[Resolved] creating links from the database

corruption

i want to create a profile system for the members of my fraternity. I'm trying to take the names out of the database and make them links. This is the code that i have...

echo "<a href=\"profile.php?id=".$id."\">"$row['fname']." ".$row['lname']"</a>";

this code produces some error because i get no output at all when i use it.

echo "<td width=100px align=center>";
        echo "<a href=profile.php?id='$id'>";
		echo $row['fname']." ".$row['lname'];
        echo "</a>";
		echo "</td>";

this code produces the resulting link that i want but when clicked it goes to
url/profile.php?id= (blank)

bubblenut

Let's see a little more code please. Can you show us everything from the query down to the end of the loop on the queries resultset please. I'm guessing that it's $id should be $row['id'] if $row is where the query result row is getting placed, but without seeing more of the code I can't say for deffinate.

Cheers
Bubble

corruption

//script to display the list of people
$sql    =    "SELECT * FROM brothers ORDER BY lname DESC";
$result    =    mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($result) > 0)
    {
    ?>
    <body bgcolor="#000000" text="#000000">
    <table width="80%" align="center" cellpadding="2px" cellspacing="0" bordercolor="#999999" border="medium" bgcolor="#CCCCCC">
    <tr>
    <?php 
    while($row = mysql_fetch_array($result)) 
    {
        echo "<td width=100px align=center>";
        echo "<a href=profile.php?id='$id'>";
		echo $row['fname']." ".$row['lname'];
        echo "</a>";
		echo "</td>";
    }
	 echo "<a href=\"profile.php?id=".$id."\">"$row['fname']." ".$row['lname']"</a>";
	?>
    </tr>
    </table>
<?php 
} 
    else
    {
    echo "No rows returned!";
    }

?>

bubblenut

And where are you getting $id from? If it's comming from the database then you will need to write it as $row['id'], as noted above.

corruption

based on your suggestion i changed that line to...

echo "<td width=100px align=center>";
        echo "<a href=profile.php?id=".$row['id'].">";
		echo $row['fname']." ".$row['lname'];
        echo "</a>";
		echo "</td>";

this seems to be giving me the correct result, now i just have to get the next page to function

corruption

next page is causing problems now... if i use the statements following i get a a sql error.

<?php
//include the connection script	
include '../connect.php';
//define the variable

$id = $_GET['$id'];

$sql  =  "SELECT * FROM brothers WHERE id=$id";
$result    =    mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($result) > 0)
    {
    ?>
    <body>
    <table>
    <tr>
    <?php 
        while($row = mysql_fetch_array($result)) 
    {
	    echo "<tr>";
        echo "<td align='center'>";
        echo $row['fname']." ".$row['lname'];
        echo "</td>";
		echo "<td align='center'>";
		echo $row['phone'];
		echo "</td>";
		echo "<td align='center'>";
		echo $row['address'];
		echo "</td>";
		echo "<td align='center'>";
		echo $row['email'];
		echo "</td>";
		echo "<td align='center'>";
		echo $row['aim'];
		echo "</td>";
		echo "</tr>";
    }
	?>

when i change the query from

$sql = "SELECT * FROM brothers WHERE id=$id";

$sql = "SELECT * FROM brothers WHERE id='$id'";

i get no results returned

bubblenut

echo the sql query before you run it just to make sure that the id's being passed accross properly. If it looks as expected then try running the query directly through MySQL and see what the result is then.

corruption

it doesn't seem to have the id # in place.

when i echo $id;
it gives me a blank page. So i'm not sure why the id isn't recieved.

when i echo $sql

i get SELECT * FROM brothers WHERE id=
with no id number. I'm not really sure why i can't get the id accross.

corruption

help

TheDefender

next page is causing problems now... if i use the statements following i get a a sql error.

What is the sql error you are receiving?

Also, what do you get if you do this? Does it give you the correct ID number here?

$id = $_GET['$id']; 
echo $id;

You may need to remove the <body> tag to show this...

Lastly, your query should look like this:

$sql = "SELECT * FROM brothers WHERE id='$id'";

so make sure it is setup this way.

Give me some feedback, and I will see what I can do to help out. The most important thing is telling us what the sql error is though. That may be the key to everything.