[英]Convert POST to Json with Java
我有简单的邮政要求
POST /savings HTTP/1.1
Host: localhost:4567
Cache-Control: no-cache
Content-Type: application/x-www-form-urlencoded
Body:
name=username&description=userdescription
我想知道如何将帖子正文name=username&description=userdescription
为Json,例如:
{
"name": "username",
"description": "userdescription"
}
任何想法,即时通讯使用http://sparkjava.com ,我需要处理内容类型为post的请求: application/x-www-form-urlencoded
post("/user", (request, response) -> {
// .. Convert request to User object
});
您可以使用request.getParameter(String param)方法从POST请求中获取参数值。
我假设您正在使用servlet,因此可以在doPost()
方法中进行以下操作:
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String name = request.getParameter("name");
String description = request.getParameter("description");
JSONObject json = new JSONObject();
json.put("name", (name==null || name.isEmpty() ? "" : name));
json.put("description", (description==null || description.isEmpty())? "" : description);
}
然后您将获得所需的json:
{
"name": "username",
"description": "userdescription"
}
我假设您是从servlet的URL中获取参数的,不是吗?
无论如何,作为Regex的练习,我会这样做:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.regex.PatternSyntaxException;
public class Any {
public static String queryStringToJasonParser(final String queryString){
try {
Pattern regex = Pattern.compile("(^|\\s+)name=(.*?)&description=(.*?)(\\s+|$)");
Matcher regexMatcher = regex.matcher(queryString);
if(regexMatcher.matches()){
final String nameValue = regexMatcher.group(2);
final String descriptionValue = regexMatcher.group(3);
return "{\"name\": \""+nameValue+"\", \"description\": \""+descriptionValue+"\"}";
}
} catch (PatternSyntaxException ex) {
//TODO: Handle it.
}
return "{\"name\": \"\", \"description\": \"\"}";
}
public static void main(String[] args){
final String resultJason = Any.queryStringToJasonParser("name=wow1&description=wow2");
System.out.println(resultJason);
final String resultJason1 = Any.queryStringToJasonParser("name=&description=wow2");
System.out.println(resultJason1);
}
}
添加任何所需的优化
问候
我个人使用Google Gson进行此操作并请求对象。
值得注意的是,如果您使用的是网址参数,则此方法将无效。
编辑:如果缺少任何参数,则UserRequest对象将为null,因为gson将无法构建该对象。
import java.io.InputStreamReader;
import com.google.gson.Gson;
import spark.Spark;
public class Handler
{
private final Gson gson = new Gson();
public Handler()
{
}
public void init()
{
Spark.post("/user", (req, resp) -> {
final UserRequest userRequest = gson.fromJson(new InputStreamReader(req.raw().getInputStream()), UserRequest.class);
//do something with user
return new Object();
});
}
private static class UserRequest
{
private String name;
private String description;
public String getDescription()
{
return description;
}
public String getName()
{
return name;
}
}
public static void main(final String[] args)
{
final Handler handler = new Handler();
handler.init();
}
}
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