我相信我如何将信息插入数据库以进行更新的查询中存在错误

Question

该程序应该使用表格来更新数据库

我删除了更新表的查询，然后出现页面，我有一个预设检查器，当我取出要更新的查询时，它会检查是否发布了ID / ISBN参数，该参数将我重定向到错误页面当我添加它时，我被重定向到页面

这是我认为有错误的代码

$sql = "UPDATE books SET ISBN=?,Title=?,PubDate=?,PubID=?,Cost = ?, Retail = ?, Category = ? WHERE books.ISBN='$isbn'";

  if($stmt = $mysqli->prepare($sql)){
        $stmt->bind_param("issiiis", $isbn, $title,$pubdate,$pubid,$cost,$retail,$category);

            // Set parameters
            $ISBN = $isbn;
            $Title = $title;
            $PubDate = $pubdate;
            $PubID = $pubid;
            $Cost = $cost;
            $Retail = $retail; 
            $Category = $category;



            // Attempt to execute the prepared statement
            if($stmt->execute())
            {
                echo "starting query";
                // Records created successfully. Redirect to landing page

                  header("location: index.php");
                exit();
            } else
            {
                echo $stmt->error; //"Something went wrong. Please try again later.";
            }


        }
        // Close statement  
        $stmt->close(); 
    }

在另一个不是这是从数据库获取信息的查询

    // Check existence of student_ID parameter before processing further
    if(isset($_GET["ISBN"]) && !empty(trim($_GET["ISBN"])))
    {
        // Get URL parameter
        $id =  trim($_GET["ISBN"]);

        // Prepare a select statement
        $sql = "SELECT * FROM books WHERE ISBN = ?";

        if($stmt = $mysqli->prepare($sql)){
            // Bind variables to the prepared statement as parameters
            $stmt->bind_param("s", $param_id);

            // Set parameters
            $param_id = $isbn;

            // Attempt to execute the prepared statement
            if($stmt->execute()){
                $result = $stmt->get_result();

                if($result->num_rows == 1){
                    /* Fetch result row as an associative array. Since the result set
                    contains only one row, we don't need to use while loop */
                    $row = $result->fetch_array(MYSQLI_ASSOC);

                    // Retrieve individual field value
                    $ISBN = $row["ISBN"];
                    $Title= $row["Title"];
                    $PubDate = $row["PubDate"];
                    $PubID = $row["PubID"];
                    $Cost = $row["Cost"];
                    $Retail = $row["Retail"];
                    $Category = $row["Category"];

                }
                else
                {
                    // URL doesn't contain valid student_ID. Redirect to error page
                    header("location: error.php");
                    exit();
                }

            }
            else
            {
                echo "Oops! Something went wrong. Please try again later.";
            }
        }

        // Close statement
        $stmt->close();

        // Close connection
        $mysqli->close();
    }  else{
        // URL doesn't contain student_ID parameter. Redirect to error page
        header("location: error.php");
        exit();
} 

我对代码的数量和我的无知深表歉意，但这非常重要

Answer 1

您的类型字符串正确吗？

您正在传递“ issiiis”，因此ISBN作为整数传递。 在where子句中，尽管您将其视为字符串，但考虑到破折号和可能的前导零，它似乎更有可能？