how to print all the keys of an array in php?

Question

how to print all the keys of an array in php?

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#1 by (-1 votes)

0

<? php

  $NombreMenu="";

  $NombreSubMenu="";

  $Menu = array();

  $consulta = 'SELECT cm.nom_menu, csm.nom_sub FROM c_submenu AS csm
               INNER JOIN c_menu AS cm ON csm.Id_sub = cm.Id_menu ORDER BY cm.nom_menu;';

  $resultMenus = $catalogo->obtenerLista($consulta);

  while ($row2 = mysql_fetch_array($resultMenus)) {     
     if (array_key_exists($row2['nom_menu'], $Menu)) {
        array_push($Menu[$row2['nom_menu']], $row2['nom_sub']);
     }else{
        $NombreMenu=$row2['nom_menu'];
        $Menu[$NombreMenu]= array($row2['nom_sub']);    
     }                          
  }

  echo '<li> <a>'. key($Menu).' </a> <h1>';
  foreach ($Menu as $menus) {
     foreach ($menus as $key => $value) {
         echo "<br>";
         echo "Submenu:".$value;                            
     }
  } 

?>

print this:

Collection

Submenu: Online Catalog
Submenu: Areas and Axes
Submenu: Subareas   Submenu: Axis / Area Types
Submenu: Crosses by Area
Submenu: Axis Crossings
Submenu: Crosses Application by Area
Submenu: Crosses Application by Axis
Submenu: Assignment of applications
Submenu: Goals and Activities   Globales
Submenu: Goals and General Activities
Submenu: Priority
Submenu: Covers
Submenu: Catalog of applications
Submenu: Menus by   Submenu application: Modules by menu
Submenu: Indicators
Submenu: Documentation of processes
Submenu: Blog Submenu: Events
Submenu: News

php array

asked by carlos becerra 20.09.2018 в 18:40

source

1 answer

Return Object C # by AJAX (Json - ASP.NET - MVC 5) Unknown class when using TextInputLayout in MainActivity

score -1 · Answer 1

Try this

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

$conn = new mysqli($servername, $username, $password, $dbname);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

    $consulta = 'SELECT cm.nom_menu AS nom_menu , csm.nom_sub AS nom_sub FROM c_submenu AS csm
                   INNER JOIN c_menu AS cm ON csm.Id_sub = cm.Id_menu ORDER BY cm.nom_menu';

$result = $conn->query($consulta);

if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc()) 
   {
        echo $row["nom_menu"];
        echo $row["nom_sub"];
    }
} else {
    echo "0 results";
}