Php不插入数据

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，但我的更新代码不起作用。请帮忙， 以下是我的php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>

我没有看到任何“表单”标签。是否缺少将“表单包装器”包装到标记中？大概是这样的：

<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>


我没有看到任何“表单”标签。是否缺少将“表单包装器”包装到标记中？大概是这样的：

<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>


您要做的第一件事是添加一个调试语句，如var_dump（$\u POST），您可以查看$\u POST['update']是否工作。因此，您要问的问题是“代码是否进入其中…”您要做的第一件事是添加一个调试语句，如var_dump（$\u POST），您可以查看$\u POST['update']是否工作。所以你要问的问题是“代码是否在那里…”