的Json返回阵列显示器
问题描述:
在下面甲笨控制器返回JSON,的Json返回阵列显示器
{"r":[{"galleryid":"1","gname":"birthday","eventdate":"2016-07-20 00:00:00","totalphoto":"250","selectedphoto" :"100","glock":"0","userid":"1"},{"galleryid":"2","gname":"anniversary","eventdate":"2016-07-14 00:00:00","totalphoto":"500","selectedphoto":"251","glock":"0","userid":"1"}]}
和从控制器代码返回:
$this->load->model('gallery_model');
$data['r'] = $this->gallery_model->gallery_data($userid);
echo json_encode($data);
但是鉴于它只是显示:未定义
视图代码:
<script type="text/javascript">
// Ajax post
$(document).ready(function() {
$("input#displaygallery").on('click', function (event) {
//var user_id = document.getElementById("userdropdown").value;
//alert(user_id);
event.preventDefault();
var user_id = document.getElementById("userdropdown").value;
jQuery.ajax({
type: "POST",
url: "<?php echo base_url(); ?>" + "admin_cont/gallery_controller/user_userdata",
dataType: 'json',
data: {userid: user_id},
success: function (res) {
//console.log(res);
$.each(res, function (idx, obj) {
alert(obj.tagName);
});
}
});
});
});
</script>
答
尝试:
的javascript:
$("input#displaygallery").on('click', function (event) {
event.preventDefault();
var user_id = document.getElementById("userdropdown").value;
jQuery.ajax({
type: "POST",
url: "<?php echo base_url(); ?>" + "admin_cont/gallery_controller/user_userdata",
dataType: 'json',
data: {userid: user_id},
success: function (res) {
var html = "<table>";
html += "<thead>";
html += "<tr>";
html += "<th>eventdate</th>";
html += "<th>galleryid</th>";
html += "<th>glock</th>";
html += "<th>gname</th>";
html += "<th>selectedphoto</th>";
html += "<th>totalphoto</th>";
html += "<th>userid</th>";
html += "<tr>";
html += "<thead>";
html += "<tbody>";
for (i = 0; i <= res.r.length - 1; i++) {
html += "<tr>";
html += "<td>" + res.r[i].eventdate + "</td>";
html += "<td>" + res.r[i].galleryid + "</td>";
html += "<td>" + res.r[i].glock + "</td>";
html += "<td>" + res.r[i].gname + "</td>";
html += "<td>" + res.r[i].selectedphoto + "</td>";
html += "<td>" + res.r[i].totalphoto + "</td>";
html += "<td>" + res.r[i].userid + "</td>";
html += "<tr>";
}
html += "</tbody>";
html += "</table>";
$("your cointainer id or class name").html(html);
}
});
});
尼斯,表现出更多的代码,显示Ajax调用方法 – rad11
的完整的代码以及如何你'view'显示它们? –
这里我想这个JSON数据如下表格作为表头gallaeryid gname evendate作为标题和其他相同的数据在tr – Jalpesh