Question

我正在尝试从mysql数据库填充我的数据表。一切正常，但是当我尝试查看我的桌子时，它说：

DataTables警告：表id = manageProductTable-无效的JSON响应。

有关此错误的更多信息，请参见http://datatables.net/tn/1

我已经检查过控制台，可以在其中看到我的列表：

connection success {
    "data": [
        ["tulum","rtert","6","12.30","dfsdfsdf"],
        ["body","bd56","8","56,30","dfsdfsdf"],
        ["dfdsf","dsfdsf","8","56,32","dsfdsfsd"],
        ["","saf","8","25.90","dsfdsfsd"],
        ["tulum","FWNWP","","","dsfdsfsd"],
        ["bluz","dd56","6","78","dsfdsfsd"],
        ["dsfdsf","sdfsdf","9","23","chicco"],
        ["atlet","ATL30","30","23,30","chicco"],
        ["BODY","56FG","6","56,30","dsfdsfsd"]
    ]
}

那我在做什么错了？

JS

$('#manageProductTable').DataTable({
    ajax: {
        'url': 'php_action/fetchProduct.php',
        'dataSrc': 'features'},
        'columns': [
            {"data":"Name"},
            {"data":"Code"},
            {"data":"Brand"},
            {"data":"Quantity"},
            {"data":"Price"},
        ]
});

PHP

$sql = "SELECT product.product_id, product.product_name, 
               product.product_code, product.brand_id,
               product.quantity, product.price, brand.brand_name 
        FROM product 
        INNER JOIN brand ON product.brand_id = brand.brand_id";

$result = $connect->query($sql);    
$output = array('data' => array());
if($result->num_rows > 0) { 
    $active = ""; 
    while($row = $result->fetch_array()) {
        $brandName = $row[6];
        $output['data'][]= array(       
            $row[1],         
            $row[2],         
            $row[4],         
            $row[5],       
            $brandName,
        );  
    }
}
$connect->close();
echo json_encode($output);

我还有一个php文件，但这只是用来构建网页。

Answer 1

在那里做什么“连接成功”？那改变了你的反应。

尝试使用php ajax mysql填充数据表

1 个答案: