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I am trying to populate results with a description from my MySQL database table interest_type

My private function is

    private function Get_Interests_Types_From_DB()
    {
        $sql = "SELECT 
                    InterestID,
                    InterestDescription
                FROM
                    interest_type";

        $result = mysqli_query($this->Con, $sql);

        while($row = mysqli_fetch_array($result))
        {
            $arrayResult[] = $row;
        }

        return ($arrayResult);
    }

The area of code that I am trying to use this function in is within a public function within the same class. The values from the form are numeric. I am trying to tie the $interests variable with the InterestID from the table and then print the InterestDescription, instead of the value of $interests.

function ProcessRegistrationForm()
    {   
        $fname = $_POST['firstname'];
        $lname = $_POST['lastname'];
        $email = $_POST['email'];
        $gender = $_POST['gender'];
        $interests = $_POST['interests'];

        if(!isset($_POST['firstname']) || !isset($_POST['lastname']) || !isset($_POST['email']) ||
            ($_POST['firstname']) == '' || ($_POST['lastname']) == '' || ($_POST['email']) == '')
        {
            echo("Please enter your first / last name and email.");
        }   
        else
        {
            echo("<h2>Results</h2>");
            echo("<div id='results'>");
            echo $fname; 
            echo("<br />");
            echo $lname; 
            echo("<br />");
            echo $email;
            echo("<br />");
            echo $gender;
            echo("<br />");

            $interestDescription = $this->Get_Interests_Types_From_DB();

            foreach($interests as $likes)
            {
                if($likes == $interestDescription['InterestID'])
                    echo $$interestDescription['InterestDescription'] . "<br />";
            }
            echo("<p style='font-weight: bold;'>Your data has been saved! We will contact you soon!</p>");
            echo("</div>");
        }

        $myClub = new Club("localhost","A340User","Pass123Word","info_club");

        $date = date("Y/m/d");

        $sql="INSERT INTO member
                    (`FirstName`,`LastName`,`Gender`,`Email`,`MemberSince`)
                VALUES
                    ('$fname','$lname','$gender','$email','$date');";

        $result = mysqli_query($this->Con,$sql);
        /*if($result == true) 
        {
            echo "Successful Insert<br />";
        }
        else
        {
            echo "Error Inserting class" . mysqli_error($this->Con) ." <br />";
        }*/

        for($i = 0; $i < sizeof($interests); $i++)
        {
            $interest = $interests[$i];

            $sql="INSERT INTO member_interests
                        (`Email`,`InterestID`)
                    VALUES
                        ('$email',$interest);";

            $result = mysqli_query($this->Con,$sql);
        }

I am getting: Notice: Undefined index: InterestID in C:\xampp-portable\htdocs\A340\Assign5\Assign5_Club_Membership_Class.php on line 287

How can I print the InterestDescription instead of the value of $interests?

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it seems that $interestDescription['InterestID'] doesnt exist... check $interestDescription with print_r(). – vlzvl May 1 at 16:58
I still get Notice: Undefined index: InterestID. I am not sure why this is. I used the same information in a previous class function that runs in the same class. I can assure you that InterestID does exist. Why is this happening? – CharlWillia6 May 1 at 17:05
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1 Answer

up vote 1 down vote accepted

there:

while($row = mysqli_fetch_array($result))
{
   $arrayResult[] = $row;
}

now the $arrayResult, which is actually your $interestDescription, contains something like:

$interestDescription[0]["InterestID"] 
$interestDescription[1]["InterestID"]
...

and not $interestDescription[InterestID"]

UPDATE:

perhaps you need something like this (assuming that $interests holds an array of IDs)

        $interestDescription = $this->Get_Interests_Types_From_DB();
        foreach($interests as $likes) {
           foreach($interestDescription as $desc) {
              if($likes == $desc['InterestID']) {
                 echo $desc['InterestDescription'] . "<br />";
              }
           }
        }
share|improve this answer
So does this mean I should use a for loop instead of foreach, so I can say $interestDescription[$i]['InterestID']? – CharlWillia6 May 1 at 17:08
Ok, so I changed my code to: $interestDescription = $this->Get_Interests_Types_From_DB(); for($i = 0; $i < sizeof($interestDescription); $i++) { if($interests = $interestDescription[$i]['InterestID']) echo($interestDescription[$i]['InterestDescription'] . "<br />"); } and all it does is populate the display with all the InterestDescription data from the table, and not just the ones by value in $interests. So if the user picked interests with the value of 2 and 5, I am still getting all six descriptions printed that are in the database – CharlWillia6 May 1 at 17:15
check my update – vlzvl May 1 at 17:16
Perfect. Thank you. I get it now. I appreciate the help. – CharlWillia6 May 1 at 17:17
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