我正在尝试使用ajax向服务器提交表单。我已经查看了各种教程,但无法弄清楚我自己的错误。因为我刚刚开始学习ajax,所以我没有尝试任何花哨的东西,而是在我的控制台中发送和显示json字符串。
这是我的控制者:
@RequestMapping(value="/create/transaction", method=RequestMethod.POST)
@ResponseBody
public String createTran(@RequestBody String json) throws IOException {
LOGGER.info("Create New Transaction");
System.out.println("--- Json: " + json);
ObjectMapper mapper = new ObjectMapper();
MyTransaction tranValue = mapper.readValue(json, MyTransaction.class);
MyTransaction tran = MyTransaction.getInstance();
tran.setName(tranValue.getName());
tran.setAmount(tranValue.getAmount());
return toJson(tran);
}
private String toJson(MyTransaction tran) {
ObjectMapper mapper = new ObjectMapper();
try {
String value = mapper.writeValueAsString(tran);
LOGGER.info("---- Value: " + value);
return value;
} catch(JsonProcessingException e) {
LOGGER.debug("----- Fail");
e.printStackTrace();
return null;
}
}
我的表格:
<form id="c-t-form" method="post" action="${pageContext.request.contextPath}/create/transaction">
<div class="modal-form">
<h3>Create New Transaction</h3>
<table class="table-form">
<tr>
<td class="c-50">
<label for="c-t-t">Title</label>
<span id="c-t-t-error" class="c-t-t-error error">Please enter the budget title.</span>
<div class="input-txt">
<input id="c-t-t" class="txt" type="text" name="name"/>
</div>
</td>
<td class="c-10">
<label for="c-t-a">Amount</label>
<span id="c-t-a-error" class="c-t-a-error error">Please enter correct budget amount.</span>
<div class="input-txt">
<input id="c-t-a" class="txt" type="text" name="amount"/>
</div>
</td>
</tr>
<tr>
<td>
<div>
<input class="submit-btn acc-btn" name="t-b" type="submit" value="Save" />
</div>
</td>
</tr>
</table>
</div>
</form>
这是我的ajax脚本:
$(document).ready(function() {
$("#c-t-form").submit(function(event) {
var tranName = $('#c-t-t').val();
var amount = $('#c-t-a').val();
var data = {"name" : tranName,"amount" : amount};
$.ajax({
url: $('#c-t-form').attr('action'),
data: JSON.stringify(data),
type: "POST",
cache: false,
beforeSend:function(xhr) {
xhr.setRequestHeader("Accept", "application/json");
xhr.setRequestHeader("Content-Type", "application/json");
},
success: function(reponse) {
alert("success");
},
error: function(xhr, status, error) {
alert(status);
}
});
return true;
});
});
我已经多次查看代码但在我的实现中找不到错误。这是我得到的错误
SEVERE: Servlet.service() for servlet [spring] in context with path [/budgetme] threw exception
com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'name': was expecting 'null', 'true', 'false' or NaN
at [Source: name=asda&amount=12&t-d=&t-b=Save; line: 1, column: 5]
我的模特:
class MyTransaction {
private int id;
private String name;
private String details;
private Bigdecimal amount;
// getters and setters
}
任何提示都会有所帮助。谢谢。
答案 0 :(得分:0)
您的功能应如下所示:
@RequestMapping(value="/create/transaction", method=RequestMethod.POST)
@ResponseBody
public MyTransaction createTran(@RequestBody MyTransaction json) throws IOException {
...
MyTransaction tran = MyTransaction.getInstance();
...
return tran;
}
无需自己处理ObjectMapper。
你可以在这里看到一个例子: http://spring.io/guides/gs/rest-service/
在本教程中,他们使用@RestController而不是@Controller。 RestController在每个函数上假设@ResponseBody。
此致