Php 为什么我的表单数据不使用input type="；插入我的数据库；隐藏的；？

这是我的数据库连接，表单页面，一致页面。我将尝试在不使用任何形式的情况下将数据插入数据库。有人请检查我的密码

db.php

<?php
$connect=mysqli_connect("localhost", "root", "", "test_form");
if(!$connect){
    die('Can not connect:'.mysqli_error());
}
else{
    echo "Connect";
}

?>

<?php
include_once 'db.php';
?>

<form action="process.php" method="post">
    <input type="text" name="f-name" placeholder="f-name"><br><br>
    <input type="text" name="l-name" placeholder="l-name"><br><br>
    <input type="text" name="bday" placeholder="bday"><br><br>
    <input type="text" name="school" placeholder="school"><br><br>
    <label><input type="radio" name="gender" value="m">Male</label>
    <label><input type="radio" name="gender" value="f">Female</label><br><br>
    <textarea name="hobby"></textarea><br><br>
    <input type="submit" value="Submit">
</form>

<?php
include_once 'db.php';
?>

<?php
$fname=$_POST['f-name'];
$lname=$_POST['l-name'];
$bday=$_POST['bday'];
$school=$_POST['school'];
$gender=$_POST['gender'];
$hobby=$_POST['hobby'];

echo "$fname"."<br>";
echo "$lname"."<br>";
echo "$bday"."<br>";
echo "$school"."<br>";
echo "$gender"."<br>";
echo "$hobby"."<br>";
?>
<button name="Sava" type="button" value="Save">Save</button>
<?php
if(isset($_REQUEST['save'])){
    $sql="INSERT INTO test_table(f_name, l_name, b_day, school, gender, hobby) VALUES ('$fname', '$lname', '$bday', '$school', '$gender', '$hobby')";
}

$query=mysqli_query($connect, $sql);
if($query){
    $msg="Successfully Inserted...";
}else{
    $msg="Not Inserted...";
}
?>
<h1><?php echo $msg; ?></h1>

index.php

<?php
$connect=mysqli_connect("localhost", "root", "", "test_form");
if(!$connect){
    die('Can not connect:'.mysqli_error());
}
else{
    echo "Connect";
}

?>

<?php
include_once 'db.php';
?>

<form action="process.php" method="post">
    <input type="text" name="f-name" placeholder="f-name"><br><br>
    <input type="text" name="l-name" placeholder="l-name"><br><br>
    <input type="text" name="bday" placeholder="bday"><br><br>
    <input type="text" name="school" placeholder="school"><br><br>
    <label><input type="radio" name="gender" value="m">Male</label>
    <label><input type="radio" name="gender" value="f">Female</label><br><br>
    <textarea name="hobby"></textarea><br><br>
    <input type="submit" value="Submit">
</form>

<?php
include_once 'db.php';
?>

<?php
$fname=$_POST['f-name'];
$lname=$_POST['l-name'];
$bday=$_POST['bday'];
$school=$_POST['school'];
$gender=$_POST['gender'];
$hobby=$_POST['hobby'];

echo "$fname"."<br>";
echo "$lname"."<br>";
echo "$bday"."<br>";
echo "$school"."<br>";
echo "$gender"."<br>";
echo "$hobby"."<br>";
?>
<button name="Sava" type="button" value="Save">Save</button>
<?php
if(isset($_REQUEST['save'])){
    $sql="INSERT INTO test_table(f_name, l_name, b_day, school, gender, hobby) VALUES ('$fname', '$lname', '$bday', '$school', '$gender', '$hobby')";
}

$query=mysqli_query($connect, $sql);
if($query){
    $msg="Successfully Inserted...";
}else{
    $msg="Not Inserted...";
}
?>
<h1><?php echo $msg; ?></h1>

男性
女性

所有数据都显示在process.php页面中，但在此页面中单击Save btn，它不会插入到数据库中

process.php

<?php
$connect=mysqli_connect("localhost", "root", "", "test_form");
if(!$connect){
    die('Can not connect:'.mysqli_error());
}
else{
    echo "Connect";
}

?>

<?php
include_once 'db.php';
?>

<form action="process.php" method="post">
    <input type="text" name="f-name" placeholder="f-name"><br><br>
    <input type="text" name="l-name" placeholder="l-name"><br><br>
    <input type="text" name="bday" placeholder="bday"><br><br>
    <input type="text" name="school" placeholder="school"><br><br>
    <label><input type="radio" name="gender" value="m">Male</label>
    <label><input type="radio" name="gender" value="f">Female</label><br><br>
    <textarea name="hobby"></textarea><br><br>
    <input type="submit" value="Submit">
</form>

<?php
include_once 'db.php';
?>

<?php
$fname=$_POST['f-name'];
$lname=$_POST['l-name'];
$bday=$_POST['bday'];
$school=$_POST['school'];
$gender=$_POST['gender'];
$hobby=$_POST['hobby'];

echo "$fname"."<br>";
echo "$lname"."<br>";
echo "$bday"."<br>";
echo "$school"."<br>";
echo "$gender"."<br>";
echo "$hobby"."<br>";
?>
<button name="Sava" type="button" value="Save">Save</button>
<?php
if(isset($_REQUEST['save'])){
    $sql="INSERT INTO test_table(f_name, l_name, b_day, school, gender, hobby) VALUES ('$fname', '$lname', '$bday', '$school', '$gender', '$hobby')";
}

$query=mysqli_query($connect, $sql);
if($query){
    $msg="Successfully Inserted...";
}else{
    $msg="Not Inserted...";
}
?>
<h1><?php echo $msg; ?></h1>

拯救
单击“保存”按钮后，页面不会重新加载。没有表单或JavaScript来完成这项工作

然而，即使您设法重新加载页面，表单中以前的
$\u POST
对于重新加载的页面也不再有效

此外，您还有一个输入错误：

if(isset($_REQUEST['save'])){ ...

但是

保存

save
应该是
Sava
您可以通过检查
isset（$\u POST['submit']）插入

或者使用
isset（$\u POST['what\u ever\u submit\u button']）


提交表单后为什么要使用按钮？您是否试图让用户查看其详细信息？如果不是这种情况，您可以直接使用Post变量来验证DB值。