[英]Convert comma separated String into Json string
String input = "Vish,Path,123456789";
預期輸出為Json字符串,並且線程安全= {“name”:“Vish”,“surname”:“Path”,“mobile”:“123456789”}
我試過用
GsonBuilder gsonBuilder = new GsonBuilder();
Gson gson = gsonBuilder.create();
但每次我創建新對象 -
MappingObject[] studentArray = new MappingObject[1];
studentArray[0] = new MappingObject("Vish","Path","123456789");
我用split()分隔了這個逗號分隔的字符串
System.out.println("JSON "+gson.toJson(studentArray));
如果您不想使用任何庫,則必須用逗號分隔字符串並創建一個新String
。
String input = "Vish,Path,123456789";
String[] values=input.split("[,]");
StringBuffer json = new StringBuffer();// StringBuffer is Thread Safe
json.append("{")
.append("\"name\": \"").append(values[0]).append("\",")
.append("\"surname\": \"").append(values[1]).append("\",")
.append("\"mobile\": \"").append(values[2]).append("\"")
.append("}");
System.out.println(json.toString());
輸出:
{“name”:“Vish”,“surname”:“Path”,“mobile”:“123456789”}
如果你想使用圖書館,那么你將獲得Jackson
這一點。 簡單地制作一個類並通過它制作json。
public class Person {
private String name;
private String surname;
private String mobile;
// ... getters and Setters
}
String input = "Vish,Path,123456789";
String[] values=input.split("[,]");
Person person = new Person(values[0],values[1],values[2]);// Assume you have All Argumets Constructor in specified order
ObjectMapper mapper = new ObjectMapper(); //com.fasterxml.jackson.databind.ObjectMapper;
String json = mapper.writeValueAsString(person);
您將不得不創建一個地圖:
Map<String,String> jsonMap = new HashMap<String,String>();
jsonMap.put("name","Vish");
jsonMap.put("surname","Path");
jsonMap.put("mobile","123456789");
然后使用com.google.gson JSONObject:JSONObject jsonObj = new JSONObject(jsonMap);
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