INNER JOIN with where

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I have the following tables: permissoes_users

  • id_user
  • id_permission

permissions  id  - Permission

I have a session variable where I have the id di user, I need to search the table for the id and those that have the id of the user to display the name of the permission ... How can I do it?

Error testing first response: 

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Notice: Undefined index: permissao in /Applications/XAMPP/xamppfiles/htdocs/DEV WEB/projetoFinalBD/permissao.php on line 51

Complete code: 

<body>
<?php
$id_user = $_GET['id_user']; 
function user_info_gamb ($id_user) {
     $sql = mysql_query("SELECT * FROM usuarios WHERE id_user = '$id_user'");
      return mysql_fetch_assoc($sql);
}
$nome = user_info_gamb($id_user);
//$permissao_pg = 1; include('funcoes/restringirpg.php'); 
?>
<div id="topo">
<?php  echo "Bem - Vindo  " . $nome["nome"]; include('funcoes/menu.php');?>
</div>

<div id="conteudo">
<table class="table-auto table-striped text-center" width="auto" border="0" align="center" >
  <tr>
    <td><strong>PERMISSÕES</strong></td>
  </tr>

  <?php
  function permissao_info ($id_permissao) {
      $sql = mysql_query("SELECT * FROM permissoes WHERE id = '$id_permissao'");
      return mysql_fetch_assoc($sql);
  }
    $aux = 0;

      $permissoes = mysql_query(" SELECT PU.ID_USER, P.PERMISSAO FROM PERMISSOES_USERS PU INNER JOIN PERMISSOES P ON PU.ID_PERMISSAO = P.ID ");
        for ($i=0; $row=mysql_fetch_assoc($permissoes); $i++){$permissao = $row["permissao"];

            echo "
            <tr>
                <td>$permissao</td>
            </tr>
            ";

        }

  ?>

</table>
    
asked by anonymous 19.11.2015 / 12:51

1 answer

5

Try this 

SELECT PU.ID_USER, P.PERMISSAO FROM PERMISSOES_USERS PU
INNER JOIN PERMISSOES P ON PU.ID_PERMISSAO = P.ID
WHERE PU.ID_USER = 1

Regarding the error

  

Notice: Undefined index: permission in / Applications / XAMPP / xamppfiles / htdocs / DEV WEB / projectFinalBD / permissao.php on line 51

This is because $permissao = $row["permissao"] is trying to access the permissao column, but the column name is PERMISSAO (upper case).

    
19.11.2015 / 12:57
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