Link Referral System

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February 4th, 2005, 07:11 AM

DKing

Contributing User

Join Date: Jul 2004

Posts: 59

Time spent in forums: 2 h 57 sec
Reputation Power: 5

Link Referral System

Here is my problem:
I am creating a links system with a hits system (in and out). The links are created wonderfully, and the "out" url (?p=referout&id=whatever) is created appropriately. But the "In" url (?p=referin&id=whatever) is always wrong: "id" is always assigned to "1", so it only updates the "in" hits for the first link inserted into the database.

I am wondering how to make it assign the correct ID number for each "in" url (?p=referin&id=CORRECT#!!!!).

Here are the scripts for the following files:
add_link.php
links.php
referin.php

add_link.php

PHP Code:


		
			
<?

$continue = $_POST['continue'];

$confirm = $_POST['confirm'];

$edit = $_POST['edit'];



if($continue)

{

    $refername = $_POST['refername'];

    $linkurl = $_POST['linkurl'];

    $buttonurl = $_POST['buttonurl'];

    

    echo "<strong>Is the following correct?</strong><br><br>";

    echo "Referer Name: <strong>" . $refername . "</strong><br>";

    echo "Link URL: <strong>" . $linkurl . "</strong><br>";

    echo "Button: <img src=\"" . $buttonurl . "\" border=\"0\" width=\"88\" height=\"31\" /><br>";

    

    echo "<form method=\"post\" action='?p=add_link&confirm'>";

    echo "<input type=\"hidden\" name=\"refername\" value=\"" . $refername . "\">";

    echo "<input type=\"hidden\" name=\"linkurl\" value=\"" . $linkurl . "\">";

    echo "<input type=\"hidden\" name=\"buttonurl\" value=\"" . $buttonurl . "\">";

    echo "<input type=\"Submit\" name=\"edit\" value=\"<< Go back\"> ";

    echo " <input type=\"Submit\" name=\"confirm\" value=\"Confirm >>\">";

    echo "</form>";

}

elseif($edit)

{

    $refername = $_POST['refername'];

    $linkurl = $_POST['linkurl'];

    $buttonurl = $_POST['buttonurl'];

    

    echo "<form method=\"post\" action='?p=add_link&continue'>";

    echo "<strong>Site Name:</strong> <input type=\"Text\" name=\"refername\" value=\"" . $refername . "\"><br>";

    echo "<strong>Web Address:</strong> <input type=\"Text\" name=\"linkurl\" value=\"" . $linkurl . "\"><br>";

    echo "<strong>Button Link:</strong>  (optional)(88x31 only) <input type=\"Text\" name=\"buttonurl\" value=\"" . $buttonurl . "\"><br>";

    echo "<input type=\"Submit\" name=\"continue\" value=\"Continue >>\">";

    echo "</form>";

}

elseif ($confirm)

{

    $refername = $_POST['refername'];

    $linkurl = $_POST['linkurl'];

    $buttonurl = $_POST['buttonurl'];



    $sql = "INSERT INTO refer (site,url,img_url) VALUES ('" . addslashes( $refername ) . "','" . addslashes( $linkurl ) . "','" . addslashes( $buttonurl ) . "')";

    $insert = mysql_query($sql);



    if($insert) {

    $result = mysql_query("SELECT * FROM `refer`");

    echo mysql_error();

    $myrow = mysql_fetch_array($result);

    $id = $myrow["id"];

    echo "Link added successfully!";

    echo "<br><br>";

    echo "<u>Please link to my site with the following URL:</u><br>";

    echo "<strong><em>http://yaggles.b7p.info/?p=referin&id=" . $id . "</em></strong>";

    }

    else

    {

    die("Cannot query database!");

    }

}

else

{

    echo "<form method=\"post\" action='?p=add_link&confirm'>";

    echo "<strong>Site Name:</strong> <input type=\"Text\" name=\"refername\"><br>";

    echo "<strong>Web Address:</strong> (With http:// included!)<input type=\"Text\" name=\"linkurl\"><br>";

    echo "<strong>Button Link:</strong>  (optional)(88x31 only) <input type=\"Text\" name=\"buttonurl\"><br>";

    echo "<input type=\"Submit\" name=\"continue\" value=\"Continue >>\">";

    echo "</form>";

