CODEWITHSUNDEEP
phpmysql
EDarrien
EDarrien

Reputation: 117

Retrieve data from two tables php

I have two tables as follows: activity_base (holds basic activity info along with activity_type) activity_email (holds specific details on email activities)

ALL tables have the activity_id that references activity_base.

An example: I am trying to output data for an email activity that gets data from other tables based on activity type.

The query will reference the activity_base to get the basic info about the activity and then if the activity_type is email, I need it to get more data from the activity_email table by running function email_activity_details.

The issue is: I cannot get $activity_details to show up comes back as undefined variable. Here is the query: Email activity Info:

function email_activity_details($activity_id){
          global $connection;
          $email_activity = "SELECT * FROM activity_email WHERE activity_id ='$activity_id'"
                    or die("Error: ".mysqli_error($connection));
          $query_email_activity = mysqli_query($connection, $email_activity);
         return $query_email_activity;
        }

Activity Details: Which calls email_activity_details():

function view_full_activity($activity_field){
            global $connection;

        $contact_id = $_REQUEST['contact_id'];
        $activity_id = $_REQUEST['activity_id'];
        $get = "SELECT * FROM activity_base WHERE activity_id = '$activity_id' "
                                    or die("Error: ".mysqli_error($connection));
        $query = mysqli_query($connection, $get);
    //Get activity base information
        while ($activity = mysqli_fetch_array($query)){
            $activity_related_to_id = $activity ['activity_related_to_id'];
            $activity_id = $activity['activity_id'];
            $activity_type_id = $activity['activity_type_id'];
    }     
    //Get detailed activity information 
      //If activity is Email
    if ($activity_type_id == "1") {
        $email_details = email_activity_details('$activity_id');
        while ( $email = mysqli_fetch_assoc($email_details)) {
          $activity_details = $email['email_message'];
        }
    }

switch ($activity_field) {
    case 'activity_id':
        return $activity_id;
        break;
    case 'activity_title':
        return $activity_title;
        break;
    default:
        # code...
        break;
}

}

Hope I am able to explain well. Thank you.

Upvotes: 1

Views: 50

Answers (1)

Evan de la Cruz
Evan de la Cruz

Reputation: 1966

Change this line

$email_details = email_activity_details('$activity_id');

to this

$email_details = email_activity_details($activity_id);

Explanation: Dollar signs inside of single-quotes are treated literally.

From the PHP manual:

variables and escape sequences for special characters will not be expanded when they occur in single quoted strings

http://php.net/manual/en/language.types.string.php#language.types.string.syntax.single

Upvotes: 1

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