ajax请求

Posted 2022-10-17 lazyli

1、比较正常的ajax请求;传递到后台的参数是id:1;形式

function alertModel(id)
        console.log("修改的id:"+id);
        $.ajax(
            url: "",
            type: "post",
            data: 
                id: id
            ,
            success:function(data)
                console.log(data);
                
        );
    

2、请求的时候，参数进行处理

function saveModel()
        var p = [];
        var temp = ;
        temp.modelStandardName = $("#modelStandardName").val();
        temp.modelFactory = $("#modelFactory").val();
        temp.modelBrand = $("#modelBrand").val();
        temp.modelCateName = $("#modelCateName").val();
        temp.modelYear = $("#modelYear").val();
        temp.modelClass = $("#modelClass").val();
        temp.modelType = $("#modelType").val();
        temp.modelCountry = $("#modelCountry").val();
        temp.modelVouchSeat = $("#modelVouchSeat").val();
        temp.modelRatedLoad = $("#modelRatedLoad").val();
        temp.modelFactoryDirectPrice = $("#modelFactoryDirectPrice").val();
        temp.remark = $("#remark").val();
        p.push(temp);
        layer.close(1);
        currTc = layer.load(currTc);
        $.ajax(
            url:"<%=basePath%>commerical/addModel",
            type:"post",
            data:
                params: encodeURI(JSON.stringify(p))
            ,
            success:function(msg)
                layer.closeAll();
                if (msg.success) 
                    layer.close(currTc);
                    layer.confirm(‘添加成功‘, 
                        btn: [‘确定‘], //按钮
                        yes:function()
                            window.location.reload();
                        
                    );
                 else 
                    currTc = 0;
                    errorTip(msg, layer,msg.msg);
                
            
        )
    

　　后台接收以String，接收参数；需要进行处理

@RequestMapping("/addModel")
    @ResponseBody
    public AjaxResult addModel(String params)
        AjaxResult ajaxResult = new AjaxResult();
        try 
            params = URLDecoder.decode(params,"utf-8" );
            JSONArray jsonArray = JSON.parseArray(params);
            System.out.println(jsonArray);
            int result = commericalModelService.addModel(jsonArray);
            if(result > 0)
                ajaxResult.setSuccess(true);
            else
                ajaxResult.setSuccess(false);
            
        catch (Exception e)
            e.printStackTrace();
            ajaxResult.setSuccess(false);
        
        return ajaxResult;