}

?>

links.php

PHP Code:


		
			
<?PHP

require_once("conf.php");

$_REQUEST['add'];

?>

<center><h1>Links</h1></center>

<BR /><BR />

<?PHP

if($add)

{

    echo "<meta refresh='refresh' content='0;?p=add_link'>";

}

else

{

    $result = mysql_query("SELECT * FROM refer ORDER BY refin DESC");

    echo mysql_error();

    while($myrow = mysql_fetch_array($result)) {

        $linkurl = $myrow["url"];

        $refername = $myrow["site"];

        $buttonurl = $myrow["img_url"];

        $referin = $myrow["refin"];

        $referout = $myrow["refout"];

        $id = $myrow["id"];

        

        echo "<div align=\"center\"><a href=\"?p=referout&id=" . $id . "\"><img src=\"" . $buttonurl . "\"";

        if($buttonurl) {

            echo "alt='". $refername . "'";

        }

        echo "border=\"0\" /></a><br>";

        echo "<a href=\"?p=referout&id=" . $id . "\">" . stripslashes($refername) . "</a><br>";

        echo "Hits In: " . $referin . " | Hits Out: " . $referout . "</div>";

        echo "<br><br><br>";

    

    }

}

?>

referin.php

PHP Code:


		
			
<?PHP

if($id)

{

    $prevcount = mysql_query("SELECT refin FROM refer WHERE id = '" . addslashes( $id ) . "'");

    $prevcount = mysql_fetch_assoc( $prevcount );

    

    $sql = "UPDATE refer SET refin = " . intval( $prevcount['refin'] + 1 ) . " WHERE id=" . addslashes( $id ) . "";

    

    $result = mysql_query( $sql ) OR die( "Couldn't perform query: " . mysql_error() );



    echo "<META HTTP-EQUIV=Refresh CONTENT=\"1; URL=?p=home\">";

    echo "<span class='text'><div align='center'><em><strong>Loading... Please wait!</strong></em></div></span>";

}

else

{

    echo "Bad ID number. But you can always go to the main site by clicking <a href='?p=home'>HERE</a>";

}

?>

Please browse through them and let me know why it might assign the id "1" for all of them!

Thanks
-Yaggles

December 26th, 2006, 10:41 AM

SKDevelopment

Contributing User

Join Date: Sep 2004

Posts: 659

Time spent in forums: 2 Weeks 4 Days 46 m 25 sec
Reputation Power: 169

These lines:

PHP Code:


		
			
 $result = mysql_query("SELECT * FROM `refer`");

    $myrow = mysql_fetch_array($result);

    $id = $myrow["id"];

Always return the id of the 1st (I mean topmost in the returned recordset) row. To return the id of the most recently inserted row you could e.g. make your id field something like

Code:

id INT AUTO_INCREMENT PRIMARY KEY

in the CREATE TABLE syntax and use e.g. the line (not checked):

PHP Code:


		
			
 $result = mysql_query("SELECT * FROM `refer` WHERE id=LAST_INSERT_ID()");

instead of

PHP Code:


		
			
 $result = mysql_query("SELECT * FROM `refer`");

__________________
PHP Programmer

December 26th, 2006, 10:51 AM

sejohnson

Contributing User

Join Date: Nov 2006

Location: Appleton, WI

Posts: 124

Time spent in forums: 4 Days 4 h 2 m 43 sec
Reputation Power: 3

looks like an old post to me.. it was posted in febuary of 2005

Comments on this post

SKDevelopment agrees: Yes, you are right ... Sorry

__________________
Tervoo

December 26th, 2006, 10:56 AM

SKDevelopment

Contributing User

Join Date: Sep 2004

Posts: 659

Time spent in forums: 2 Weeks 4 Days 46 m 25 sec
Reputation Power: 169

Oh... Sorry!!! there was a spamming post in the thread (probably will be deleted by the moderators shortly), after which the thread got to the top. And I answered without looking at the date of the original post.

Will have to pay more attention in the future.

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