[SOLVED] dont know mysql sytnax at all

[thewooleymammoth]

i need to run a query to tell me the type of account the person has

 

the only thing i can think of is...

"SELECT type FROM users WHERE  email = '$username'"

and it returns '0'

although there is a type for each one and i can see there is on phpmyadmin

 

and i cant find the proper syntax on any tutorials or anything.

 

the correct code would really help thanks

[dropfaith]

is username defined anywhere??

 

$Name = mysql_escape_string($_GET['Name']);
$query = "SELECT * FROM Art  WHERE Name = '$Name'";

[revraz]

Chances are $username is not coming over, so you are sending a bad query.  One way to tell is to put the query in a variable and echo it.

 

Are you sure $username is the right variable?  Since you are checking the email field, is the email the same as the $username?

 

Other than that, your syntax is correct.

[Maq]

Paste the code before this, where you actually assign $username its value.

[thewooleymammoth]

$username=  mysql_escape_string(strip_tags($_POST['username']));
$password=  mysql_escape_string(strip_tags($_POST['password']));
$q = "SELECT email FROM musicusers WHERE email = '$username' AND password = '$password'";
$q2="SELECT type FROM musicusers WHERE email = '$username'";
if (mysql_num_rows(mysql_query($q)))
  {
  $_SESSION['type']=mysql_query($q2);
  $_SESSION['username'] = $username;
  $type=$_SESSION['type'];
  
  echo  $type;
  }

 

this is the login page, i changed it around a little bit and im not getting '0' anymore im getting Resource id #8

 

ive included a print screen from php my admin with the data included in the table

 

 

 

[attachment deleted by admin]

[dropfaith]

if im not mistaken or missing some other code why are you running two querys for this just select * from musicusers WHERE email = '$username' AND password = '$password'"; will get all fields or just seperate fileds with a comma

email, type

 

$username=  mysql_escape_string(strip_tags($_POST['username']));
$password=  mysql_escape_string(strip_tags($_POST['password']));
$q = "SELECT * FROM musicusers WHERE email = '$username' AND password = '$password'";
if (mysql_num_rows(mysql_query($q)))
  {
  $_SESSION['type']=mysql_query($q);
  $_SESSION['username'] = $username;
  $type=$_SESSION['type'];

  echo  $type;
  }

[thewooleymammoth]

now my code looks like this

$username=  mysql_escape_string(strip_tags($_POST['username']));
$password=  mysql_escape_string(strip_tags($_POST['password']));
//$q = "SELECT email, type FROM musicusers WHERE email = '$username' AND password = '$password'";
$q="SELECT * FROM musicusers WHERE email = '$username' AND password = '$password'";
//$q2="SELECT type FROM musicusers WHERE email = '$username'";
if (mysql_num_rows(mysql_query($q)))
  {
  $_SESSION['type']=mysql_query($q);
  $_SESSION['username'] = $username;
  $type=$_SESSION['type'];
  echo  $type."<br>";
  echo $username."<br>";
}

 

and the output is

Resource id #8

5wooley4@gmail.com

 

[thewooleymammoth]

by the way im trying to get $type to = 'artist' idk if i made that clear

[thewooleymammoth]

for whatever reason (after a ton of playing around and searching)

 

this works

 

$q="SELECT * FROM musicusers WHERE email = '$username' AND password = '$password'";
$q2="SELECT type FROM musicusers WHERE email = '$username'";
if (mysql_num_rows(mysql_query($q)))
  {
  $last=mysql_query($q2);
  $type=mysql_fetch_assoc($last);
  $_SESSION['username'] = $username;
  $_SESSION['type']=$type;
  foreach($type as $b)
  {
  if ($b != '')
  {
   $_SESSION['type']=$b;
  }
  }

 

which makes $_SESSION['type']=to the account type