Question

我想使用php在数据库表中保存输入选择选项。我想将它保存在具有现有属性的表中我知道如何在使用单独的表时添加它，但我想在现有的表中添加它，只有很少的其他属性

ps第一个问题是熊我

<?php 

if(isset($_POST['signup'])) {

    $name = $_POST['name'];
    $email = $_POST['email'];
    $phone = $_POST['phone'];
    $password = $_POST['password'];
    $department = $_POST['deptlist'];

    echo $name;

    $connection = mysqli_connect('localhost', 'root', '', 'filetracking');    

    if($connection) {
        echo "We are connected";
    } else {
        die("Database connection failed");
    }

    $query = "INSERT INTO faculty(name,email,contactno,password,office) ";
    $query .= "VALUES ('$name', '$email', '$phone', '$password', '$department')";

    $result = mysqli_query($connection, $query);
    if(!$result) {
        die('Query FAILED' . mysqli_error($connection));
    } else {
        echo "Record Create"; 
    }
}

?>

<form method="post" action="signup.php">
    <div class="form-group">
        <label for="name" class="control-label">Name</label>
        <input type="text" name="name" id="Name" class="inputmd textinput form-control required" placeholder="Full Name" style="margin-bottom: 10px" required="true">
    </div>

    <div class="form-group">
        <label for="email" class="control-label">Email</label>
        <input type="email" name="email" id="Email" class="inputmd textinput form-control required" placeholder="Email" style="margin-bottom: 10px" required="true">
    </div>

    <div class="form-group">
        <label for="phone" class="control-label">Phone</label>
        <input type="number" name="phone" id="Phone" class="inputmd textinput form-control required" placeholder="Phone No" style="margin-bottom: 10px" required="true">
    </div>

    <div class="form-group">
        <label for="password" class="control-label">Password</label>
        <input type="password" name="password" id="Password" class="inputmd textinput form-control" placeholder="Password" style="margin-bottom: 10px" required="true">
    </div>

    <div class="form-group">                        
        <label for="deptlist" class="control-label">Department</label>
        <select name="deptlist" form="deptform" class="inputmd form-control" style="margin-bottom: 10px">
            <option value="DCSE">Department of Computer Systems Engineering</option>
            <option value="DME">Department of Mechanical Engineering</option>
            <option value="DEE">Department of Electrical Engineering</option>
            <option value="DCS">Department of Computer Science</option>
        </select>

    <div class="form-group">                    
        <input type="submit" name="signup" id="Singup" value="Sign Up" class="btn btn-info col-md-4 col-md-offset-4">
    </div>

</div>

Answer 1

您错误地提到了名为deptform <select name="deptlist" form="deptform"的表单，但没有这样的表单。这可以防止脚本接收此数据。

我想使用php保存数据库表中的输入选择选项

1 个答案: