Multi Table Query

[jaZon]

So I have this query:

select brand.brand, brand.price, flavor.flav_name, style.style_name from brand, cig, flavor, style WHERE cig.brand_id = brand.brand_id and cig.flav_id = flavor.flav_id and cig.style_id = style.style_id

It works fine in the DB, but, of course it will not work when I try to do it through php as that will not return a valid array. At least it won't the way I have it now. What can I do?? I just want to be able to run the query(and similar variations) through a php site.

[kevinsequeira]

post the code

reg
kevin

[stolzyboy]

Originally posted by jaZon
What can I do??

you could post some code you are using to let us know what you
have already tried

[jaZon]

$query = "select brand.brand, brand.price, flavor.flav_name, style.style_name from brand, cig, flavor, style WHERE cig.brand_id = brand.brand_id and cig.flav_id = flavor.flav_id and cig.style_id = style.style_id";
	$result = mysql_query($query);


while($row=mysql_fetch_array($result)){

	echo("Doesn't matter what I echo, doesn't work.  The array is invalid.");

	}

Error =
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/virtual/site69/fst/var/www/html/test.php on line 8

[stolzyboy]

put this

echo $query;

after your

$query = "";

and before your

$result = ;

line and see what it holds in spot of your vars to ensure they are getting into the query line

[jaZon]

That prints just query exactly as I entered it

select brand.brand, brand.price, flavor.flav_name, style.style_name from brand, cig, flavor, style WHERE cig.brand_id = brand.brand_id and cig.flav_id = flavor.flav_id and cig.style_id = style.style_id

[stolzyboy]

change this

$result = mysql_query($query);

to this

$result = mysql_query($query) or die(mysql_error());

[jaZon]

yea, i forgot to mention i had the ordie earlier.......just came up with a blank page. no error. like i said, the query runs in the db.

[stolzyboy]

$query = "select brand.brand, brand.price, flavor.flav_name, style.style_name from brand, cig, flavor, style WHERE cig.brand_id = brand.brand_id and cig.flav_id = flavor.flav_id and cig.style_id = style.style_id;";

added a semicolon, i am running out of ideas, it should work if it works like in phpmyadmin or in command line or whatever

[jaZon]

🙁 still nothing

[kevinsequeira]

what about the code for your database connections ?

reg
kevin

[stolzyboy]

try putting this before your while...fetch loop and see if it returns
an error also

if (mysql_num_rows($result) != 0)
{
echo "it does work";
}
else
{
echo "it doesn't work";
}

[jaZon]

hmm....that pulls a doesn't

what would my connection code have to do with anything

[stolzyboy]

what would my connection code have to do with anything

---

what do you mean by that?

[kevinsequeira]

well like stolzyboy said, if you are copying and pasting the output of echo $query directly into the database and it works there, there is no reason why it shouldnt work here.
check to see if you are connecting to the same database / table. if the username and password combo has access to all the tables.

reg
kevin

[jaZon]

kevinsequeira

what about the code for your database connections ?

[stolzyboy]

are you certain you are connected to mysql, are you certain you are connected to the correct db, etc..

maybe post that code for us as well, obviously it isn't in the code you have posted already if the query works elsewhere

[jaZon]

ok, i'm an idiot 😃

I had an include db function where I didn't type the db name correctly....sorry you guys. And THANKS for the help.

[stolzyboy]

np

[kevinsequeira]

what about the code for your database connections ?

😃 :p
reg
kevin