Trying to display a link

[AMulligan]

Hello,
I have a db table: page: *page_id, page_name, page_link.
What I am trying to do is display records from the table showing the page_name and using the information in page_link to create a link.
Hopefully, this is making some sense.

This is the code I have:

while ($add_info = mysql_fetch_array($get_page_res)) 
    {
        $page = $add_info[page_id];
        $page_name = $add_info[page_name];
        $page_url = $add_info[page_link];
        $display_block .= "<li><a href=\"$page_url\">$page_name</a></li>";
    } 

Everything runs fine but when I click on the link it goes to the index.htm of my site not the address from the db.

So, how do you do this or can it even be done. I figure that if anyone knows how it will be someone on this forum.

[bfsog]

echo "<a href=\"$page_url\">link text</a>"; 

is what I use and it works fine.

Tell me, when you hover over the link is the URL http://youraddress.com ?
If so, then $page_url is set to that.

Alternatively instead of assigning the value to a variable you could go

echo "<a href=\"$add_info[column]\">link text</a>"; 

[towerboy]

Try this and see if it works:

$result = mysql_query("SELECT * FROM (Table Name) ORDER BY page_id ASC") or die ("Invalid query: SELECT");
while ($data = mysql_fetch_array($result))
  {
  extract($data);
  $n++;
  print "<li><a href=\"$page_url\">$page_name</a></li>";
  } 

[AMulligan]

Thank you both but I must admit that I am totally confused.

bfsog - I tried using your echo... script but kept getting a parse error...no doubt I used it wrong. I admit to not being sure where the echo should have gone.

$page_url = echo "<a href=\"$add_info[column]\">link text</a>"; 

To answer your other question. When I look at the view source for the page the html has www.domain.com not the page I am trying for.


towerboy - I am not sure where I would would put your code. Is it to replace my original $get_page_result or is it to replace the code I originally posted.

Maybe more explanation is needed for you to help me. This is a page after a login. If the username and password is valid then the user selects a page from a list of pages they are allowed to access. The display_block shows the allowed list of pages. I am still fumbling through the learning process and often miss the obvious. Thank you for the help.

[bfsog]

Hey there.

My original code would go after you pull or get the data from the db.. so..

while ($add_info = mysql_fetch_array($get_page_res))
    {
        $page = $add_info[page_id];
        $page_name = $add_info[page_name];
        // $page_url = $add_info[page_link]; commented out as we include directly
        echo "<a href=\"$add_info[page_link]\">link text</a>";
    } 

Just to show heres some code from a file of mine:

    $query="SELECT * FROM users where username='$user'";
    $result = mysql_query($query) or die("Query failed: ".mysql_error());
    $numrows = mysql_num_rows($result);
    if ($numrows == "0") {
        echo "Sorry. We do not have a user with that id number.<br /><br />";}
    
    else {
    while($row = mysql_fetch_array($result))
    {
        echo "Url: <a href=\"$row[url]\">Visit website</a>";
        echo "<br /><br />";
$numrows++;
    }} 

Hope this helps.. it can be done though so we should soon have you sorted.

[towerboy]

I would say either way (bfsog or mine) should work. Mine would be an alternative way of doing it, but either way would work.

[AMulligan]

Sorry to be so slow to catch on. And I really do appreciate the help. I made the changes but the links are still going to index.htm.

So here is all the code (forgive me for what you are about to see): :nervous:

<?php
if (($_POST['username'] =="") || ($_POST['password'] ==""))
{
header("Location: userlogintest.htm");
exit;
}
$display_block = "<p><h2>Please select the page you wish to work on:</h2></p>";
$conn = mysql_connect("localhost", "dbname", "dbpass") or die(mysql_error());
mysql_select_db("db",$conn)  or die(mysql_error());

$sql = "select user_id, username, password from user
where username='$_POST[username]' and password='$_POST[password]'";
$sql_res = mysql_query($sql) or die(mysql_error());

if (mysql_num_rows($sql_res) > 0) 
{
    //if authorized, get the values of f_name l_name
    $user_id = mysql_result($sql_res, 0, 'user_id');
}
    
//$display_block .= "<p>Hi there.</p>";
$get_page = "select page_id, page_name, page_link from page";
$get_page_res = mysql_query($get_page) or die(mysql_error());

$get_page_info = "select user_page.user_id, user_page.page_id, page.page_id, 
        page.page_name FROM page, user_page WHERE user_page.user_id = $user_id
        AND page.page_id=user_page.page_id";
    
$get_page_res = mysql_query($get_page_info)or die(mysql_error());

if (mysql_num_rows($get_page_res) != 0) 
{
    $display_block .= "<P><strong>page(s):</strong><br>
        <ul>";

    while ($add_info = mysql_fetch_array($get_page_res)) 
    { 
        $page = $add_info[page_id]; 
        $page_name = $add_info[page_name]; 
        // $page_url = $add_info[page_link]; commented out as we include directly 
        $display_block .= "<a href=\"$add_info[page_link]\">$page_name</a>"; 
    } 

    $display_block .= "</ul>";
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<title>User login</title>
</head>
<body>
<?php echo $display_block; ?>
</body>
</html>

[AMulligan]

Oops I have found my first two big mistakes. 1. A duplicate $get_page_res and in the second which the code uses forgot to add page.page_link to the query.

[AMulligan]

Thank you for all the help...it works...it really works. Now, since I have been working at this from the time I got up this morning (6 1/2 hours ago) I will go enjoy some breakfast. I really appreciate the help and patience. Thank you again.