Question

I have previously asked a question to populate a page depending on which link the user clicked on this post, found here:

How to populate 1 php page differently depending on link clicked?

However I want to try and take it further now as I could make it so that other people are allowed to data therefore I want always be able to create the if statement for them.

So I was thinking something along the lines of populating an array which links to the company name field in the database (which I have already done it to display their name but have coded it manually instead of putting in an array which depends on user input).

Here is my kind of idea or train of thought:

'Declare Array/s

for (user click - use array item relating to the same name as the user clicked link -> [link]==COMPANYNAME-Which is in the database) {

    Display other info relating to that company'

My attempt:

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

    $companyarray[] = $row["company"]; // Declare array to store list of company names inserted to database by each company

    $varcompany[] = $row["company"]; //runs through the company column and populates the array varcompany with those names
    $varwebsite[] = $row["website"]; //runs through the website column and populates the array varwebsite with those names
    $varstory[] = $row["story"]; //runs through the story column and populates the array varstory with the text
}

for ($_GET['link'] == '$companyarray[i]') { // Thanks to Johnny Craig, Crashspeeder and Kolink for the help on this part (I have this working but by manually inserting the company names and creating a seperate if statement for each company) I want to be able to automatically populate this list with each new company added

    echo "<div id='companyname'><a href='http://$varwebsite[0]' />" . $varcompany[0] . "</a></div>";// Displays companies name with link to their website
    echo "<div id='website'><a href='http://$varwebsite[0]' />" . $varwebsite[0] . "</a></div>";// Displays companies website with link
    echo "<img src='images/example.jpg' class='profilePic' />";// At the moment manually entering image link (hopefully will be automatic in future)
    echo "<div id='story'>" . $varstory[0] . "</div>";// Displays a text field from database'

Hope this explains my problem.

********EDITED*********

while($row = mysql_fetch_array($result)) {
    $varcompany[] = $row["company"]; //runs through the company column and populates the array varcompany with those names
    $varwebsite[] = $row["website"]; //runs through the website column and populates the array varwebsite with those names
    $varstory[] = $row["story"]; //runs through the story column and populates the array varstory with the text
}

if($_GET['link']=='miiniim'){
    //print company1 details on single.php page
    echo "<div id='website'><a href='http://$varwebsite[0]' />" . $varwebsite[0] . "</a></div>"; //MIINIIM 1st Company in database
    echo "<div id='companyname'><a href='http://$varwebsite[0]' />" . $varcompany[0] . "</a></div>"; //MIINIIM 1st Company in database
    echo "<img src='images/example.jpg' class='profilePic' />";
    echo "<div id='story'>" . $varstory[0] . "</div>"; //MIINIIM 1st Company in database
}elseif($_GET['link']=='other'){
//print company1 details on single.php page
    echo "<div id='companyname'><a href='http://$varwebsite[1]' />" . $varcompany[1] . "</a></div>"; //MIINIIM 1st Company in database
    echo "<div id='website'><a href='http://$varwebsite[1]' />" . $varwebsite[1] . "</a></div>"; //MIINIIM 1st Company in database
    echo "<img src='images/example.jpg' class='profilePic' />";
    echo "<div id='story'>" . $varstory[1] . "</div>"; //MIINIIM 1st Company in database

********EDITED AGAIN**********

$result = mysql_query("SELECT * FROM ddcompanies WHERE company = {$_GET['link']}");

while($row = mysql_fetch_array($result)) {
    $varcompany = $row["company"];
$varwebsite = $row["website"];
$varstory = $row["story"];
}
print_r($result);
print_r($varcompany);
print_r($varwebsite);
print_r($varstory);

echo "<div id='website'><a href='http://$varwebsite' />" . $varwebsite . "</a></div>";
echo "<div id='companyname'><a href='http://$varwebsite' />" . $varcompany . "</a></div>";
echo "<img src='images/example.jpg' class='profilePic' />";
echo "<div id='story'>" . $varstory . "</div>";

Answer 1

(QUERY = SELECT * FROM [table] WHERE company = {$_GET['link']};)
while($row = mysql_fetch_array($result)) {
    $companyarray = $row;
}
echo "<div id='companyname'><a href='http://" . $companyarray['website']  . "' />" . $companyarray['company'] . "</a></div>";
echo "<div id='website'><a href='http://" . $companyarray['website']  . "' />" . $companyarray['website'] . "</a></div>";
echo "<img src='images/example.jpg' class='profilePic' />";
echo "<div id='story'>" . $companyarray['story'] . "</div>"

How I understood you correctly.