mysql_num_rows error

[pedromsouza]

hi!

 

I need a query that checks if typed login is new or already in use... when I run it...

[27-Jan-2011 20:13:59] PHP Warning:  mysql_num_rows() expects parameter 1 to be resource, null given in /Applications/MAMP/htdocs/recebe_dados_cadastro_usuario.php on line 30

 

Please help me (newbie)!

 

<?php

include "include_conecta_bd.inc";

$nome = $_POST["nome"];

$cpf = $_POST["cpf"];

$cnpj = $_POST["cnpj"];

$login = $_POST["login"];

$email = $_POST["email"];

$ddd = $_POST["ddd"];

$tel = $_POST["tel"];

$sexo = $_POST["sexo"];

$senha1 = $_POST["senha1"];

$senha2 = $_POST["senha2"];

$erro=0;

if(empty($nome))

{echo "Favor digitar um nome.<br>"; $erro=1;}

elseif(strlen($nome)<15)

{echo "Favor digitar nome completo.<br>"; $erro=1;}

elseif(empty($login) or strlen($login)<5)

{echo "O login precisa ter 5 caracteres no mínimo.<br>"; $erro=1;}

elseif((($senha1)!==($senha2)) or (strlen($senha2)<2))

{echo "As senhas não são idênticas.<br>"; $erro=1;}

else

{echo "Dados digitados corretamente!<br>";}

{echo "<p>NOME:$nome<br>CPF:$cpf<br>LOGIN:$login<br>EMAIL:$email<br>TEL:$ddd.$tel<br>SEXO:$sexo<br></p>";}

 

$verif = mysql_query ("select * from usr where login = $login");

$chk = mysql_num_rows ($verif);

if($chk != 0)

{break; echo "Este LOGIN já está em uso!<br>"; }

elseif($verif == 0)

{$resultado = mysql_query ("insert into usr (nome,cpf,login,sexo,email,ddd,tel,senha)

values ('$nome','$cpf','$login','$sexo','$email','$ddd','$tel','$senha1')") or

die(mysql_error());

echo mysql_error();

$linhas = mysql_affected_rows ();

mysql_close ($conexao);

echo "$linhas linha(s) afetada(s)!";}

?>

[l4nc3r]

As the error implies, $verif is not a MySQL resource as expected, but is instead null.  Try printing mysql_error() after defining $verif and see what happens.

[jcbones]

Your query is failing.  You need to find out why.  Try echo'ing the query, or just echo mysql_error on failure.

$sql = "select * from usr where login = $login";
$verif = mysql_query ($sql) or die(mysql_error() . ' <br /> in ' . $sql);
$chk = mysql_num_rows ($verif);

 

[litebearer]

I believe it is due to the variable $login NOT being encased by single quotes

[jcbones]

I thought so to, litebearer.  But we shall see.

[blew]

I believe it is due to the variable $login NOT being encased by single quotes

 

yep

$verif = mysql_query ("select * from usr where login = $login")

 

shall be

$verif = mysql_query ("select * from usr where login = '$login'")

[pedromsouza]

That's it! I put quotes and mysql_error()... now it (query) seems to be working but a new error shows up...

 

[27-Jan-2011 20:44:38] PHP Fatal error:  Cannot break/continue 1 level in /Applications/MAMP/htdocs/recebe_dados_cadastro_usuario.php on line 33

 

I wanted it to do the following:

1. if $login is already in use, BREAK to avoid duplicate line...

2. if $login is new, insert into....

 

Thanks a lot for replying!

[sasa]

just remove break

[pedromsouza]

It works!

 

"break" is unnecessary because of the "if" clause...

 

If it was a loop, then I should use break, correct?

 

Thanks!

 

PS: how do I set this topic "solved"?

 

[sasa]

If it was a loop, then I should use break, correct?

yes