HTML form sends the wrong database ID PHP - php

Question

I made a user system for my website. If you open it you well see a list of all users and you can lock or unlock them by changing "n_gesperrt" in the database to 1 or 0. 1 means locked.
The form you see here is always sending the last user id (n_id) to the php script on the top. So if I want to change something it only works with the last user shown on the list.
How can I solve that problem?
These are the code pieces:
THE FORM:
while ($zeile = mysqli_fetch_assoc($res))
{
echo "<form name=\"nutzer_sperren\" method=\"POST\" action=\"index.php?page=nutzer_sperren\">";
echo "<input type=\"hidden\" name=\"n_id\" value=\"".$zeile['n_id']."\">";
echo $zeile['n_id'];
if($zeile['n_gesperrt'] == 0){
echo "<input type =\"submit\" name=\"sperren\" value=\"Nutzer sperren\">";
}
else {
echo "<input type =\"submit\" name=\"entsperren\" value=\"Nutzer entsperren\">";
}
echo '</form">';
THE EDITING IN DATABASE:
if(isset($_POST['sperren']))
{
$sperrung = mysqli_real_escape_string($verbinde, 1);
mysqli_query($verbinde, 'UPDATE nutzer SET n_gesperrt ="'.$sperrung.'" WHERE n_id ='.$_POST["n_id"].'') or die(mysqli_error($verbinde));
echo "Daten wurden erfolgreich geändert. <br>Bitte aktualisieren Sie die Seite!";
echo $_POST["n_id"];
echo "<hr>";
}
if(isset($_POST['entsperren']))
{
$entsperren = mysqli_real_escape_string($verbinde, 0);
mysqli_query($verbinde, 'UPDATE nutzer SET n_gesperrt ="'.$entsperren.'" WHERE n_id ='.$_POST["n_id"].'') or die(mysqli_error($verbinde));
echo "Daten wurden erfolgreich geändert. <br>Bitte aktualisieren Sie die Seite!";
echo $_POST["n_id"];
echo "<hr>";
}

Answer 1

There are few things you need to change in your code, such as:
... The form you see here is always sending the last user id (n_id) to the php script on the top. So if I want to change something it only works with the last user shown on the list.
That's because you've assigned the same name attribute for all of your hidden input elements. Change your name attribute in the following way,
... name='n_id[]' ...
Change the name attribute of submit elements to just submit, because in this way you don't have to use two separate if blocks to handle user's response.
Remove the double quote from echo '</form">';, otherwise your form would break.
So your while() loop should be like this:
while ($zeile = mysqli_fetch_assoc($res)){
echo "<form method='POST' action='index.php?page=nutzer_sperren'>";
echo "<input type='hidden' name='n_id[]' value='".$zeile['n_id']."'>";
if($zeile['n_gesperrt'] == 0){
echo "<input type ='submit' name='submit' value='Nutzer sperren'>";
}
else {
echo "<input type ='submit' name='submit' value='Nutzer entsperren'>";
}
echo '</form>';
}
Furthermore, as I said above, you don't have to use two separate if blocks, just use one if block to check whether the form is submitted or not and toggle the n_gesperrt column value in your UPDATE operation, like this:
if(isset($_POST['submit'])){
mysqli_query($verbinde, 'UPDATE nutzer SET n_gesperrt = IF(n_gesperrt = 1, 0, 1) WHERE n_id ='.$_POST["n_id"][0]) or die(mysqli_error($verbinde));
echo "Daten wurden erfolgreich geändert. <br>Bitte aktualisieren Sie die Seite!";
}

Answer 2

bro use name as array
like below and you will get your result in array
while ($zeile = mysqli_fetch_assoc($res))
{
echo "<form name=\"nutzer_sperren\" method=\"POST\" action=\"index.php?page=nutzer_sperren\">";
echo "<input type=\"hidden\" name=\"n_id[]\" value=\"".$zeile['n_id']."\">";
echo $zeile['n_id'];
if($zeile['n_gesperrt'] == 0){
echo "<input type =\"submit\" name=\"sperren[]\" value=\"Nutzer sperren\">";
}
else {
echo "<input type =\"submit\" name=\"entsperren[]\" value=\"Nutzer entsperren\">";
}
echo '</form">';
on submission page get value like
$name=$_POST['entsperren'];
print_r($name);