PHP - Linking 2 Mysql Databases.

Okay so I have 2 tables in my database. One called user and one called messages. A user logs in to the message board and leaves a message (eg nice website). They write in the author name and the message then after the message is posted it says

"Nice website"
Posted by (author) on (date).

All is good so far. It works. However if you look at my code you will see I have a session started. This session is storing the username of the logged in user. From the column username in the users table. (This table has has an id for each user). Ive played around with the code trying to make it so the user doesnt have to fill in the author box. I want rid of that box So the logged in user just leaves a message then it says "posted by (username) on (date).

Im missing something from my code.
Can anyone tell me what? Please?

<?php
session_start();
mysql_connect("*************", "*****************", "***************");
mysql_select_db("***********************");
$time = time();



//this checks to see if the $_SESSION variable has been not set 
//or if the $_SESSION variable has been not set to true
//and if one or the other is not set then the user gets
//sent to the login page
if (!isset($_SESSION['username'])) {
    header('Location: http://***************.com/login.php');
}

$query = "INSERT INTO messages VALUES( NULL, '". mysql_real_escape_string($_POST['message']) ."', '". mysql_real_escape_string($_POST['username']) ."', '$time' )";if( $result = mysql_query($query) ) {
        if(mysql_affected_rows() > 0 ) {
                echo "Message Posted.<br><a href='messageboard.php'>Return</a>";
        } else {
                echo 'There was an error posting your message. Please try again later.';
        }
} else {
        echo "There was a database error.";
        // comment out next line for live site.
        echo "<br>Query string: $query<br>Returned error: " . mysql_error() . '<br>';
}


;

im finding it hard to link PHP, MySQL and dreamweaver together.

when php document is alaunched within the localhost all the data in the database is present how ever

when the php coding is entered into the html file via dreamweaver no data is present which i think connection to the mysql database cannot be establish

why is this can someone help me please, the php document is attached with all the coding


[attachment deleted by admin]

Hi,

I'm trying to make a mysql output to a link so the name will be a link so when you hit this link you will get the full information of this mysql input.

Can someone point me in the correct direction?

Here is how i get the output from my mysql database into my table.
Code: [Select]
<td>";
echo $row['name'];
echo "</td>

Hi,

Currently on my website I have a section where you can select an article held within a database, I still want this to happen but I wouldn't a different word to be linked to the article.
Currently it reads 'Click here to view this entry' underneath the title of the article, I want he title of all the different articles to link to the right article is there anyway to do this?

My code currently reads:

Code: [Select]
<?php

$blog_postnumber = 5;

if(!isset($_GET['page'])) {
$page = 1;
}
else {
$page = (int)$_GET['page'];
}
$from = (($page * $blog_postnumber) - $blog_postnumber);

$sql = "SELECT * FROM cms_article ORDER BY timestamp DESC LIMIT $from, $blog_postnumber";

$result = mysql_query($sql) or print ("Can't select entries from table cms_article.<br />" . $sql . "<br />" . mysql_error());

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

    $date = date("l F d Y", $row['timestamp']);

    $title = stripslashes($row['title']);
    $entry = stripslashes($row['entry']);
$id = $row['id'];

if (strlen($entry) > 0) {
    $entry = substr($entry, 0, 0);
    $entry = "$entry<a href=\"journal.php?id=" . $id . "\">Click here to view this entry.</a>";
}

    ?>

I have a few tables in a project and i'm linking pages together via PK/FK's.

i have a main table with client names,
ID, CLIENT NAME, ETC

i have a table with communications
COMMID, ID, CONTACT NAMES, ETC

and another of communication history.
COMMHISTORYID, COMMID, DATE, TIME, ETC

I'm not sure how to write the query so that i can view/update information from all 3 tables on the same webpage page using php/mysql. I'd appreciate any assistance you may provide. thanks.

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

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

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?

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 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

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

Each database is different, and may return different information.

Hello again guys!

What if i have 2 databases, db1 and db2. i want to read raw data from db1 and match it's content with data from db2, then produce some information. How can i do it at the same time? I know that to be able to access mysql data, we have to use the mysql_connect() and mysql_select_db(). mysql_select_db() only allows to connect to single database. what should i do?

thanks for your help in advance.

Hi!
I have two similar  tables on two different databases and i need to check if there are any entries on the table1, for example, that there aren't on the second table and print if there are i need to print them.
Any suggestions?
Thank you...

I have done select statements where you loop through all the records.
But what I need to do is pull record from one database than insert parts of that record
into the second. The issue I have is no knowing how to select a different db with out
stopping the loop of the first.

Thanks in advance.