How to show news?It is error in line 55

[miha]

I would like to read all articles in rows.

Do you know why it gives me error
Warning: link(): Permission denied in ...read_article.php on line 55
Line 55 is:

echo '<td><A HREF="'.link("article.php", "ID=".$_GET['id']).'"><B>'.$rec[$i].'</B></A></td>';

code is:

<?  

# Param 1 : MySQL Host Name
# Param 2 : MySQL Username
# Param 3 : MySQL Password
# Param 4 : MySQL Database
# Param 5 : SQL Statement (SELECT)

show_table("localhost","...","...","...","SELECT * FROM articles");

function show_table($hostName,$userName,$passWord,$dataBase,$sqlQuery)
{
    # Connect to MySQL
    $conn=mysql_connect("...", "...", "...");
    # Select Database
    mysql_select_db($dataBase,$conn);
    # Validate SQL Statement
    $array=explode(" ORDER",$sqlQuery);
    $sqlQuery=$array[0];
    if(!strstr($sqlQuery,"SELECT"))
        die("Invalid Query : SQL statement should be a SELECT statement.");
    # ORDER records by requested column
    if($_GET['order'])
        $sqlQuery=$sqlQuery." ORDER BY ".$_GET['order'];
    # Execute SQL query
    $result=mysql_query($sqlQuery) or die("Invalid Query : ".mysql_error());
    $row=mysql_fetch_array($result);
    # Check whether NULL records found
    if(!mysql_num_rows($result))
        die("No records found.");

echo "<table border=1><tr>";
# Make the row for table column names
while (list($key, $value) = each($row))
{
    $i++;
    if(!($i%2))
       echo "<td><b><a href='?order=$key'>$key</a></td>";
}
echo "</tr>";
$result=mysql_query($sqlQuery);

// Make rows for records
while($rec=mysql_fetch_array($result))
{
    echo "<tr>";
    for($i=0;$i<count($rec);$i++)
    {
        if($rec[$i])
        	echo '<td><A HREF="'.link("article.php", "ID=".$_GET['id']).'"><B>'.$rec[$i].'</B></A></td>';

    }
    echo "</tr>";
}
echo "</table>";
}
?>

[Weedpacket]

Looks like you don't have permission to do whatever it is you're doing on line 55.

[miha]

I need on line 55 link which will be shown for each row (article) like: www.mywebsite.com/article.php?id=1.php

[justsomeone]

The link function is used to create a symbolic link. This is a *nix feature which allows the same instance of a file to exist under two different locations on the file system. It has nothing to do with creating web page hyperlinks which seems to be what you want.

In this case, the link is trying to create a new symbolic link on the host filesystem, and the chances are that the apache(?) user doesn't have permission to do stuff like that.

In summary your web oage doesn't have permission to use the link function, but that's ok, cause I don't think you actually want to use it.

[justsomeone]

You just want to create a hyperlink... which isn't what the link() function does. just echo the html you want to create.

echo '<td><A HREF="article.php?id='.$_GET['id'].'"><B>'.$rec[$i].'</B></A></td>'; 

[miha]

I have solved and it works now. Code is:

	echo '<td><a href="article.php?ID='.$_GET['id'].'"><B>'.$rec[$i].'</B></A></td>';  

I have put while sentence :

// Make rows for records
while($rec=mysql_fetch_array($result))
{
   echo "<tr>";
    for($i=0;$i<count($rec);$i++)
    {
        if($rec[$i])


$query = "SELECT ID FROM articles where ID = '$rec[$i]' ";
$s = mysql_query($query);
echo(mysql_error($s));

$result = mysql_fetch_assoc($s);
$ID=$result['ID'];
//echo the article
//echo("<p>".$result['ID']."</p>");


    	echo '<td><a href="article.php?ID='.$ID.'"><B>'.$rec[$i].'</B></A></td>';
    }
    echo "</tr>";
}
echo "</table>";
}

It gives me error:
Warning: mysql_error(): supplied resource is not a valid MySQL-Link resource in .../read_news.php on line 59

Line 59 is:
echo(mysql_error($s));

Do you know how is correct to get ID number in this while sentence? Need help