我使用 YUI 将 json 从页面发送到服务器以及如何从 servlet 中的 json 获取数据。我使用杰克逊库 2.4。谢谢!!!
var user = {
userName: username,
password: password,
customerId: customerId
};
new Y.IO().send("http://localhost:7778/MyController", {
method: 'POST',
data: user
});
实际上,当您提出这样的请求时
<html>
<body>
<script src="http://yui.yahooapis.com/3.14.1/build/yui/yui-min.js"></script>
<script>
var user = {
userName: 'x1',
password: 'y2',
customerId: 'z3'
};
YUI().use('io-form', function (Y) {
new Y.IO().send("http://localhost:8080/web/MyController", {
method: 'POST',
data: user
});
});
</script>
</body>
</html>
您不会将 JSON 对象发送到 servlet。相反,您正在创建一个 http 请求,其中 javascript 对象的每个属性(以 JSON 格式表示)都作为 http POST 属性发送,因此您可以像这样检索这些值
package mine;
import java.io.IOException;
import java.util.Map;
import java.util.Map.Entry;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
/**
* Servlet implementation class MyController
*/
@WebServlet("/MyController")
public class MyController extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public MyController() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
Map<String, String[]> map = request.getParameterMap();
for(Entry<String,String[]> entry:map.entrySet()){
System.out.println(entry.getKey());
System.out.println(entry.getValue()[0]);
}
}
}
这反过来打印了这个
userName
x1
password
y2
customerId
z3
因此,您不需要 Jackson 来解析您从页面检索到的数据。