form表单提交数组数据

数组数据可以通过ajax提交给后台，但是如果要跳转页面的话ajax是无法跳转的，要到success做location.href的跳转，

先定义一个class是接受数组类型的

public class TicketCart {

    private int count;
    private String name;



    public int getCount() {
        return count;
    }

    public void setCount(int count) {
        this.count = count;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }


}

再定义一个对象来接受form表单提交的数组

public class TicketCarts {
    private List<TicketCart> ticketCarts;

    public List<TicketCart> getTicketCarts() {
        return ticketCarts;
    }

    public void setTicketCarts(List<TicketCart> ticketCarts) {
        this.ticketCarts = ticketCarts;
    }
}

html代码  i代表数组下标

<form action="xxx" method="post">  
<input  name="ticketCarts[i].count"     />
 <input name="ticketCarts[i].name" />

<button type="submit">提交</button>
<form>

后台接收：

public ModelAndView sellTicket(TicketCarts ticketCarts）{

//把数据拿出来用list装着
List<TicketCart> ticketCart=ticketCarts.getTicketCarts();
//定义两个数组提取数据出来方便操作
int[] count=new int[ticketCart.size()];
String[] name=new String[ticketCart.size()];
//遍历list把数据赋给数组
for (int i=0;i<ticketCart.size();i++){
    count[i]=ticketCart.get(i).getCount();
    name[i]=ticketCart.get(i).getName();
}

}

这样就可以得到前端提交的两个数组。

 

其实可以直接用ajax更加方便

ajax代码

var count=new Array();

$.ajax({
    type:"post",
    url:"xxx",
    data:{deleteNum:deleteNum},
   
    success:function(data){
        if(data.success){
            deleteNum = [];
        }
    }
});

参考文章：https://www.cnblogs.com/yunspider/p/6337872.html