Php不插入数据

Question

我的代码没有在我的php上插入任何数据，我使用的表单将显示值，但我在更新中的代码不起作用。请帮帮忙，以下是我在php中的代码：

if (isset($_POST['update'])) {
    $landowner_id = $_POST['landowner_id'];
    $firstname = $_POST['firstname'];
    $middlename = $_POST['middlename'];
    $lastname = $_POST['lastname'];
    $municipality = $_POST['municipality'];
    $barangay = $_POST['barnagay'];
    $areacovered = $_POST['areacovered'];
    $sex = $_POST['sex'];

    mysqli_query($db, "UPDATE info SET firstname='$firstname', middlename='$middlename', lastnamename='$lastname', municipality='$municipality', barangay='$barangay', areacovered='$areacovered', sex='$sex' WHERE landowner_id=$landowner_id");
    $_SESSION['message'] = "Address updated!"; 

}

这是我的html

<div class="form-wrapper">
                    <input type="number" id = "check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $landowner_id;?>"  required>
                </div>
                <div class="form-wrapper">
                    <input type="text" id = "check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $firstname;?>"  required>
                </div>
                <div class="form-wrapper">
                    <input type="text"  name="middlename" placeholder="Middle Name" class="input-field"  value="<?php echo $middlename;?>">
                </div>
                <div class="form-wrapper">
                    <input type="text"   name="lastname" placeholder="Last Name" class="input-field"  value="<?php echo $lastname;?>" required>
                </div>
                <div class="form-wrapper">
                    <input type="text"  name="municipality" placeholder="Municipality" class="input-field"   value="<?php echo $municipality;?>" required>
                </div>
                <div class="form-wrapper">
                    <input type="text"  name="barangay" placeholder="Barangay" class="input-field" value="<?php echo $barangay;?>" >
                </div>
                <div class="form-wrapper">
                    <input type="text" id = "check" name="areacovered" placeholder="Area Covered" class="input-field" value="<?php echo $areacovered;?>" required>
                </div>
                <div class="form-wrapper">
                    <input type="text" id = "check" name="sex" placeholder="Sex" class="input-field" value="<?php echo $sex;?>" required>
                    <br>
                    <button class="btn" type="submit" name="update"  >Update</button>
                </div>

Answer 1

我没有看到任何"form“标签。您是否错过了将“表单包装器”封装到标记中？如下所示：

<form action="" method="post">
    <div class="form-wrapper">
        <input type="number" id="check" name="firstname" placeholder="First Name" class="input-field" value="<?php echo $landowner_id; ?>" required>
    </div>

    <!-- Other inputs -->

    <div class="form-wrapper">
        <input type="text" id="check" name="sex" placeholder="Sex" class="input-field" value="<?php echo $sex; ?>" required>
        <br>
        <button class="btn" type="submit" name="update">Update</button>
    </div>
</form>

其他需要考虑的重要事项：

• 从不将来自外部的普通输入发送到数据库，而不对其进行清理！否则，您将接受SQL注入。使用prepare-statements来解决这个问题。
• 而不是mysqli_query，我建议你使用PDO。你可以非常容易地准备语句。在这里您可以看到一个用法示例：https://stackoverflow.com/a/60988740/3454593