Question

好吧，我一直在努力，存储用户在输入文本中写入的内容，并存储用户点击的复选框的ID。我实现了存储两个东西，但每一个分开，更好的说在其他PHP文件。当我尝试仅使用一个php文件来处理这两件事时出现问题。这是我的代码

HTML

<label for="Teachers_Name">Teacher's Name</label>
<input type="text" name="Teachers_Name" id="Teachers_Name" placeholder="Name" required>
<br>
<br>
<label for="School_Name">School Name</label>
<input type="text" name="School_Name" id="School_Name" placeholder="School Name" required>
<br>
<br>
<label for="Implementation_Quality">Write here your 
Implementation Quality score :</label>
<input type="text" name="Implementation_Quality" id="Implementation_Quality" placeholder="Implementation Quality" required>
</div>
<input type="submit" name="submit" id="submit" value="Send" 
    style="margin-left: 50px; margin-bottom: 20px; margin-top: 
    20px;background:#0774D9; color: #fff; font-size: 20px; 
    border-radius: 10px; font-family: 'Verdana';padding: 5px 
    15px 5px 20px">
<p>BOOKS</p>
<div class="checkbox">
    <br>
    <input type="checkbox" value="20" id="CourseBooks" class="get_value" style="display: none;">
    <label for="CourseBooks">CourseBooks</label>
    <input type="checkbox" value="20" id="PracticeBooks" class="get_value"  style="display: none;">
    <label for="PracticeBooks">PracticeBooks</label>
    <br>
</div>
<h4 id="result"></h4>

Jquery的

<script>
    $(document).ready(function() {
        $('#submit').click(function() {
            var insert = [];
            $('.get_value').each(function() {
                if ($(this).is(":checked")) {
                    insert.push($(this).attr("id"));
                }
            });
            insert = insert.toString();
            $.ajax({
                url: "insert1.php",
                method: "POST",
                data: {
                    insert: insert
                },
                success: function(data) {
                    $('#result').html(data);
                }
            });
        });
    });
</script>

复选框的Php

if(isset($_POST['insert'])) {
    $conn = mysqli_connect("localhost", "root", "", "datos1");
    $query = "INSERT INTO BOOKS(name) VALUES('" . $_POST["insert"] . "') ";
    $result = mysqli_query($conn, $query);
    echo "Data Inserted Succesfully";
}

PHP文本

$conexion = new mysqli("localhost", "root", "");

if(!$conexion) {
    echo "Conexión no exitosa";
} else {

    $base = mysqli_select_db($conexion, "datos1");
    if(!$base) {
        echo "No se pudo conectar a la base de datos";
    }
}
//LLAMAMOS LAS VARIBALES
$Teachers_Name = "";
$Teachers_Name = isset($_POST['Teachers_Name']) ? $_POST['Teachers_Name'] : '';
$Teachers_Name = empty($_POST['Teachers_Name']) ? $_POST['Teachers_Name'] : '';
$Teachers_Name= $_POST['Teachers_Name'] ?? '';
$School_Name = "";
$School_Name = isset($_POST['School_Name']) ? $_POST['School_Name'] : '';
$School_Name = empty($_POST['School_Name']) ? $_POST['School_Name'] : '';
$School_Name= $_POST['School_Name'] ?? '';
$Implementation_Quality = "";
$Implementation_Quality = isset($_POST['Implementation_Quality']) ? $_POST['Implementation_Quality'] : '';
$Implementation_Quality = empty($_POST['Implementation_Quality']) ? $_POST['Implementation_Quality'] : '';
$Implementation_Quality= $_POST['Implementation_Quality'] ?? '';
//Guarda las variables
$sql = "INSERT INTO datos_1 (Teachers_Name, School_Name, 
     Implementation_Quality) VALUES(?, ?, ?)";

$sth = mysqli_prepare($conexion, $sql);
mysqli_stmt_bind_param($sth, 'sss', $Teachers_Name, $School_Name, $Implementation_Quality);
$ejecutar = mysqli_stmt_execute($sth);

if(!$ejecutar) {
    echo "Hubo algun error";
} else {
    echo "Datos guardados correctamente<br><a href='index.php'>Volver</a>";
}

为什么我的代码不会将变量存储在mySQL中？

0 个答案: