[英]Spring MVC and jquery post request with array of POJO
I have a simple POJO Java class (getters and setters is not shown) 我有一个简单的POJO Java类(未显示getter和setter)
public class VacationInfo {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Temporal(TemporalType.TIMESTAMP)
private Date vacationFrom;
@Temporal(TemporalType.TIMESTAMP)
private Date vacationTo;
, Spring MVC controller with next method ,Spring MVC控制器的下一种方法
@RequestMapping(value = "updateVacations", method = RequestMethod.POST)
public String updateVacations(@RequestParam VacationInfo[] vacationInfos) {
...
}
and jQuery post request 和jQuery发布请求
$.ajax({
type: "POST",
url: "updateVacations",
dataType: 'json',
data: vacationInfos
});
where "vacationInfos" is a array with JSON objects, which represent VacationInfo class: 其中“ vacationInfos”是带有JSON对象的数组,这些对象表示VacationInfo类:
[
{
vacationFrom: "01-01-2013",
vacationTo: "01-01-2013"
},
{
vacationFrom: "01-01-2013",
vacationTo: "01-01-2013"
}
]
But when I do request - i got a HTTP 400 error. 但是当我请求时-我收到了HTTP 400错误。
jquery method:: jQuery方法::
$('#testButton').click(function(){
var testList= [];
$('.submit').filter(':checked').each(function() {
var checkedFrom= $(this).closest('form');
var testPojo= checkedFrom.serializeObject();
testList.push(testPojo);
});
$.ajax({
'type': 'POST',
'url':"testMethod",
'contentType': 'application/json',
'data': JSON.stringify(testList),
'dataType': 'json',
success: function(data) {
if (data == 'SUCCESS')
{
alert(data);
}
else
{
alert(data);
}
}
});
});
whereas jquery provide two level's of serialization like serialize() and serializeArray().But this is custome method for serialize a 而jquery提供了两个级别的序列化,如serialize()和serializeArray()。但这是用于序列化一个的定制方法
User defined Object. 用户定义的对象。
$.fn.serializeObject = function() {
var o = {};
var a = this.serializeArray();
$.each(a, function() {
if (o[this.name]) {
if (!o[this.name].push) {
o[this.name] = [o[this.name]];
}
o[this.name].push(this.value || '');
} else {
o[this.name] = this.value || '';
}
});
return o;
};
in spring controller 在弹簧控制器中
@RequestMapping(value = "/testMethod", method = RequestMethod.POST)
public @ResponseBody ResponseStatus testMethod(HttpServletRequest request,
@RequestBody TestList testList)
throws Exception {
..............
}
where TestList is an another class which is written to handle form post with a array of Test in mvc controller for test method 其中TestList是另一个类,用于处理带有MVC控制器中用于测试方法的Test数组的表单发布
public class TestList extends ArrayList<Test> {}
try using @RequestBody
instead of @RequestParam
尝试使用@RequestBody
而不是@RequestParam
@RequestMapping(value = "updateVacations", method = RequestMethod.POST)
public String updateVacations(@RequestBody VacationInfo[] vacationInfos) {
...
}
The @RequestBody
method parameter annotation indicates that a method parameter should be bound to the value of the HTTP request body, which is the JSON data in your case. @RequestBody
方法参数注释表示方法参数应绑定到HTTP请求正文的值,在您的情况下为HTTP数据。
I answered my question. 我回答了我的问题。 Form client I send Date as timestamp. 表单客户端我发送日期作为时间戳。 Because server should not know anything about what time zone is the client and should not depend on a specific date format(it's one of the best practice). 因为服务器不应该知道客户端在哪个时区,也不应该依赖于特定的日期格式(这是最佳做法之一)。 And after that I'm add JsonDeserializer annotation on VacationInfo date fields and this is work. 之后,我在VacationInfo日期字段上添加了JsonDeserializer注释,这是可行的。
public class VacationInfo {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@JsonDeserialize(using=DateDeserializer.class)
@Temporal(TemporalType.TIMESTAMP)
private Date vacationFrom;
@JsonDeserialize(using=DateDeserializer.class)
@Temporal(TemporalType.TIMESTAMP)
private Date vacationTo;
Ajax POST request Ajax POST请求
[
{
vacationFrom: "1359197567033",
vacationTo: "1359197567043"
},
{
vacationFrom: "1359197567033",
vacationTo: "1359197567043"
}
]
If you need to send Date as string in specific format("mm-dd-yyyy" for example) - you need to define own JsonDesiarilizer(org.codehaus.jackson.map package in Jackson ) class, which extends frpm JsonDeserializer class and implement your logic. 如果您需要以特定格式(例如“ mm-dd-yyyy”)以字符串形式发送Date-您需要定义自己的JsonDesiarilizer( Jackson中的 org.codehaus.jackson.map包)类,该类扩展了frpm JsonDeserializer类并实现你的逻辑。
You can create your own formatter to parse incoming request. 您可以创建自己的格式化程序来解析传入的请求。 Read here . 在这里阅读。 The code below is a little trimmed. 下面的代码略有修饰。
public class LinkFormatter implements Formatter<List<Link>> {
@Override
public List<Link> parse(String linksStr, Locale locale) throws ParseException {
return new ObjectMapper().readValue(linksStr, new TypeReference<List<Link>>() {});
}
}
Jquery: jQuery:
$.ajax({
type: "POST",
data: JSON.stringify(collection)
...
});
Spring controller: 弹簧控制器:
@RequestParam List<Link> links
And don't forget to register it in application context: 并且不要忘记在应用程序上下文中注册它:
<bean id="conversionService" class="org.springframework.format.support.FormattingConversionServiceFactoryBean">
<property name="formatters">
<set>
<bean class="LinkFormatter"/>
</set>
</property>
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