Variable not working

[Anon]

I want to change $sql depending on what $user is given. I found a replace function to change it. I enter that variable in and i get an error
I used
$table_name_fetched1 = 'select * from user_link_types where';
$pat2 = "user";
$repl2 = "$user";
$table_name_fetched2 = ereg_replace( $pat2, $repl2, $table_name_fetched1 );

$db=connect_db();
$db_name=db_name();
mysql_select_db($db_name,$db);

$sql="$table_name_fetched2 lt_id != 0 order by lt_name";
$result=mysql_query($sql,$db);
while($myrow=mysql_fetch_array($result))

I get an error Warning: 0 is not a MySQL result index on the line "while($myrow=mysql_fetch_array($result))"

[Anon]

You should echo your resulting sql query so you
can see exactly what it is expanding to. The query
itself is failing; the error message comes from calling
mysql_fetch_array() using the result id ($result) when
the query itself failed (and returned a value of 0 which
is never a valid result id.)

It might help to simplify your code. Am I correct in
assuming that you could replace all that regex stuff
with the following?

$qry = 'select * from ' . $user . '_link_types where lt_id != 0 order by lt_name';

[Anon]

Yes that would replace all that is needed. I'll try that code

Kirk Parker wrote:

You should echo your resulting sql query so you
can see exactly what it is expanding to. The query
itself is failing; the error message comes from calling
mysql_fetch_array() using the result id ($result) when
the query itself failed (and returned a value of 0 which
is never a valid result id.)

It might help to simplify your code. Am I correct in
assuming that you could replace all that regex stuff
with the following?

$qry = 'select * from ' . $user . '_link_types where lt_id != 0 order by lt_name';

[Anon]

I changed everything to this format
$sql_links="select * from ' . $user . '_links where lt_id=$lt_id";
$result_links=mysql_query($sql_links,$db);
while($myrow=mysql_fetch_array($result_links))

And i still get the error Warning: 0 is not a MySQL result index in on the line "while($myrow=mysql_fetch_array($result_links))"

[Anon]

Now is when you be sure to <b>echo</b> the query
string so you can see the exact contents. Also,
isn't this mysql you are using? It should be displaying
some kind of error message for the call to mysql_query(),
too, giving you some idea of <i>where</i> it doesn't like
the syntax of your statement.

[Anon]

After your $result = line...have this

echo mysql_error();

That'll give you a better idea of what is going wrong..rather than it's just not working.

Always echo your sql statement and mysql_error while you're debugging...

---John Holmes...