PHP - Connecting 2 Databases On Different Hosts

Hi guys, hoping you can help.

Iv recently moved a site from one server to another. Needless to say this site encountered a few errors when the move was done. I believe i have corrected most of these errors but now it is as though the mysql data base isnt connecting correctly with the site. I can see the category's on the site but when i click on a category the products do not display. Its almost as though the info is not being pulled through from the data base. Im at a complete loss!

It was originally on a site running MYSQL5.0.92 old and the new site is running 5.1.56 new.Not sure if that makes a difference. Any help greatly needed/appreciated

Hello,

I am looking for a push in the right direction. I would like to be able to view a website that is duplicated on many servers by specifying the IP address. Similar to how one would use the hosts file.

I can do this using netcat
nc 1.2.3.4 80
Host: example.com
GET /contact-us.html
<CR><CR>

Can someone tell me how I would go about porting this to PHP? I would just like to set up a small page on my webserver where I can enter the domain and IP address and have the page displayed back to me.

Thanks Colin

Hi guys.

This is my first question on phpFreaks (the other 900+ posts I've made were always an attempt to help someone else)

Here's the issue: when using gethostbyaddr($_SERVER['REMOTE_ADDR']) on a local machine (current using MAMP on OS X) the name returned is not the first entry in my /etc/hosts file.
(On php.net there's another person also commenting on this and saying that apparently gethostbyaddr() selects a random entry from the file.)

This is not really an issue, since everything will work fine on my FreeBSD production server where each IP address has only one DNS record (I'm building the company's intranet, so I know exactly what's in our DNS tables) but I'm curious to know if anyone has had the same issue/problem and if there's a way around it. (Logic/Common sense tells me the first record should be the one returned)

Currently I have the following information in my /etc/hosts file:
Code: [Select]
127.0.0.1 localhost
127.0.0.1       iSy
255.255.255.255 broadcasthost
::1             localhost
fe80::1%lo0 localhost
# Other names for testing purposes
127.0.0.1 name1.blablabla.com
127.0.0.1 name2.blablabla.com
127.0.0.1 name3.blablabla.com
127.0.0.1 name4.blablabla.com
127.0.0.1 name5.blablabla.com
127.0.0.1 name6.blablabla.com
127.0.0.1 name7.blablabla.com
127.0.0.1 name8.blablabla.com
the value returned is always line 8: name2.blablabla.com. If I comment out that line, then it moves 'randomly' to line 12: name6.blablabla.com and sticks with it even if I shut down the servers and then boot up again.

any ideas?

cheers.

Hello,

I'm attempting to write a PHP script to query WMI for an IP address of a host. The reason I'm doing this instead of getbyhostname is because NAT translations on the network can cause inaccurate results. The question is two-fold:

1) My Script is as follows
<?

$obj = new COM ( 'winmgmts://localhost/root/CIMV2' );
$wmi_network = $obj->ExecQuery("Select * From Win32_NetworkAdapterConfiguration");

foreach ( $wmi_network as $wmi_call )
{
    $ipaddr = $wmi_call->IPAddress;
    echo $ipaddr;
}
?>


I get the following error when running the script:
Quote

Catchable fatal error: Object of class variant could not be converted to string



2) How do I go about querying a remote host on the network with valid credentials?


Thank you

Hi,

I'm thinking about finding a free host and using a free sub domain to build and test the php-database stuff... Along with learning php and mysql stuff  I've already got a lot of the php stuff setup but my file paths are all wrong. Is there anything I should do while I'm building this part of my site to make sure it still works when I move it to a permanent home ?

So far I know that I shouldn't use the full path but instead something like assets/php/reallycomplex.php

Once its complete and I have all of my info in the database my plan is to move to a real host and get a real domain name instead of the sub domain.


Are there any other things I should know?

Thanks for the advice.

Just wondering, which is better php databases or sql? I have phpMyAdmin and I've that you can convert the database into php scripts?

ok , here is my mysql code to get all posts from the posts table .
Code: [Select]
$query = mysql_query("SELECT id,to_id,from_id,post,type,state,date FROM posts WHERE state='0' ORDER BY id DESC LIMIT 50"); and here is the code to display the users friends...
Code: [Select]
$sqlArray = mysql_query("SELECT friend_array FROM myMembers WHERE id='" . $logOptions_id ."' LIMIT 1");
while($row=mysql_fetch_array($sqlArray)) { $iFriend_array = $row["friend_array"]; }
$iFriend_array = explode(",", $iFriend_array);
if (in_array($id, $iFriend_array))see now i got as far as , if(in_array($id, $iFriend_array))
How would i put these togeather to where it would get the posts from the posts table that there friends posted?

I have just read my upcoming modules for my final year at uni and 'multimedia databases' is one of them. I am just wondering if any of you had any clue on what a multimedia database is? I am guessing it's a database populated with directory data, but that would be to simple...

