jquery ajax 调用后台函数
1 var res; 2 $.ajax({ 3 type: "POST", 4 url: "fast_index_overview.aspx/GetOverViewZongLunContent", 5 data: "{city:‘" + regionCityName + "‘,time:‘" + time + "‘}", 6 contentType: "application/json", 7 dataType: "json", 8 async: false, 9 success: function (result) { 10 try { 11 res = eval(‘(‘ + result.d + ‘)‘)[0]; 12 13 } catch (e) { 14 15 } 16 }, 17 error: function(err) { 18 res = "错误:" + err.d; 19 } 20 }); 21 22 console.log("获取数据:" + res);
时间: 2024-10-23 09:47:14