Question

我正在尝试查询mysql并获取要返回的结果，但是在尝试检查返回的结果数是否大于零时遇到了以下错误。这是我的代码：

    <body>

            <?php
                $servername = "localhost";
                $username = "root";
                $password = "";
                $dbname = "cd";

                 // Create connection
                $conn = new mysqli($servername, $username, $password, $dbname);
                // Check connection
                if ($conn->connect_error) {
                    die("Connection failed: " . $conn->connect_error);
                } 
                /*

                $sql = "SELECT * FROM cdlist WHERE cdlist.title =" . $_POST["cd_name"]." OR cdlist.Artist=" .$_POST["artist_name"]. " OR cdlist.Price=" . $_POST["cd_price"] . " OR cdlist.Year=" . $_POST["year"];
                */
                $sql = "SELECT * FROM cdlist WHERE cdlist.Title=" . $_POST["cd_name"];

                //$sql = "SELECT * FROM cdlist";
                $result = mysqli_query($conn, $sql);
                $json_array = array();
                if (mysqli_num_rows($result)>0){
                    while($row = mysqli_fetch_assoc($result)){
                        $json_array[] = $row;
                    }

                    echo json_encode($json_array);
                }
                $conn->close();
            ?>

    </body>

php中的任何查询是否存在问题？我只能使它与SELECT * FROM cdlist一起使用，但这不是我要使用的查询。我知道cd_name也是变量的正确名称。感谢您能为我提供的任何帮助

收到错误-警告：mysqli_num_rows（）期望参数1为mysqli_result

0 个答案: