MYSQL : Selecting from Multiple Databases (an error occured) [duplicate] - php

Question

This question already has answers here:
mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
(31 answers)
Closed 9 years ago.
I have three databases in the same server.. They are margoplatinum, detos and plaza.
They also have the same structure of tables. One of the table is called movie.
Here is the table structure of movie table:
id_movie int(11) PK, AI;
movie_title varchar(100);
status varchar(30);
Data from margoplatinum in movie table are:
id_movie | movie_tittle | status
--------------------------------------------------
1 | Frozen | Now Playing
2 | The-Amazing-Spidey2 | Coming Soon
Data from detos in movie table are :
id_movie | movie_tittle | status
--------------------------------------------------
1 | Comic8 | Now Playing
2 | Godzilla | Coming Soon
and data from plaza in movie table are :
id_movie | movie_tittle | status
--------------------------------------------------
1 | The Killer | Now Playing
2 | Godzilla | Coming Soon
So what i want to do is to select from the three databases above (in my php code and convert it into JSON) for table movie where the status = "Now Playing"..
Here is my php code :
<?php
$db_host = "127.0.0.1";
$db_uid = "root";
$db_pass = "";
$db_con = mysql_connect($db_host,$db_uid,$db_pass) or die('could not connect');
$q = mysql_query('SELECT DISTINCT a.id_movie, b.id_movie, c.id_movie, a.movie_tittle,
b.movie_tittle, c.movie_tittle FROM `margoplatinum`.movie a
INNER JOIN `detos`.movie b ON a.id_movie = b.id_movie
INNER JOIN `plaza`.movie c ON b.id_movie = c.id_movie
WHERE a.status = "Now Playing"');
$v = '{"info" : [';
while($r=mysql_fetch_array($q))
{
$ob = array("<br>","<b>","</b>");
if(strlen($v)<12)
{
$v .= '{"id_movie" : "'.$r['id_movie'].'", "movie_tittle" :
"'.$r['movie_tittle'].'"}';
}
else
{
$v .= ',{"id_movie" : "'.$r['id_movie'].'", "movie_tittle" :
"'.$r['movie_tittle'].'"}';
}
}
$v .= ']}';
echo $v;
echo mysql_error();
mysql_close();
?>
But when I execute the php above, an error occured..Here is the error message :
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\CinemaInfo\list_movie.php on line 17
{"info" : []}Unknown column 'a.status' in 'where clause'
Is my query above correct? or i did something wrong with my query..
Can anyone help me to write the correct query based on my case above?
I really appreciate any help from you.
Thank you :)

Answer 1

I think that what you need is a UNION not a JOIN, a join operation will find movies with same ID but different names.
So, I would go for:
As you mentioned that you only want two columns, here is what you need.
select id_movie, movie_tittle
from
(
select id_movie, movie_tittle, status
from `margoplatinum`.movie
UNION
select id_movie, movie_tittle, status
from `detos`.movie
UNION
select id_movie, movie_tittle, status
from `plaza`.movie
) as a
where a.status = 'Now Playing'

Answer 2

First make sure your query is working by running it in mysql query editor directly.
If it is working then there is something wrong with the way query is executed using php.
The below query worked for me
select a.id from `test1`.app_user a
inner join `test2`.app_user r on r.id=a.id
inner join `test`.app_user f on a.id=f.id
where a.Status="ACTIVE"
So, i'm guessing the column name might be different or something like that

Answer 3

Try changing your query for the following:
$q = mysql_query("SELECT DISTINCT a.id_film, b.id_film, c.id_film, a.judul_film,
b.judul_film, c.judul_film FROM `margoplatinum`.film a
INNER JOIN `detos21`.film b ON a.id_film = b.id_film
INNER JOIN `depokplaza21`.film c ON b.id_film = c.id_film
WHERE a.status = 'Now Playing'");
Strings in SQL use simple ' not double "

Answer 4

It would be better to have a single database to handle this. The table structures could be:
CREATE TABLE `movies` (
id_movie INT AUTO_INCREMENT,
movie_title VARCHAR(50),
PRIMARY KEY( id_movie )
);
CREATE TABLE `locations` (
id_movie_fk INT,
location VARCHAR(50),
STATUS VARCHAR(25)
);
I've set up an sqlFiddle example.