PHP - Multi DB Connection suddenly not working - php

Question

I have 2 databases:
news_db. having table : t_news
branding. having table : t_branding
I have db connection:
$con1 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('news_db', $con1);
$con2 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('branding', $con2);
My codes:
$dataNews = mysql_fetch_assoc(mysql_query("SELECT * FROM t_news",$con1));
echo $dataNews['title']; /* it is working, showing "Test Title" */
$dataBrand = mysql_fetch_assoc(mysql_query("SELECT * FROM t_branding",$con2));
echo $dataBrand['title']; /* it is not working, nothing to show */
But if i reverse the query like :
$dataBrand = mysql_fetch_assoc(mysql_query("SELECT * FROM t_branding",$con2));
echo $dataBrand['title']; /* it is working, showing "Test Brand Title */
$dataNews = mysql_fetch_assoc(mysql_query("SELECT * FROM t_news",$con1));
echo $dataNews['title']; /* it is not working, nothing to show" */
Can anybody help me, why the sudden php and mysql as though I can only run one connection, but yesterday all of connections can walk and coding nothing has changed. Thanks

Answer 1

The issue is that you're reusing the connection:
Taken from the manual for mysql_connect():
If a second call is made to mysql_connect() with the same arguments,
no new link will be established, but instead, the link identifier of
the already opened link will be returned. The new_link parameter
modifies this behavior and makes mysql_connect() always open a new
link, even if mysql_connect() was called before with the same
parameters. In SQL safe mode, this parameter is ignored.
You should parse the fourth (true which is $new_link) parameter to the second connect like below:
$con1 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('news_db', $con1);
$con2 = mysql_connect('127.1.0.0', 'root', 'root', true);
mysql_select_db('branding', $con2);
As stated, you should avoid using mysql_* functions as the api IS depreciated.
Alternatively you should be looking into PDO or Mysqli Prepared Statements.
Some More Notes
You should turn on error reporting when developing to ensure you don't run into any issues:
ini_set('display_errors', 1);
error_reporting(E_ALL);
And since you're using mysql, you should look at mysql_error() which probably would've thrown a message something like:
The table TABLENAME doesn't exist in DATABASE

Answer 2

With the code-
$con1 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('news_db', $con1);
$con2 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('branding', $con2);
Or,
$con1 = mysql_connect('127.1.0.0', 'root', 'root');
$con2 = mysql_connect('127.1.0.0', 'root', 'root'); // this line is redundant, both lines are same
mysql_select_db('news_db', $con1);
mysql_select_db('branding', $con2); // this will be selected
the effect will be that- the database branding will be selected, news_db wont be selected, coz its written after news_db!
So, your bnoth the queries will look for the database branding and will throw error "Table ...... not found" if you check with mysql_error()

Answer 3

$con1 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('news_db', $con1);
$dataNews = mysql_fetch_assoc(mysql_query("SELECT * FROM t_news",$con1));
echo $dataNews['name'];
$con2 = mysql_connect('127.1.0.0', 'root', 'root');
mysql_select_db('branding', $con2);
$dataBrand = mysql_fetch_assoc(mysql_query("SELECT * FROM t_branding",$con2));
echo $dataBrand['title'];
You can use this code.I think it will be helpful for you.