Ajax方法回调函数中遍历JSON数据类型问题

if (userid != null) {

jq.ajax({

"url" : "ShowShoppingCarServlet",

"type" : "post",

async : false,

"data" : "userid=" + userid,

"dataType" : "JSON",

"success" : function(data4) {

var car = ""

var str=""

for (var m = 0m <data4.lengthm++) {

car += "<li><div class=\"img\"><a href=\"#\"><img src=\"/imgUrl"+data4[m].fileName+"\" width=\"58\" height=\"58\"></a></div><div class=\"name\"><a href=\"#\">"+ data4[m].names+ "</a></div><div class=\"price\"><font color=\"#ff4e00\">￥"+ data4[m].price+ "</font>X"+ data4[m].goodsnum + "</div></li>"

str+="<tr class=\"car_tr\"><td><div class=\"c_s_img\"><img src=\"/imgUrl"+data6[m].fileName+"\" width=\"73\" height=\"73\" /></div>"+data6[m].names+"</td><td align=\"center\">颜色：灰色</td><td align=\"center\"><div class=\"c_num\"><input type=\"button\" value="" onclick=\""+jianUpdate1(jq(this))+"\" class=\"car_btn_1\" /><input type=\"text\" value=\"1\" name="" class=\"car_ipt\" /><input type=\"button\" value="" onclick=\""+addUpdate1(jq(this))+"\" class=\"car_btn_2\" /></div></td><td align=\"center\" style=\"color:#ff4e00\">￥"+data6[m].price+"</td><td align=\"center\">26R</td><td align=\"center\"><a href=\"#\">删除</a><a href=\"#\">加入收藏</a></td></tr>"

}

jq("#cars").append(car)

jq("#showcar").append(str)

}

})

通过JSON方式返回数据，然后在success中遍历数据。 *** 作json

$.ajax({

type: 'post',

url: "maintain_findRoomByBuildingId.shtml",

cache: false,

data: {"buildingId":buildingId},

dataType: 'json',

success: function(data){

jQuery.each(data.roomList, function(i,item){

alert(item.id+","+item.name)

})

},

error: function(){

return

}

})

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