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php - 如何在mysql表中插入html表值

问题描述

使用以下代码，我已从 html 表中的 mysql 表中获取Student ID和获取。Student name在第三列中，Obtained Marks我正在通过输入框获取学生分数。现在我想在新表 testrecord 中插入所有三列（学生 ID、学生姓名和获得的分数）。

<html>
<body>
<?php
$connection = mysqli_connect ('localhost', 'admin', 'password', 'db');
if (!$connection) {
    echo 'Not connected to server';
}

$select_db = mysqli_select_db($connection, 'db');
if (!$select_db) {
    echo 'Not connected to database';
}

$SelectClass = $_POST ['selectclass'];
$sql= "SELECT * FROM students WHERE class = '$SelectClass'";
$query = mysqli_query($connection, $sql);
if (!$query) {
    die ('SQL Error: ' . mysqli_error($connection));
}

mysqli_close($connection);
?>
 <body>
    <div class="container">
    <form class="well form-horizontal" action="insert_marks.php" method="post">
    <h1><strong>Please enter marks of each student for subject</strong></h1>
    <form action="" method="post">
    <table id = "result" class="data-table">
        <caption class="title"></caption>
        <thead>
            <tr>    
                <th>Sr.No.</th>
                <th>Student ID</th>
                <th>Student Name</th>
                <th>Marks Obtained</th>
            </tr>
        </thead>
        <tbody>
        <?php
        $no     = 1;
        $total  = 0;
        while ($row = mysqli_fetch_array($query)) {
            $stu  = $row['stu_id'] == 0 ? '' : number_format($row['stu_id']);
            echo '<tr>
                    <td>'.$no.'</td>
                    <td>'.$row['student_id'].'</td>
                    <input type="hidden" name="student_id" value='.$row['student_id'].'>
                    <td>'.$row['student_name'].'</td>
                    <input type="hidden" name="student_name" value='.$row['student_name'].'>
                    <td>
                        <div class="search-block clearfix">
                            <input name="obtmarks" placeholder="" type="number">
                        </div>
                    </td>
                </tr>';
            $total += $row['stu_id'];
            $no++;
        }
        ?>
        </tbody>
    </table>
    <button type="submit" class="btn btn-warning" value="insert" align="right">Update<span class="glyphicon glyphicon-send"></span></button>
    </form>
</div>
</body>
</html>

插入标记.php：

<html>
<body>
<?php
$connection = mysqli_connect ('localhost', 'admin', 'password', 'db');
if (!$connection) {
    echo 'Not connected to server';
}

$select_db = mysqli_select_db($connection, 'db');
if (!$select_db) {
    echo 'Not connected to database';
}
//***********Form Submit Goes Here***********//
while 
if($_POST) {
    $student_id     =   $_POST['student_id'];
    $student_name   =   $_POST['student_name'];
    $student_marks  =   $_POST['obtmarks'];

    $sql= "INSERT INTO testrecord (student_id,student_name,obtained_marks) VALUES ('$student_id','$student_name','$student_marks')";
    if (mysqli_query($connection, $sql)) {
    echo "Marks added successfully.";
    echo "<br>";
    echo "<br>";
    } else {
        echo "Error: " . $sql . "<br>" . mysqli_error($connection);
    }
}
mysqli_close($connection);
?>
</body>
</html>

表中有 20 个条目。在文本框中为每个学生插入标记后，上面的编码仅在 mysql 表“testrecord”中插入最后一条记录。您能否更正 insert_marks.php 代码。

标签： php

解决方案

Html 表格 tbody 部分的更改：

<tbody>
    <?php
    $no     = 1;
    $total  = 0;
    while ($row = mysqli_fetch_array($query)) {
        $stu  = $row['stu_id'] == 0 ? '' : number_format($row['stu_id']);
        echo '<tr>
                <td>'.$no.'</td>
                <td>'.$row['student_id'].'</td>
                <input type="hidden" name="student_id[]" value='.$row['student_id'].'>
                <td>'.$row['student_name'].'</td>
                <input type="hidden" name="student_name[]" value='.$row['student_name'].'>
                <td>
                    <div class="search-block clearfix">
                        <input name="obtmarks[]" placeholder="" type="number">
                    </div>
                </td>
            </tr>';
        $total += $row['stu_id'];
        $no++;
    }
    ?>
</tbody>

在insert_marks.php变化的同时部分：

//***********Form Submit Goes Here***********//
while 
if($_POST) {
    $student_id     =   $_POST['student_id'];
    $student_name   =   $_POST['student_name'];
    $student_marks  =   $_POST['obtmarks'];

    for($i = 0; $i < count($student_id); $i++){
        $sql= "INSERT INTO testrecord (student_id,student_name,obtained_marks) VALUES ('$student_id[$i]','$student_name[$i]','$student_marks[$i]')";
        if (mysqli_query($connection, $sql)) {
             echo "Marks added successfully.";
             echo "<br>";
             echo "<br>";
        } else {
             echo "Error: " . $sql . "<br>" . mysqli_error($connection);
        }
    }


}

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