Question

我试图通过ajax调用将一些变量通过JSON从JSP传递给servlet。但我在servlet端获得null值。请有人帮助我找出我犯错的地方/我错过的地方

＆＃13;

package com.example;

import org.kie.api.KieServices;
import org.kie.api.runtime.KieContainer;
import org.kie.api.runtime.KieSession;

public class SalaryTest {

    public static final void main(String[] args) {
        try {
            // load up the knowledge base
            KieServices ks = KieServices.Factory.get();
            KieContainer kContainer = ks.getKieClasspathContainer();
            KieSession kSession = kContainer.newKieSession("ksession-dtables");

            Salary a = new Salary();
            a.setAmount(600);

            kSession.insert(a);
            kSession.fireAllRules();


        } catch (Throwable t) {
            t.printStackTrace();
        }
    }

    public static class Salary{
        private String message;
        private int amount;
        private int status;

        public String getMessage() {
            return message;
        }
        public void setMessage(String message) {
            this.message = message;
        }
        public int getAmount() {
            return amount;
        }
        public void setAmount(int amount) {
            this.amount = amount;
        }
        public int getStatus() {
            return status;
        }
        public void setStatus(int status) {
            this.status = status;
        }

    }
}

＆＃13;

//JSON	
		var masterdata = new Object();
	    masterdata.grn = $('#grn').val();
	    masterdata.pono = $('#pono').val();
	    masterdata.podt = $('#podt').val();
      
//call the servlet to insert the data only when error = 0
		if (error != 1){
			$.ajax({
				url : 'insertserv',
				type: 'POST',
				dataType: 'json',
		        data: {test : JSON.stringify(masterdata)},
		        contentType: 'application/json',
		        mimeType: 'application/json',
				success : function(data) {
							alert('Hi');
						}			
		          });
		    }
		else{
			alert("Save cannot be performed. Please check the entered data!");
		}
	  	});

＆＃13;

Answer 1

用我的代码替换你的ajax代码......

//JSON	
		var masterdata = new Object();
	    masterdata.grn = $('#grn').val();
	    masterdata.pono = $('#pono').val();
	    masterdata.podt = $('#podt').val();
      
//call the servlet to insert the data only when error = 0
		if (error != 1){
			$.ajax({
				url : 'insertserv',
				type: 'POST',
				dataType: 'json',
		        data: JSON.stringify({"test" :masterdata}),
		        contentType: 'application/json',
		        mimeType: 'application/json',
				success : function(data) {
							alert('Hi');
						}			
		          });
		    }
		else{
			alert("Save cannot be performed. Please check the entered data!");
		}
	  	});

在servlet中获取数据

 BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
      String json = "";

            if (br != null) {
                json = br.readLine();
            }
 JSONObject wholedata= new JSONObject(json);

现在对象wholedata有你的json ..

如果你正在使用JSON.stringify（），那么你必须使用 servlet中的BufferedReader，当您在servlet的网址中传递数据时，可以在servlet中使用request.getparameter。

Answer 2

如果您的后端使用json内容进行回复，则只有dataType:"json"有效。尝试更改响应类型：

response.setContentType("application/json");

因为你的ajax期望来自后端的json dataType:"json",。

反之亦然。

将dataType更改为text dataType:"text",，因为响应标头显示为response.setContentType("text/plain");。但在这种情况下，您必须使用JSON.parse()来解析json字符串。

//JSON  
var masterdata = new Object();
masterdata.grn = $('#grn').val();
masterdata.pono = $('#pono').val();
masterdata.podt = $('#podt').val();

//call the servlet to insert the data only when error = 0
if (error != 1) {
  $.ajax({
    url: 'insertserv',
    type: 'POST',
    dataType: 'text', //<------change this
    data: {
      test: JSON.stringify(masterdata)
    },
    contentType: 'application/json',
    mimeType: 'application/json',
    success: function(data) {
      alert('Hi');
    }
  });
} else {
  alert("Save cannot be performed. Please check the entered data!");
}
});

无法在servlet中获取JSON字符串

2 个答案: