这是我的HTML
<table id="dt">
<thead>
<tr>
<th>make</th>
<th>model</th>
<th>serial</th>
<th>status</th>
<th>User</th>
<th>dept</th>
<th>location</th>
</thead>
<tbody>
</tbody>
<tfoot></tfoot>
</table>
这是我的JS
<script type="text/javascript" language="javascript" >
$(document).ready(function() {
$.post("json.php",function(data){
$('#dt').DataTable( {
"aaData": data,
"aoColumns": [
{"mDataProp": "make"},
{"mDataProp": "model"},
{"mDataProp": "serial"},
{"mDataProp": "status"},
{"mDataProp": "user"},
{"mDataProp": "dept"},
{"mDataProp": "location"}
]
});
});
} );
</script>
这里是 json.php
$data = Array ();
$data[] = array("make" => "Hp", "model" => "jhbh", "serial" => "kjkhn", "status" => "active", "user" => "John Doe", "dept" => "Manufacturing Services", "location" => "Bindura");
$data[] = array("make" => "Dell", "model" => "Vostro", "serial" => "kjkhn", "status" => "active", "user" => "Percy Holdin", "dept" => "Manufacturing Services", "location" => "Kwekwe");
echo json_encode($data,JSON_PRETTY_PRINT);
我已经对这个问题进行了编辑,因为现在我想获取动态数据。
错误:
DataTables warning: table id=dt - Requested unknown parameter 'make' for row 0, column 0. For more information about this error, please see http://datatables.net/tn/4
但是我没有从那个帮助链接中找到任何有用的东西
最佳答案
JSON 格式应如下例所示:https://datatables.net/examples/data_sources/ajax.html
{
"data": [
[
"Hp",
"jhbh",
"kjkhn",
"active",
"John Doe",
"Manufacturing Services",
"Bindura"
]
]
}
关于如何在 php 中像这样格式化 json 的示例(好吧,一种方式):
$data = (object) [
'data' => [[
'test',
'test',
'test'
]]
];
echo json_encode($data);
还有实例:http://sandbox.onlinephpfunctions.com/code/632c288c6c743da25e49958c204a8d4e0a936b54
关于javascript - DataTables 不显示 JSON 数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46682669/