Creating a like system

[mdmartiny]

I am in the process of creating a like system for a personal website that I am working on. I am trying to pull information from one database and use that to place info into another database. When I do I get resource error #15.

 

This is my code that I have so far

 

function like_add() {
    $last = $_GET['l'];
    $first = $_GET['f'];
    $ip = $_SERVER['REMOTE_ADDR'];

    $like_sql = "Select * from `counter` where first = ' $first ' and last = '$last '";
    $like_result = mysql_query($like_sql);

    $likes = '<p class="like_button"> Likes ';
    while ($like_row = mysql_fetch_assoc($like_result)) {
        $like = $like_row['likes'];

        $likes .= " $like";
    }
    $likes .= '</p>';
    echo "$likes";
    echo '<form name="like_add_form" action="" method="POST" class="like_add">
<input type="submit" name="like_add" class="like_add_button" value="Like Me" />
</form>';

    if (isset($_POST['like_add'])) {
        $page_id = "SELECT `page_id` FROM `counter` WHERE first = '$first' AND last = '$last'";
        $page_id_result = mysql_query($page_id);
        echo $page_id_result; <----- Here I am trying to pull information from a field in my database to use as a variable and to place in the second database

        $voted = mysql_query("Select * FROM `liked_ip` where ip ='$ip' AND page_id='$page_id_result'"); <----- Here I want to use the information from the first table to place in the second one

        if (mysql_num_rows($voted) != 0) {
            echo "You have all ready liked this post";
        } else {
            mysql_query("INSERT into `liked_ip`(id,page_id,ip) VALUES ('$page_id_result','$ip')");
        }
        if (mysql_num_rows($voted) == 0) {
            mysql_query("UPDATE counter SET `likes` = `likes` + 1 where first = '$first' AND last = '$last'");
        }
    }
}

 

I know that I have to be doing something so simple that I am just not seeing it this late at night. I may even be completely wrong in all of this. If someone could please help me out I would appreciate it very much.

[trq]

mysql_query returns a result resource not an integer.

[mdmartiny]

So I have to use a while loop to get the resource into an integar. Or is there some other way?

[trq]

No, you don't need any loop. Just pass the resource to mysql_fetch_assoc which will give you an array. This array will have your data in it.

[mdmartiny]

Let's see if I hot this right. From what I read from  foing some searching. I run the query and it its everything in a array. Using the mysql_fetch_assoc makes it an associative one. Will that work if I only want info from one row on table

[lonewolf217]

you always have to push the results of the query into an array of some sort before you can read the data, whether is 1 row or 1 million rows

[mdmartiny]

So then I need to assign am variable to the array. Something like

 

$page_id = mysql_fetch_assoc($result);

 

Right?

[lonewolf217]

<?php
$row = mysql_fetch_array($result);
$page_id = $row['page_id'];
?>

[mdmartiny]

The makes sense. I will have to try it when I get in front of a computer. Thanks for the help.