Retrieving database info based on user

[Nothingman0]

Hello, everyone.  I'm still quite new to php and mysql but I have been unable to figure this out.  I'm attempting to display a specific row and column from a mysql database based on the user.

 

 

$result = mysql_query("SELECT fbleague FROM users WHERE username = $_SESSION[username]");

 

fbleague is the column

users is the table

username is the primary key column

$_SESSION[username] is the username variable

 

Then later I try to display it with

 

echo "You are in league $result ";

 

I know $_SESSION[username] is set correctly because I can display that with no problems.  Any help would be greatly appreciated.

 

 

[genericnumber1]

you need to put quotes around string variables in a mysql query, I also like to put {} around variables inside double quotes...

 

$result = mysql_query("SELECT fbleague FROM users WHERE username = '{$_SESSION['username']}'");

[DeathStar]

why not use it like this:

 

$username = $_SESSION['username'];
$result = mysql_query("SELECT fbleague FROM users WHERE username='$username'");

 

thats how I work I just use $_SESSION['userid'].

[Barand]

According to the manual you only need the {} when using 2D (or more) array values such as $_SESSION['x']['y'] in a string.

 

However, I agree with GC1 and use them whenever I insert an array element in a string, or avoid the issue competely as deathstar suggests.

[Nothingman0]

Thanks for the quick responses.  That seemed to work, sort of.  It didn't used to display anything but now it displays

 

Resource id #6

 

for the value.  The value is actually 0.

[DeathStar]

try this:

if (!$username){
echo "None";}
else{
echo "$username";}

[Nothingman0]

try this:

if (!$username){
echo "None";}
else{
echo "$username";}

 

Maybe I'm not understanding what you want me to check with your code suggestion but;

 

echo "Welcome, <b>$_SESSION[username]</b>.;

 

does return the username correctly.

 

[Barand]

$result is a result set and not the selected value, you need to extract that from $result.

 

With a single value as you are selecting the easiest way is

 

$fbleague = mysql_result ($result, 0, 0);

 

which gets the first field of the first row.

[Nothingman0]

That worked!  Thanks a million, Barand and everyone.