我是jQuery的新手,并尝试使用Ajax调用Web服务。这似乎是相当简单的代码,但不知何故无法理解其失败的原因。我尝试了所有可能的方式(我能想到的),但不知何故没有让它发挥作用。
Jsp代码:
function tmpData() {
var dataObject = JSON.stringify({
'empfname': "First Name",
'emplname': "Last Name"
});
alert("dataObject=" + dataObject);
$.ajax({
url:"http://localhost:8080/OnlineStore/kmsg/grocery/tmpinfo",
type:"POST",
contentType: 'application/json',
data: dataObject,
done: setData,
fail: displayError()
});
}; // end of function
控制器:
@RestController
@RequestMapping("/kmsg/grocery")
public class TmpSvcImpl implements TmpSvcInt {
@Override
@RequestMapping(value = "/tmpinfo", method = RequestMethod.POST, headers = "Accept=application/json")
public @ResponseBody Map<String, Object> setData(@RequestBody final Emp employee1) throws Exception {
System.out.println("employee1=" + employee1);
String fname = employee1.getEmpfname();
String lname = employee1.getEmplname();
System.out.println("fn=" + fname ) ;
System.out.println("ln=" + lname ) ;
return null;
}
}
模特课:
public class Emp implements Serializable {
private static final long serialVersionUID = 1L;
String empfname;
String emplname;
public String getEmpfname() {
return empfname;
}
public void setEmpfname(String empfname) {
this.empfname = empfname;
}
public String getEmplname() {
return emplname;
}
public void setEmplname(String emplname) {
this.emplname = emplname;
}
public Emp(String fn, String ln){
this.empfname = fn ;
this.emplname = ln ;
}
@Override
public String toString() {
return "Emp {empfname=" + empfname + ", emplname=" + emplname + "}" ;
}
}
答案 0 :(得分:0)
在您描述的块中:
@RestController
@RequestMapping("/kmsg/grocery")
public class TmpSvcImpl implements TmpSvcInt {
@Override
@RequestMapping(
value = "/tmpinfo",
method = RequestMethod.POST,
headers = "Accept=application/json")
public @ResponseBody Map<String, Object> setData(@RequestBody final Emp employee1) throws Exception {
System.out.println("employee1=" + employee1);
String fname = employee1.getEmpfname();
String lname = employee1.getEmplname();
System.out.println("fn=" + fname ) ;
System.out.println("ln=" + lname ) ;
return null;
}
}
为此用例指定“Accept”标头并不正确。如果您希望此方法仅在content-type为'application / json'时响应,则您必须添加'application / json'的consumemes属性。正如评论所指出,这并非绝对必要。添加'application / json'的produce属性将确保Spring将尝试将返回的内容封送为JSON结构。
@RestController
@RequestMapping("/kmsg/grocery")
public class TmpSvcImpl implements TmpSvcInt {
@Override
@RequestMapping(
value = "/tmpinfo",
method = RequestMethod.POST,
consumes = "application/json",
produces = "application/json")
public @ResponseBody Map<String, Object> setData(@RequestBody final Emp employee1) throws Exception {
System.out.println("employee1=" + employee1);
String fname = employee1.getEmpfname();
String lname = employee1.getEmplname();
System.out.println("fn=" + fname ) ;
System.out.println("ln=" + lname ) ;
return null;
}
}
答案 1 :(得分:0)
将默认构造函数添加到Emp
类,以便Jackson
能够创建它的实例,如下所示:
public class Emp implements Serializable {
private static final long serialVersionUID = 1L;
public String empfname;
public String emplname;
public String getEmpfname() {
return empfname;
}
public void setEmpfname(String empfname) {
this.empfname = empfname;
}
public String getEmplname() {
return emplname;
}
public void setEmplname(String emplname) {
this.emplname = emplname;
}
public Emp(String fn, String ln) {
this.empfname = fn;
this.emplname = ln;
}
/**
* default constructor
*/
public Emp() {
}
@Override
public String toString() {
return "Emp {empfname=" + empfname + ", emplname=" + emplname + "}";
}
}