Questions about Notice Undefined variable

Question

what this warning means Notice: Undefined variable: login in C: xampp htdocs Techphp profile.php on line 7 detail the login exist

if(isset($_POST['login'])){ 
$login = $_POST['login']; 
} else { 
echo "Login Vazio."; 
} 

$sql = mysql_query( "SELECT * FROM usuario WHERE login='{$login}' "); 
$linha = mysql_fetch_array($sql); 

Answer 1

I would do it this way:

if(isset($_POST['login'])){ 
    $login = $_POST['login'];
    $sql = mysql_query( "SELECT * FROM usuario WHERE login='{$login}' ") or print mysql_error(); 
    $linha = mysql_fetch_array($sql);  
} else { 
    echo "Login Vazio."; 
} 

This way we will be able to know if the login really exists and we will also be able to know if there is any mysql error in the query.

Based on this, we will know where to fix the error, if there is still.

Answer 2

The error occurs because you are probably sending a form without method="POST" or the attribute name="" in the input is wrong.

To run the <form> must be something like:

<form method="POST" action="perfil.php">
   <input type="text" name="login" value="">
   <input type="submit" value="Buscar">
</form>

However by your query I suspect that you are not using a form but a GET method, probably a link with <a href="perfil.php?login=usuario">

Links do not use POST method, they use GET method, so the correct is this:

$login = NULL; //Declaramos a variavel

if(false === empty($_GET['login'])){//Se existir define em login e não for vazia
    $login = $_GET['login'];
} else {//Se não existir ou for vazia emite um erro
    echo "Login Vazio."; 
}

if ($login) {//Se a variável existir executa a query
    $sql = mysql_query( "SELECT * FROM usuario WHERE login='{$login}'") or print mysql_error(); 
    $linha = mysql_fetch_array($sql);
}