Show Check saved in Database

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Show Check saved in Database

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I am saving the values of a checkbox in mysql but I need that when I show it if I keep 1 it appears activated but if it keeps 0 it does not appear checked

So I keep it

$id = $_POST['mod_id'];
$name = $_POST["mod_name"];
$code = $_POST["mod_code"];
$active = isset($_POST["mod_active"]) ? 1 : 0;
$user_update = $_SESSION['user_id'];
$date_update = date("Y-m-d H:i:s");

var_dump($code, $name, $user_update, $date_update, $active);

if (isset($_POST['active']) == 1) {

  $sql = "UPDATE way SET code=\"$code\",name=\"$name\", user_update=\"$user_update\", date_update=\"$date_update\" where id = $id";
} else {

  $sql = "UPDATE way SET code=\"$code\",name=\"$name\", active=\"$active\", user_update=\"$user_update\", date_update=\"$date_update\" where id = $id";
}

This is how I want to show it, but it does not work

if (isset($active) && $active == 1) {
  $is_active = "Activo";
}
elseif(isset($active) && $active == 0) {
  $is_active = "Inactivo";
}

mysql html php checkbox

asked by Julián Cordoba 20.07.2018 в 19:34

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1 answer

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score 0 · Answer 1

isset () is used to know if a variable is defined or is null, in your case you want to know if the variable has value 1 or different from 1.

If you need to validate if the variable is defined, first validate with the isset and verify that the value has your variable.

if ($_POST['active'] == 1) {
   $sql = "UPDATE way SET code=\"$code\",name=\"$name\",
   user_update=\"$user_update\", date_update=\"$date_update\" where id = $id";
} else {
  $sql = "UPDATE way SET code=\"$code\",name=\"$name\", active=\"$active\", user_update=\"$user_update\", date_update=\"$date_update\" where id = $id";
}