根据链接显示数据的 PHP 数组 - 跟进

Question

我之前问过一个问题，根据用户点击这篇文章的哪个链接来填充页面，可以在这里找到:

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

但是我现在想尝试更进一步，因为我可以做到这一点，以便允许其他人访问数据，因此我希望始终能够为他们创建 if 语句。

所以我在考虑填充一个数组，该数组链接到数据库中的公司名称字段(我已经这样做是为了显示他们的名字，但是手动编码而不是放入一个数组，这取决于用户输入)。

这是我的想法或思路:

'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'

我的尝试:

'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'

希望这能解释我的问题。

*********已编辑*********

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

*********再次编辑************

$result = mysql_query("SELECT * FROM ddcompanies WHERE company = {$_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>"

我如何正确理解你。