PHP Query Error

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PHP Query Error

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#1 by (0 votes)
#2 by (-3 votes)

-1

I'm trying to make a file in php, which refers to the age of the person and then shows, however, the $ login appears to me but the person's age does not. I would like to know what I am doing wrong.

<?php
    include ("sistema/validar_session.php");
    include("config.php");

    $login = "Goncalo";
    $select = mysql_query("SELECT * FROM dados_usuarios WHERE Login = '$login'");
    $consulta = mysql_fetch_array($select);
    $idade = $consulta['idade'];

    echo "Olá $login a sua idade é de $idade";
 ?>

Here's the table structure :

php

asked by anonymous 18.06.2015 / 05:54

2 answers

What attribute do you use for database fields that work with numbers from 0 to 5? Doubt with foreach [closed]

score 0 · Answer 1

Hello, your query is probably not bringing results.

<?php 
include ("sistema/validar_session.php"); 
include("config.php");

$login = "Goncalo";


$select = mysql_query("SELECT * FROM dados_usuarios WHERE Login = '$login'");

if( mysql_num_rows ( $select ) == 0 )
   echo 'O login '.$login.' não foi encontrado';
else
{
   while ($row = mysql_fetch_assoc($select))
     echo "Olá ".$login." a sua idade é de ".$row['idade'];
}
?>

score -3 · Answer 2

The best thing would be to place:

<?php include ("sistema/validar_session.php"); include("config.php");

$login = "Goncalo";


$sql= mysql_query("SELECT * FROM dados_usuarios WHERE Login = '$login'");

$row= mysql_fetch_array($sql);

$idade = $row['idade'];

echo "Olá $login a sua idade é de $idade"; ?>