Ajax响应无法显示
问题描述:
这是我从下面的链接Ajax响应无法显示
function jsonCall1() {
$.post("http://tallentex.com/phpwebservices/feedbackapp/index.php/hostel_service/syncData", { id: 0 }, function (data) {
var tbl=$("<table/>").attr("id","mytable");
$("#dv").append(tbl);
for (var i = 0; i < data.length; i++) {
var tr = "<tr>";
var td1 = "<td>" + data[i]["id"] + "</td>";
var td2 = "<td>" + data[i]["fno"] + "</td>";
var td3 = "<td>" + data[i]["attn_no"] + "</td></tr>";
$("#mytable").append(tr + td1 + td2 + td3);
}
console.log(data);
});
}
提取数据这是console.I这个函数的输出要显示在HTML表格,也该输出javascript函数想用数据表 发送给服务器在此回应,我想显示HTML表阵列中的数据,也希望这个阵列中的数据发送到服务器
{status: 1, data: Array(4)}
data:Array(4)
0:{id: "1", fno: "16078134", hostel_id: "12345", attn_no: "0010146998", create_date: "2017-08-21 10:31:02"}
1:{id: "2", fno: "16078134", hostel_id: "12345", attn_no: "0010146998", create_date: "2017-08-21 10:31:02"}
2:{id: "3", fno: "16078134", hostel_id: "12345", attn_no: "0010146998", create_date: "2017-08-21 10:31:02"}
3:{id: "4", fno: "16078134", hostel_id: "12345", attn_no: "0010146998", create_date: "2017-08-21 10:31:02"
答
function jsonCall1() {
$.post("http://tallentex.com/phpwebservices/feedbackapp/index.php/hostel_service/syncData", { id: 0 }, function (data) {
var tbl=$("<table/>").attr("id","mytable");
$("#dv").append(tbl);
for (var i = 0; i < data.length; i++) {
var str = `<tr>
<td>${data[i].id}</td>
<td>${data[i].fno}</td>
<td>${data[i].attn_no}</td>
</tr>`;
$("#mytable").append(str);
}
console.log(data);
});
}
你有你的问题r代码此刻? –
不,完全没有,它正在成功运行 – user6180198