Question

我正在使用DataTables，并希望使用Ajax加载我的数据。但是加载页面时出现此错误：

DataTables警告：表id = example-请求的未知参数'DocentID'用于第0行第0列。有关此错误的更多信息，请参见http://datatables.net/tn/4

我不知道我在做什么错，这是我的数据表脚本：

$(document).ready(function() {
  $("#example").DataTable({
    "ajax": {
      "url": "assets/includes/ajax.php",
      "dataSrc": ""
    },
    "columns": [
        {"data": "DocentID"}
    ]
  });
});

我的ajax.php文件及其回声

include_once('DB_Handler.php');

$result = mysqli_query($mysqli, "SELECT DocentID FROM gebruiker");

$rows = array();
while ( $row = mysqli_fetch_array($result))
{
    $rows[] = json_encode($row);
}

echo json_encode($rows);

数组：

["{\"0\":\"0\",\"DocentID\":\"0\"}","{\"0\":\"67\",\"DocentID\":\"67\"}","{\"0\":\"0\",\"DocentID\":\"0\"}","{\"0\":\"0\",\"DocentID\":\"0\"}"]

Answer 1

尝试删除第一个@Table(name = "invoice") public class InvoiceEntity { @OneToMany(mappedBy = "futureInvoice", fetch = FetchType.LAZY, cascade = CascadeType.ALL) private List<CorrectionEntity> corrections; } @Table(name = "correction") public class CorrectionEntity extends BaseTimestampedEntity { @ManyToOne(fetch = FetchType.LAZY) @JoinColumn(name = "future_invoice_id") private InvoiceEntity invoice; }。无需对其进行两次编码。

json_encode()

dataTable不显示正确的ajax数据

1 个答案: