MySQL SELECT

[ATLien]

$id = mysql_query("select user_id from sv_users where user_name='$username'") or die ("Mysql error: " . mysql_error());

I get this when I echo $id that statment.

Resource id #6

Why does that happen? am I doing something wrong with trying to echo a SQL query?

[stolzyboy]

if you want to see the actual query do this:

$query = "select user_id from sv_users where user_name='$username';";
$id = mysql_query($query) or die(mysql_error());
echo $query;

and then if you want to see the info from that, you need to use [man]mysql_fetch_array[/man], [man]mysql_fetch_object[/man], or others to, like it says, fetch the results

[ATLien]

Originally posted by stolzyboy
if you want to see the actual query do this:

$query = "select user_id from sv_users where user_name='$username';";
$id = mysql_query($query) or die(mysql_error());
echo $query;

and then if you want to see the info from that, you need to use [man]mysql_fetch_array[/man], [man]mysql_fetch_object[/man], or others to, like it says, fetch the results

What is the benifit of laying it out the way you said to? Also I am just trying to get one thing out of the database that is a num value like 1 so would I use mysql_fetch_object?

[BlueMeteor]

Do you want this?

$id = mysql_query("select user_id from sv_users where user_name='$username'") or die ("Mysql error: " . mysql_error());
if ($rs=mysql_fetch_array($id))
{
    echo $rs['user_id'];
}

hth

[tekky]

because of the way mysql works

you would need to do

$row = mysql_fetch_array($id);
echo $row[user_id]; //or replace with another selected column name

[stolzyboy]

Originally posted by ATLien
What is the benifit of laying it out the way you said to? Also I am just trying to get one thing out of the database that is a num value like 1 so would I use mysql_fetch_object?

like i stated, you can use object, array method, whatever, just as long as you fetch the array somehow

[ATLien]

Originally posted by stolzyboy
like i stated, you can use object, array method, whatever, just as long as you fetch the array somehow

The fetch object didn't work.

[stolzyboy]

$query = "select user_id from sv_users where user_name='$username';";
$result = mysql_query($query) or die(mysql_error());

$row = mysql_fetch_array($result);

echo $row['user_id'];

[ATLien]

Originally posted by stolzyboy
$query = "select user_id from sv_users where user_name='$username';";
$result = mysql_query($query) or die(mysql_error());

$row = mysql_fetch_array($result);

echo $row['user_id'];

That is how i ended up doing it. Why do you use 2 varibles for the query instead of just one? What is the benifit of that?

[stolzyboy]

in case you need to echo $query; to make sure your variables come through correctly, otherwise, if you combine it with the mysql_query, it returns the results array right away, and that is when you get the Resource id #2 things, i just like them seperate so i can see what is REALLY getting to the query, if i need to

[ATLien]

Originally posted by stolzyboy
in case you need to echo $query; to make sure your variables come through correctly, otherwise, if you combine it with the mysql_query, it returns the results array right away, and that is when you get the Resource id #2 things, i just like them seperate so i can see what is REALLY getting to the query, if i need to

So if I did

$query = "SELECT user_id FROM sv_users WHERE user_name='$username';"

then echoed $query would i get the value of that? lets say the user_id was 2 would i get 2 by echoing that?

[stolzyboy]

no, if you echo'd this:

$query = "SELECT user_id FROM sv_users WHERE user_name='$username';"

and $username was stolzyboy, it would look like this in the browser

SELECT user_id FROM sv_users WHERE user_name='stolzyboy'

see, so i make sure that $username is REALLY getting into the query