I'm trying to connect to two databases and I'm having problems with the following code.  I googled to come up with this but can't figure out the errors I'm getting.
Code: [Select]
$connection="localhost";
$username="user";
$password="password";
$database1="dbone";
$database2="dbtwo";

$db1 = mysql_connect($connection,$username,$password) or die(mysql_error());
$sel1 = mysql_select_db($database1, $db1);

$query1 = "SELECT * FROM TBLUSERS";
$result1 = mysql_query($query1, $db1);
while($nt1 = mysql_fetch_array($result1, $db1))
{

}

$db2 = mysql_connect($connection,$username,$password) or die(mysql_error());
$sel2 = mysql_select_db($database2, $db2);
$query2 = "SELECT * FROM TBLPD20101101";
$result2 = mysql_query($query2, $db2) or die(mysql_error());
while($nt2 = mysql_fetch_array($result2, $db2))
{

}
The error I get is
Quote

Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 13

Warning: mysql_fetch_array() expects parameter 2 to be long, resource given in C:\xampp\htdocs\HighVisibility\DashBoard2.php on line 22

I have a site where I need to have lets call it image1 displayed, then I want to change this image based on a php if statement, for instance:

if $var == $var2
change the image
....blah blah

so I was also going to have the names of my images stored in my database, i.e. image1.jpg and image2.jpg in my database.

the image is in its own div tag set as the background image of the div tag if that makes any diference.

Thanks

Hi,

I am in the procress of creating discussion system however I am a bit puzzled about the best way to go about it.

I am starting the discussion by creating an ID number and then match the answer to the initial ID number.  However, I dont know whether if is best to put the responses into a different database.

I'm a bit puzzled how ID matching systems works.

Lets say:

Question 1 =    ID1
Question 2 =    ID2
Question 3 =    ID3
Question 1 Answer 1    =   ID4   (How is this matched to ID1)
Question 2 Answer 1    =   ID5   (How is this matched to ID2)

is this based on preg_match?

What would be the fastest way to search 2+ databases with the same search information?

Each database is different, and may return different information.

right i did this code
<table width="400" border="1" align="center" cellpadding="2" cellspacing="0" bordercolor="#000000" class=thinline>
        <tr class="header" background="includes/grad.jpg">
          <td height="20" colspan="3" background="includes/grad.jpg"  class=header><div align="center" class="header">Last 10 Kills</div></td>
        </tr>
        <tr class="header" background="includes/grad.jpg"> 
          <td width="166" height="20" background="includes/gradgrey.jpg">Name</td>
          <td width="157" height="20" background="includes/gradgrey.jpg">Rank</td>
          <td width="157" height="20" background="includes/gradgrey.jpg">Killed Time</td>
        </tr>
      <?
   $c=mysql_query("SELECT * FROM attempts  WHERE outcome='Dead' ORDER BY id DESC LIMIT 10");
   while($d=mysql_fetch_object($c)){
   echo "<tr><td><a href='profile.php?viewuser=$d->target'>$d->target</a></td><td>$d->rank</td><td>$d->date</td></tr>";
   
   
   
   }
   ?>

but you see the $d->rank

is in another database called users

How can i link the databases so the last 10 kills shows the users rank

hi all,

I am trying to link 2 mysql tables and display some information from each of them.

I have a list of all the possible items for sale in table1 and I am trying to count the number of rows in the other table2 where the items exist.

E.g. 'table1'
manufacturer       model
      man1                  item1
      man1                  item2
      man1                  item3

'table2'
    id                 model
         1                    item3
         2                    item3
         3                    item2

And the result would show:
item1(0)
item2(1)
item3(2)
It would list all the items from table1 and show next to it how many rows are related to that item from table2.

I have inserted a quote where I have tried many times to enter something similar to that show in the note below - but I can not get it to work - it just shows the total number of models in table1 for a given manufacturer.

The php I have made so far is:


<?php
    case 'manufacturer':
        $query = "
            SELECT * 
            FROM table1";
        $query .= "
            WHERE manufacturer = '".$data."' ";
        $query .= "
            ORDER BY model ";

        $result = mysqli_query($cxn,$query);
        $returnData[''] = "Select a Model...";
        while($row = mysqli_fetch_assoc($result)){
// I THINK I NEED TO INSERT SOMETHING LIKE $query2 = "SELECT * FROM table2 WHERE model = table1.model";
    $k=$row['model'];
    $k2=$row2['model'];
            $counter[$k]+=1;
            $returnData[$k]=$k;
        }
        
        foreach($counter as $k => $row)
        {
            $returnData[$k] .= " ($row)";
        }

        break;
?>

What's the best way for putting actual quotations into a database?

I was using...

$quote = htmlspecialchars(mysqli_real_escape_string($dbc, $_POST['quote']));

Should I be?