form表單提交數組數據

數組數據可以通過ajax提交給后台，但是如果要跳轉頁面的話ajax是無法跳轉的，要到success做location.href的跳轉，

先定義一個class是接受數組類型的

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;
    }


}

再定義一個對象來接受form表單提交的數組

public class TicketCarts {
    private List<TicketCart> ticketCarts;

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

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

html代碼  i代表數組下標

<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）{

//把數據拿出來用list裝着
List<TicketCart> ticketCart=ticketCarts.getTicketCarts();
//定義兩個數組提取數據出來方便操作
int[] count=new int[ticketCart.size()];
String[] name=new String[ticketCart.size()];
//遍歷list把數據賦給數組
for (int i=0;i<ticketCart.size();i++){
    count[i]=ticketCart.get(i).getCount();
    name[i]=ticketCart.get(i).getName();
}

}

這樣就可以得到前端提交的兩個數組。

 

其實可以直接用ajax更加方便

ajax代碼

var count=new Array();

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

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