如何在ajax成功后刷新DataTable？

Question

我试过不同的方式来刷新我的数据表后AJAX成功，但没有运气。我尝试了draw（）和.ajax.reload（）函数，但仍然没有运气。你有什么想法如何刷新它？

这里是我的代码

HTML

<table id="giacenza" class="table table-striped table-bordered"> 
    <thead> 
     <tr> 
     <th>Modello</th> 
     <th>Descrizione</th> 
     <th>Colore</th> 
     <th>Taglia</th> 
     <th>Barcode</th> 
     <th>Quantità</th> 
     </tr> 
    </thead> 
    <tbody> 
    </tbody> 
</table> 

的JavaScript

$(document).ready(function() { 
    function getRecords(){ 
     $.ajax({ 
      type: "POST", 
      url: "functions/getImported.php", 
      success: function(result) { 
       var table = $('#giacenza').DataTable({ 
        dom: 'Bfrtip', 
        buttons: [ 
         'copy', 'csv', 'excel', 'pdfHtml5', 'print' 
        ], 
        order: [[ 1, "asc" ]], 
       }); 
       var data = JSON.parse(result); 
       table.rows.add(data).draw(); 
      }, 
      error: function(result) { 
       alert("error: "+result); 
      } 
     }); 
    } 
    getRecords(); 
    $("#scarica-articolo").submit(function(event) { 
     event.preventDefault(); 
     var codbarra = $("#codbarra").val(); 
     $("#scarica-btn").attr("disabled",true); 
     $.ajax({ 
      type: "POST", 
      url: "functions/scaricaArticolo.php", 
      data: {codbarra:codbarra}, 
      success: function(result) { 
       if(result = "OK"){ 
        new PNotify({ 
         title: 'Articolo scaricato!', 
         text: 'L\'articolo è già stato scaricato dalla tabella seguente!', 
         type: 'success', 
         styling: 'bootstrap3' 
        }); 
        $("#scarica-btn").attr("disabled",false); 
        getRecords(); 
       }else if(result == "KO"){ 
        new PNotify({ 
         title: 'Oh No!', 
         text: 'La query non è andata a buon fine. Contattare l\'assistenza.', 
         type: 'error', 
         styling: 'bootstrap3' 
        }); 
        $("#scarica-btn").attr("disabled",false); 
       } 
      }, 
      error: function(result) { 
       alert("error: "+result); 
      } 
     }); 
    }); 
}); 

PHP

<?php 
include("db_config.php"); 
$mysqli->set_charset("utf8"); 
$codbarra = $mysqli->real_escape_string($_POST["codbarra"]); 

$query = "UPDATE records SET PARESTALLA = PARESTALLA-1 WHERE CODBARRA = ".$codbarra.";"; 
$results = array(); 
$result = $mysqli->prepare($query); 
$result->execute(); 
if($result->affected_rows > 0){ 
    echo "OK"; 
}else{ 
    echo "KO"; 
} 
?> 

编辑： 我更新了我的JavaScript脚本，但我可以没有解决我的问题。现在，我仍然收到类似的错误：“数据表警告：表ID = giacenza - 不能重新初始化数据表中有关此错误的详细信息，请参阅http://datatables.net/tn/3”

Answer 1

要成功地重新加载表

table.ajax.reload();//where the table variable is the variable that holds the DataTable itself 

例

var table = $('#giacenza').DataTable({ 

//ajax call that fetches data from the database. 

}); 

另一个想法是有一个JavaScript函数从数据库中获取记录。在从数据库删除成功后，再次调用该函数。

例

function getRecords(){ 
//this gets the full list from the database on document ready 
//make Ajax call to retrieve from database 
} 

if(result == 'OK'){ 
//its been successfully deleted 
getRecords();//call the js function that gets from database again 
} 

编辑 您正在使用的mysqli。该API支持预准备语句。请不要使用..

$query = "UPDATE records SET PARESTALLA = PARESTALLA-1 WHERE CODBARRA = ?"; 

$result = $mysqli->prepare($query); 
$result->bind_param('s', $codbarra); 
$result->execute(); 
echo $result->affected_rows > 0 ? 'OK' : 'KO';