我正在使用jquery.validate.js验证表单,我需要识别重复的条目,为此我使用自定义方法,即:
jQuery.validator.addMethod("uniqueName", function(name, element) {
var response;
$.ajax({
type: "POST",
dataType : "json",
url:"${pageContext.request.contextPath}/company/getDuplicate",
data:"name="+name,
async:false,
success:function(data){
response = data;
},
error: function (data) {
alert(request.responseText);
}
});
}, "Name is Already Taken");
在规则部分:
rules : {
name : {
required : true,
uniqueName : true
}
},
errorElement : "span",
messages : {
name : {
required : "Name Is Required"
}
}
这是我的JSP代码:
<label>Name:</lable>
<form:input path="name"></form:input>
它正在命中指定的url,但Json将null值发送给方法
这是我的控制器方法:
@RequestMapping(value = "/company/getDuplicate", method = RequestMethod.POST, headers = "Accept=*/*")
public @ResponseBody void getTitleList(HttpServletRequest request, HttpServletResponse response) {
JSONObject json = new JSONObject();
String data = ((String)json.get("name"));
List<Company> matched = companyService.getDuplicate(data);
if(matched != null && !"".equals(matched)){
json.put("name", "present");
System.out.flush();
}
else{
json.put("name", "notPresent");
System.out.flush();
}
}
我想要的是: 1.如何将文本框的值发送给控制器(Json在我的情况下发送null)。 2.在上面的方法中,我不认为'if语句有写条件',因为当'name'在数据库中不存在时,'matched'变量显示如下=&gt; []
请帮我解决这个问题。提前致谢。
答案 0 :(得分:0)
修改您的代码如下:
$.ajax({
type: "POST",
dataType : "json",
url:"${pageContext.request.contextPath}/company/getDuplicate",
data:{"name":name},
async:false,
success:function(data){
response = data;
},
error: function (data) {
alert(request.responseText);
}
});
并将控制器处理程序修改为
请注意方法签名
@RequestParam(value="name") String name
@RequestMapping(value = "/company/getDuplicate", method = RequestMethod.POST, headers = "Accept=*/*")
public @ResponseBody void getTitleList(@RequestParam(value="name") String name,HttpServletRequest request, HttpServletResponse response) {
//JSONObject json = new JSONObject();
//String data = ((String)json.get("name"));
List<Company> matched = companyService.getDuplicate(name);
if(matched != null && !"".equals(matched)){
json.put("name", "present");
System.out.flush();
}
else{
json.put("name", "notPresent");
System.out.flush();
}
}