Warning: file_get_contents(/data/phpspider/zhask/data//catemap/2/jquery/78.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Jquery 如何在spring控制器中获取参数_Jquery_Spring - Fatal编程技术网

Jquery 如何在spring控制器中获取参数

Jquery 如何在spring控制器中获取参数,jquery,spring,Jquery,Spring,如何发送数组到ajax请求。下面是我的代码技巧是数组 var url= "/ajax/gDirectory/saveskills.htm"; $.ajax({ type: "POST", data: {skills:skills}, url: url, async: true,

如何发送数组到ajax请求。下面是我的代码技巧是数组

var url= "/ajax/gDirectory/saveskills.htm";
                 $.ajax({
                    type: "POST",
                    data: {skills:skills},
                    url: url,
                    async: true,
                    success :skillsUI
                  }); 
Spring控制器如何在控制器中获取参数

 @RequestMapping(value="/ajax/gDirectory/saveskills.htm" ,method=RequestMethod.POST)
        public @ResponseBody String saveskills(HttpServletRequest request,@RequestParam String skills) {}

您可以使用@RequestBody注释代替@RequestParam

假设您有一个名为“Skill”的类,该类具有与“Skill”json对象相同的属性

@RequestMapping(value="/ajax/gDirectory/saveskills.htm" ,method=RequestMethod.POST)
        public @ResponseBody String saveskills(HttpServletRequest request, @RequestBody List<Skill> skills) {}
@RequestMapping(value=“/ajax/gDirectory/saveskills.htm”,method=RequestMethod.POST)
public@ResponseBody字符串保存技能(HttpServletRequest请求,@RequestBody列表技能){}

该技能对象在js中的外观如何?它是普通数组。我有数组,因此如果有其他方法,请让我知道..它不是json字符串您可以将数组转换为json字符串并传递它,请查看此线程-感谢工作正常…:)