Question

在我的代码中，我试图使代码在数据库中的两个表之间切换，具体取决于用户从下拉列表中选择的内容。如果用户选择“学生”，那就绝对没问题，并且可以毫无问题地从CSV上传数据。如果用户选择“Staff”它不起作用，但在仔细检查数据库和变量中的名称，尝试不同的CSV文件，并重新编写代码后，我真的想不起为什么。这是我的代码：

<?php

if(ISSET($_POST['submit'])) {

    $hostname = "localhost";
    $db_username = "root";
    $db_password = "";
    $database = "project_cherrypicker";

    $db_connection = mysqli_connect($hostname, $db_username, $db_password) or die ("Unable to connect to database." . mysqli_error($db_connection));

    mysqli_select_db($db_connection, $database) or die ("Unable to connect to database." . mysqli_error($db_connection));

    $db_choice = $_POST['database_choice'];

    $file = $_FILES['file']['tmp_name'];
    $handle = fopen($file, "r");

    $count = 0;

    $skippedfirstline = false;

    while(($csvdata = fgetcsv($handle, 1000, ",")) !== FALSE) {

        if(!$skippedfirstline) {$skippedfirstline = true; continue;}

        switch($db_choice) {

            case "students":

                $student_id = $csvdata[0];
                $forename = $csvdata[1];
                $surname = $csvdata[2];
                $dob = $csvdata[3];
                $course = $csvdata[4];
                $department = $csvdata[5];
                $campus = $csvdata[6];
                $level = $csvdata[7];
                $tutor = $csvdata[8];

                $sql = "INSERT INTO students (student_id, forename, surname, dob, course, department, campus, level, tutor)
                VALUES ('$student_id', '$forename', '$surname', '$dob', '$course', '$department', '$campus', '$level', '$tutor')";
                break;

            case "staff":

                $staff_id = $csvdata[0];
                $forename = $csvdata[1];
                $surname = $csvdata[2];
                $course = $csvdata[3];
                $department = $csvdata[4];
                $number_of_students = $csvdata[5];
                $campus = $csvdata[6];
                $level = $csvdata[7];
                $a_cherries = $csvdata[8];
                $b_cherries = $csvdata[9];
                $c_cherries = $csvdata[10];

                $sql = "INSERT INTO staff (staff_id, forename, surname, course, department, number_of_students, campus, level, a_cherries_complete, b_cherries_complete, c_cherries_complete)
                VALUES ('$staff_id', '$forename', '$surname', '$course', '$department', '$number_of_students', $campus', '$level', '$a_cherries', '$b_cherries', $c_cherries')";
                break;

            }

        $query = mysqli_query($db_connection, $sql);

        $count = $count + 1;

    }

    if($query) {

        echo "<script>alert('Data uploaded succesfully.')</script>";

    }

    else {

        echo "<script>alert('An error has occured. Data not uploaded.')</script>";

    }

}

＆GT;

<head>

    <title>Project Cherrypicker</title>

    <!-- Bootstrap & CSS -->
    <link rel = "stylesheet" href = "https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css">
    <link rel = "stylesheet" href = "../CSS/home-page.css">
    <link rel = "stylesheet" href = "../CSS/sidebar.css">

</head>

<body>

    <div class = "row reduce-margin">

        <!-- Sidebar -->
        <div class = "col-md-2">

            <div class = "container-fluid">

                <div class = "sidebar">

                    <h4>PROJECT CHERRYPICKER</h4>

                    <!-- Search Box -->
                    <input type = "text" name = "menu-search" class = "search-bar" placeholder = " Search Databases">

                    <!-- Divider -->
                    <hr>

                    <!-- Navigation Buttons -->
                    <a href = "home-page.php"><button name = "btnHome" class = "menu-button">Home</button></a>
                    <hr>
                    <a href = "students-database.php"><button name = "btnStudents" class = "menu-button">Students</button></a>
                    <a href = "staff-database.php"><button name = "btnStaff" class = "menu-button">Staff</button></a>
                    <button name = "btDepartments" class = "menu-button">Departments</button>
                    <button name = "btnUpload" class = "menu-button">Upload</button>
                    <hr>
                    <button name = "btnHelp" class = "menu-button">Log Out</button>

                </div>

            </div>

        </div>

        <!-- Main Content -->
        <div class = "col-md-10">

            <div class = "container-fluid margin">

                <!-- Banner Display -->
                <div class = "jumbotron">

                    <h4>UPLOAD .CSV FILE</h4>
                    <p> Select the table you wish to update and upload the .csv file.</p>

                    <form name = "upload-form" method = "POST" role = "form" enctype = "multipart/form-data"> <!-- enctype needed to upload files -->
                        <div>
                            <select name = "database_choice">
                                <option value = "students">Students</option>
                                <option value = "staff">Staff</option>
                            </select>
                        </div>

                        <br>

                        <div>
                            <input type = "file" name = "file" id = "file" size = "150"/>
                            <span><input type = "submit" name = "submit" value = "Submit" /></span>
                        </div>
                    </form>

                </div>

            </div>

        </div>

    </div>

</body>

Answer 1

$sql案例的staff变量缺少$campus和$c_cherries的几个引号。

更改

$sql = "INSERT INTO staff (staff_id, forename, surname, course, department, number_of_students, campus, level, a_cherries_complete, b_cherries_complete, c_cherries_complete)
                VALUES ('$staff_id', '$forename', '$surname', '$course', '$department', '$number_of_students', $campus', '$level', '$a_cherries', '$b_cherries', $c_cherries')";
                break;

到

$sql = "INSERT INTO staff (staff_id, forename, surname, course, department, number_of_students, campus, level, a_cherries_complete, b_cherries_complete, c_cherries_complete)
                VALUES ('$staff_id', '$forename', '$surname', '$course', '$department', '$number_of_students', '$campus', '$level', '$a_cherries', '$b_cherries', '$c_cherries')";
                break;

另外，我建议你使用准备好的陈述。您的用户通过电子表格上传这一事实也不会使其安全。

请阅读manual如何实施准备好的陈述

将CSV上传到数据库错误

1 个答案: