从复选框只有最后一个值插入的许多值插入数据库中

Question

I need to insert many values from checkbox but only the last value is inserting to database.

HTML CODE:

 <?php

  $conn = mysqli_connect("localhost", "root", "", "fix_in_time");
  $result = mysqli_query($conn, "SELECT * FROM `material` WHERE id > 0");

  while($row = mysqli_fetch_assoc($result)):?>
  <input type="checkbox" value="<?php echo $row['tipo'];?>" name="Tipo" id="Tipo"><label><?php echo $row['tipo'];?></label><br>
  <?php endwhile;?>

There is the PHP CODE:

    $Sala = mysqli_real_escape_string($conn, $_POST['Sala']);
    $Descricao = mysqli_real_escape_string($conn, $_POST['Descricao']);
    $Tipo = mysqli_real_escape_string($conn, $_POST['Tipo']);
    $Data = date("d-m-Y H:i:s", strtotime('-1 hour'));

    if(empty($_POST['Sala']) || empty($_POST['Descricao'])|| 
    empty($_POST['Tipo'])){
    echo"<script language='javascript' type='text/javascript'>alert('Por favor 
    preencha os campos!');window.location.href='../index.php';</script>";
    exit();
  }

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

        if (!empty($_POST['Tipo'])) {

            foreach ((array)$Tipo as $Tipo) {

                $query = "INSERT INTO `relatorios` (Data, Sala, Descricao, Tipo) VALUES ('$Data', '$Sala', '$Descricao', '$Tipo')";

                     mysqli_query($conn, $query);
        }
    }
}

Answer 1

In your html code change input tag to: <input type="checkbox" value="<?php echo $row['tipo'];?>" name="Tipo[]">. If you append tag name with [], then this post variable in php will be an array. I've removed tag id because it should be unique, not sure if you need it, but if you do, then make it unique.

And your php code should look something like this:

if(isset($_POST['submit'])) {
    if (!empty($_POST['Tipo']) && is_array($_POST['Tipo'])) {
        foreach ($_POST['Tipo'] as $Tipo) {
             $TipoEscaped = mysqli_real_escape_string($conn, $Tipo);
             $query = "INSERT INTO `relatorios` (Tipo) VALUES ('$TipoEscaped')";
             $success = mysqli_query($conn, $query);
             if (!$success) {
                 //handle mysql error
             }
        }
    }
}