ajax传递数组和参数

Question

现在在页面取了数组存放在数组里面，同时还需要传递另外一个参数，代码如下
---------------------------------------------------------------------
var promActId=$("#promActId").val()
 var selectedIds = jQuery("#infoListTable").jqGrid("getGridParam","selarrrow");
             
                var pricingPlanIds  = new Array();  
                  if(selectedIds.length) {
                     for(var i=0;i<selectedIds.length;i++) {
                          var order = jQuery('#infoListTable').jqGrid('getRowData',selr[i]);
                               pricingPlanIds.push(pricingPlanIds);
                           }
                       }
                       $.ajax({
                       url:"<%=path%>/promotionTAction.do?method=addPricInfo",
                       type:"post",
                       date:{
                       
                       
                       }-----------------------------------------------------
现在想把pricingPlanIds  和promActId传递到后台Action.~怎么处理好，后台怎么接收~
求CSDn牛人给指点下ajaxjquerydate

Answer 1

var jsonData = {
         regionId:"1004",
         promActName:$("#promActName").val(),
         pricingPlanName:$("#pricingPlanName").val(),
         pricingPlanId:$("#pricingPlanId").val()
         };
         $.ajax({
         url:"<%=path%>/promotionTAction.do?method=queryPromotionActivity",
type:"POST",
data:jsonData,
dataType:"json",
cache:false,
success:function(data){
    alert(data.length);
    alert(JSON.stringify(data));
    },
error:function (XMLHttpRequest, textStatus, errorThrown){
alert(errorThrown);
alert(textStatus);
}我只用过这种。

Answer 2

按照data的格式写就行了，至于数组，你可以连接成特殊符号分隔的字符串，在服务端获取后解析一下就可以了

Answer 3

$.ajax({
 url:"<%=path%>/promotionTAction.do?method=addPricInfo",
type:"post",
data:{promActId:promActId,pricingPlanIds:pricingPlanIds}

Answer 4

$("#addBtn").bind("click",function(){
           var promActId=$("#promActId").val();
            var selectedIds = jQuery("#InfoListTable").jqGrid("getGridParam","selarrrow");
             
                var pricingPlanIds  = new Array();  
                  if(selectedIds.length) {
                     for(var i=0;i<selectedIds.length;i++) {
                          var order = jQuery('#infoListTable').jqGrid('getRowData',selectedIds[i]);
                               pricingPlanIds.push(pricingPlanIds);
                           }
                       }
                      $.ajax({
                          url:"<%=path%>/promotionTAction.do?method=addPricInfo",
                          type:"post",
                          data:{promActId:promActId,pricingPlanIds:pricingPlanIds},
                          dataType:"json",
                          cache:false,
                          success:function(data){
      alert("增加成功！");
     
    },
                         
                      
                      }); 
                      
                     
});我这样写报js错误。。
-----------------------------------------------------------------------------------------
}, getRowData:function (e) {
var i = {}, j, c = false, f, k = 0;
this.each(function () {
var n = this, a, r;
if (typeof e == "undefined") {
c = true;
j = [];
f = n.rows.length;
} else {
r = n.rows.namedItem(e);
if (!r) {
return i;
}
f = 1;
}
for (; k < f; ) {
jquery.jqgrid.min.js里面  r = n.rows.namedItem(e); 报错了

Answer 5

var order = jQuery('#infoListTable').jqGrid('getRowData',selectedIds[i]);
                               pricingPlanIds.push(pricingPlanIds);
修改了一下
------------------------------------
var pricingPlanId = jQuery('#infoListTable').jqGrid('getRowData',selectedIds[i],'PRICING_PLAN_ID');
                               pricingPlanIds.push(pricingPlanId);
还是报同样的错误

Answer 6

jqgrid就不知道了。。
 success:function(data){
      alert("增加成功！");
     
    },这里不要多“,”，要不ie会报错。。firefox，chrome没问题可以这样写

Answer 7

 jQuery("#InfoListTable").jqGrid("getGridParam","selarrrow"); 我这里写错了InfoListTable是table名字~谢谢了啊