Javascript 如何使用onclick从另一个ajax请求重新加载数据表_Javascript_Php_Ajax_Datatables

Javascript 如何使用onclick从另一个ajax请求重新加载数据表

javascript php ajax datatables

Javascript 如何使用onclick从另一个ajax请求重新加载数据表,javascript,php,ajax,datatables,Javascript,Php,Ajax,Datatables,我尝试使用datatables是因为它很简单，但现在我遇到了在单击后从ajax请求中重新加载和获取datatables中新数据的问题。有关详细信息，请参阅下面的我的脚本： index.html <html> <head> <title>Datatables</title> <link rel="stylesheet" type="text/css" href="../plugins/DataTables/styles.css"/>

我尝试使用datatables是因为它很简单，但现在我遇到了在单击后从ajax请求中重新加载和获取datatables中新数据的问题。有关详细信息，请参阅下面的我的脚本：

index.html

<html>
<head>
  <title>Datatables</title>
  <link rel="stylesheet" type="text/css" href="../plugins/DataTables/styles.css"/>
  <script type="text/javascript" src="../plugins/DataTables/datatables.min.js"></script>
  <script type="text/javascript" src="../plugins/DataTables/javascript.js"></script>
  <script src="../lib/js/jquery/jquery.min.js"></script>
</head>
<body>
  <table id="tablenya" class="datatable responsive nowrap" style="width:100%">
   <thead>
     <tr>
       <th>No</th>
       <th>PO Date</th>
       <th>WO No</th>
       <th>PO No</th>
       <th>Customer</th>
       <th>Size</th>
       <th>Status</th>
       <th>Option</th>
     </tr>
    </thead>
    <tbody>
    </tbody>
  </table>
</body>
</html>

然后process.php节获取数据库中的数据

<?php
require 'connect.php';
$action = '';
$id  = '';
if(isset($_GET['action'])){
  $action = $_GET['action'];
  if($action == 'result'){
    if (isset($_GET['id'])){
        $id = $_GET['id'];
        if (!is_numeric($id)){
            $id = '';
        }
     }
   } else {
    $action = '';
   }
}

$mysqli_data = array();

if ($action == 'result'){
  $query = "SELECT * FROM workorder WHERE status='1' ORDER BY id DESC";
  $sql = $connect->query($query);
    if (!$sql){
      $result  = 'error';
      $message = 'query error';
    } else {
        $result  = 'success';
        $message = 'query success';
        $no = 1;
        while($row = $sql->fetch_array()){

            $functions  = '<div class="function_buttons"><ul>';
            $functions .= '<li class="function_view"><a data-id="'.$row['id'].'" data-name="'.$row[$dataName].'" title="View details"><span>View details</span></a></li>';
            $functions .= '</ul></div>';

            $mysqli_data[] = array(
                "no"        => $no++,
                "date_po"   => $row['date_po'],
                "no_spk"    => $row['no_spk'],
                "no_po"     => $row['no_po'],
                "customer"  => $row['customer'],
                "size"      => $row['size'],
                "status"    => $status,
                "functions" => $functions
            );
        }
    }
}

mysqli_close($connect);

$data = array(
    "result"  => $result,
    "message" => $message,
    "data"    => $mysqli_data
);

$json_data = json_encode($data);
print $json_data;
?>

这是另一个.php部分

<?php
require 'connect.php';
$action = '';
$id  = '';
if(isset($_GET['action'])){
  $action = $_GET['action'];
  if($action == 'result'){
    if (isset($_GET['id'])){
        $id = $_GET['id'];
        if (!is_numeric($id)){
            $id = '';
        }
     }
   } else {
    $action = '';
   }
}

$mysqli_data = array();

if($action == 'result'){

    if ($id == ''){
        $result  = 'erro';
        $message = 'ID missing';
    } else {

        $idx = mysqli_real_escape_string($connect, $id);
        $query = "SELECT * FROM workorder LEFT JOIN workorder_process ON workorder.id_fk = workorder_process.id_fk WHERE workorder.id = '".$idx."'";
        $sql = $connect->query($query);
        $get = $sql->fetch_array();

        if($get['delivery_type'] == '2'){
            $result  = 'success';
            $message = 'query success';
            $no = 1;

            while($row = $sql->fetch_array()){
                $functions  = '<div class="function_buttons"><ul>';
                $functions .= '<li class="function_edit"><a data-id="'.$row['id'].'" data-name="'.$row[$dataName].'" title="Edit"><span>Edit</span></a></li>';
                $functions .= '</ul></div>';

                $mysqli_data[] = array(
                    "no"        => $no++,
                    "date_po"   => $row['date_po'],
                    "date_spk"  => $row['date_spk'],
                    "no_spk"    => $row['no_spk'],
                    "no_po"     => $row['no_po'],
                    "customer"  => $row['customer'],
                    "size"      => $row['size'],
                    "qore"      => $row['qore'],
                    "roll"      => $row['roll'],
                    "material"  => $row['material'],
                    "ingredient"  => $row['ingredient'],
                    "send_qty"  => $row['send_qty'],
                    "volume"    => $row['volume'],
                    "annotation" => $row['annotation'],
                    "functions" => $functions
                );
            }

        } else {
            $result  = 'error';
            $message = 'ID missing';
        }
    }
}

mysqli_close($connect);

$data = array(
    "result"  => $result,
    "message" => $message,
    "data"    => $mysqli_data
);

$json_data = json_encode($data);
print $json_data;
?>

那么，在使用ajax请求单击函数视图之后，如何重新加载并获取新数据呢？在注释行javascript.js中

$(document).ready(function(){
  var tablenya = $('#tablenya').dataTable({
    "ajax": "process.php?action=result",
    "columns": [
      { "data": "no" },
      { "data": "date_po" },
      { "data": "no_spk"},
      { "data": "no_po" },
      { "data": "customer"},
      { "data": "size"},
      { "data": "status"},
      { "data": "functions","sClass": "functions" }
    ]
  });
});

//重新加载数据表并显示新请求

可能吗？请给我一些建议 $document.readyfunction{ $country.onchange，函数{ var country_id=$this.val；如果国家/地区id{ $.ajax{ 类型:POST,， url:location_ajax.php，数据：{ countryid:country\u id， }, 成功：functionhtml{ $state.htmlhtml； } }; }否则{ $state.htmlSelect Country； } }; }; 地点：选择一个国家

您只需要这一行来重新加载数据表，您还可以创建一个函数并调用它。试试看

tablenya.ajax.reload(null,false);

在onclick处理程序中，另一个解决方案是销毁表，然后重新创建。要销毁，请使用$'tablenya'.DataTable.destroy。然后，您可以像以前一样，在代码的前面初始化表格。

请尽量少提问题并使其完整！我已经搞定了帖子，但我不知道如何传达更多细节，先生。总之，我希望在单击datatables索引中选项列中的view（查看）后重新加载并获取新数据。htmlonly reload datatables sir（重新加载数据表）而不是从ajax请求中替换并获取新数据，请给出更多建议

tablenya.ajax.reload(null,false);