警告：mysqli_infected_rows（）要求参数1为mysqli，布尔值在第23行的C:\xampp\htdocs\ajax complete\live-table-post.php中给出

当我试图从数据库中获取显示警告错误的数据时

警告：mysqli\u受影响的\u行希望参数1为mysqli，布尔值在第23行的C:\xampp\htdocs\ajax complete\live-table-post.php中给出

我试图从数据库中获取数据，它显示错误我也使用了musqli\u affedted\u行而不是mysqli\u num\u行，但它显示了相同的错误

<?php
    $host     = "localhost"; $username = "root"; $password = "";$db_name  = "ajax_complete";
    $conn = mysqli_connect( $host, $username, $password, $db_name ) or die("cannot connect");
    $output = '';
    $sql    = "SELECT * FROM tbl_live-crud ORDER BY id DESC";
    $result  = mysqli_query($conn,$sql);
    $output.= '
        <div class="table-responsive">
            <table class="table table-bordered">
                <tr>
                    <td width="10%;"></td>
                    <td width="20%;"></td>
                    <td width="20%;"></td>
                    <td width="20%;"></td>
                    <td width="10%;"></td>
                </tr>';
                if(mysqli_num_rows($result) > 0){
                    while($row = mysqli_fetch_array($result)){
                        $output .= '
                            <td>'.$row['id'].'</td>
                            <td class="fname" data-id1="'.$row['id'].'" contenteditable>'.$row['fname'].'</td>
                            <td class="lname" data-id2="'.$row['id'].'" contenteditable>'.$row['lname'].'</td>
                            <td class="email" data-id3="'.$row['id'].'" contenteditable>'.$row['email'].'</td>
                            <td><button name="delete" id="delete" data-id4="'.$row['id'].'">X</button></td>';
                            }

                        $output = '
                            <tr>
                                <td id="fname" contenteditable></td> 
                                <td id="lname" contenteditable></td> 
                                <td id="email" contenteditable></td> 
                                <td id="btn_add" name="btn_add" class="btn btn-xs btn-success"><button>+</button></td> 
                            </tr>';
                }else{
                    $output = '
                        <tr>
                            <td colaspan="4"> Data not Found</td>
                        </tr>';
                }
                $output .='</table>
                </div>';

?>

试试这个

$host     = "localhost"; 
$username = "root"; 
$password = "";
$db_name  = "ajax_complete";

$conn = mysqli_connect( $host, $username, $password, $db_name );
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$output = '';
$sql    = "SELECT * FROM tbl_live_crud ORDER BY id DESC";
$result = $conn->query($sql);

$output.= '
<div class="table-responsive">
    <table class="table table-bordered">
        <tr>
            <td width="10%;"></td>
            <td width="20%;"></td>
            <td width="20%;"></td>
            <td width="20%;"></td>
            <td width="10%;"></td>
        </tr>';
        if($result->num_rows > 0){
            while($row = $result->fetch_assoc()) {
                $output .= '
                            <td>'.$row['id'].'</td>
                            <td class="fname" data-id1="'.$row['id'].'" contenteditable>'.$row['fname'].'</td>
                            <td class="lname" data-id2="'.$row['id'].'" contenteditable>'.$row['lname'].'</td>
                            <td class="email" data-id3="'.$row['id'].'" contenteditable>'.$row['email'].'</td>
                            <td><button name="delete" id="delete" data-id4="'.$row['id'].'">X</button></td>
                        ';
            }

            $output = '
                    <tr>
                        <td id="fname" contenteditable></td> 
                        <td id="lname" contenteditable></td> 
                        <td id="email" contenteditable></td> 
                        <td id="btn_add" name="btn_add" class="btn btn-xs btn-success"><button>+</button></td> 
                    </tr>
            ';
        }else{
                $output = '
                        <tr>
                            <td colaspan="4"> Data not Found</td>
                        </tr>
                ';
        }
    $output .='</table>
</div>';

---

这两个代码都不工作Source:注意：在第26行的C:\xampp\htdocs\ajax complete\live-table-post.php中尝试获取非对象的属性Yeah table name不应该有-在该行中。因此，请在重命名表后重试。您的表包含-in name tbl_live-crud使用tbl_live_crud重命名它。请参阅编辑的代码。这对我来说很有效。是的，这可以改变，但它的热显示数据和ajax响应状态是200 ok。。。