我正在尝试创建一个动态表来获取数据，我可以编辑任何我想要的值

I can fetch the data but the submit button is not working. I am confused with the update query and storing data in array. Here is the code for fetching and showing data in table.

     `<?php
           while ($rows=mysqli_fetch_assoc($result)){ ?>
               <tr>
               <td align="center">
               <?php $id[]=$rows['id']; ?><?php echo $rows['id']; ?>
               </td>
               <td align="center">  
               <input name="name[]" type="text" id="name" value="<?=$rows['name']; ?>">
               </td>
               <td align="center">
               <input name="lastname[]" type="text" id="lastname" value="<? =$rows['lastname']; ?>">
               </td>
               <td align="center">
               <input name="email[]" type="text" id="email" value="<?=$rows['email']; ?>">
               </td>
               </tr>

`

Here is the code for submit button

if ($_SERVER["REQUEST_METHOD"] == "POST" && $_POST["Submit"] != ""){
    $count = mysqli_num_rows($result);
        for($i=0;$i<$count;$i++){
            $sql2="UPDATE test_mysql SET name='".$_POST["name"][$i]."',lastname='".$_POST["lastname"][$i]."', email='".$_POST["email"][$i]."' WHERE id='$id[$i]'";
            $result1=mysqli_query($con, $sql2);
        }
header("Location: update-multiple-2.php");

}

If you handle POST after outputting form (to set variable $id) and variable $test_mysql is set to existing table in your database and all columns exists and you are only one who can add or remove row (because it will change the $id variable) then it should work.

Maybe this could help:

if(!$result1=mysqli_query($con, $sql2)){
    echo 'Error: '.mysqli_error();
}

You need to debug it somehow:

if ($_SERVER["REQUEST_METHOD"] == "POST" && $_POST["Submit"] != ""){
    $count = mysqli_num_rows($result);
    $error = '';
    for($i=0;$i<$count;$i++){
        $sql2="UPDATE $test_mysql SET name='".$_POST["name"][$i]."',lastname='".$_POST["lastname"][$i]."', email='".$_POST["email"][$i]."' WHERE id='$id[$i]'";
        if(mysqli_query($con, $sql2) === FALSE){
            $error .= 'SQL query failed, SQL: '.$sql2.', Error: '.mysqli_error() . "
";        
        }
    }

    if(!$error){
        header("Location: update-multiple-2.php");
    }else{
        echo $error;
        exit();
    }
}