dcAge = new DataColumn("Age")

DataColumn dcCity = new DataColumn("City")

dt.Columns.Add(dcName)

dt.Columns.Add(dcAge)

dt.Columns.Add(dcCity)

for (int i = 0i <配散 10i++)

{

DataRow dr = dt.NewRow()

dr[0] = "Name" + i

dr[1] = "Age" + i

dr[2] = "City" + i

dt.Rows.Add(dr)

}

json = JsonConvert.SerializeObject(dt)

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