How can I connect two mysql using while?

Question

I would like to know why am I having some errors doing this:

 $query = "SELECT * FROM servers ";
      while($rows = mysql_fetch_array(mysql_query($query)))
    {
    $database1 = array();

 $database1['host']     = $rows['db_host'];
 $database1['user']     = $rows['db_user'];
 $database1['password'] = $rows['db_pass'];
 $database1['table']    = $rows['db_user'];

 $con1 = @mysql_connect($database1['host'], $database1['user'], $database1['password']) or die($connect_error);
 if(!$con1) {
       exit;
 }


$connect_error = 'Sorry, there are some connection problems.';

mysql_select_db($database1['table']) or die($connect_error);


$info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"));
 echo "No servidor ".$rows['name']." existem um total de ".$info['total']. " contas      registradas, alem disso";

    }

Ok. Lest explain: I have two differents database's, In the fisrt one that I wrote this query:

   query = "SELECT * FROM servers ";

I have in the table servers a mysql connections.So I have to connect in diferrent mysqls to get mysql host, user and password.When I get these things, I have to get a value for each one connection and echo this value using COUNT as a variable, when I did that:

 $info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"));
echo "No servidor ".$rows['name']." existem um total de ".$info['total']. " contas      registradas, alem disso";

I hope that u guys undestand because I dont speak english very well ): and Thank u for your attention..

Two different mysql databases on the same mysql database server or on different database servers? — hakre
– hakre, Commented Aug 24, 2013 at 21:39

Vahid Hallaji · Accepted Answer · 2013-08-24 21:36:54Z

1

You should run mysql_query($query) 1 time not in loop. Your code makes running mysql_query in every loop.

Remove from:

while($rows = mysql_fetch_array(mysql_query($query)))
                          //HERE^

Do:

<?php
$query = "SELECT * FROM servers";
$result = mysql_query($query);

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

   // do more stuff

edited Aug 24, 2013 at 21:36

answered Aug 24, 2013 at 21:31

Vahid Hallaji

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Comments

jnelson · Accepted Answer · 2013-08-24 22:12:07Z

Looking at your script there are still some improvements you could make. For starters your variable declaration ($connect_error) is created AFTER it is called.

Though you should adopt the mysqli PHP API instead as the MySQL functions you use have been marked as deprecated.

# Where query results are stored
$database1 = array();

# Error Message
$connect_error = 'Sorry, there are some connection problems.';

# mySQL Loop No.1
while($rows = mysql_fetch_array(mysql_query("SELECT (name,db_host,db_user,db_pass) FROM servers"))) {
    foreach($rows as $key=>value) {
       $database1[$key] = $value;   
    }   
}

# Since this is a duplicate of existing row duplicate
$database1['table'] = $rows['db_user'];

# Connect to new db using details
$con1 = @mysql_connect($database1['host'], $database1['user'], $database1['password']) or die($connect_error);
if(!$con1) {
       exit;
}

# Select 2nd DB
mysql_select_db($database1['table']) or die($connect_error);

# mySQL Loop No.2
while($info = mysql_fetch_assoc(mysql_query("SELECT COUNT(username) as total FROM authme"))) {
    $total = $info['total'];
}

#Lastly with all data gotten you can show string
echo "No servidor ".$database1['name']." existem um total de ".$total. " contas registradas, alem disso";

Please see for further reading:

http://www.php.net/manual/en/mysqli.quickstart.dual-interface.php

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