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Home/ Questions/Q 9218701
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T02:55:28+00:00 2026-06-18T02:55:28+00:00

I don’t understand this at all. I have a class which is roughly the

  • 0

I don’t understand this at all.

I have a class which is roughly the following:

<?php
class pageData
{
    private $Bookmark;
    private $Program;
    private $Agency;

    //With appropriate setters/getters
}
?>

Then I try to create a new object, pass it around a bit, and eventually end up with:

mysql_query("INSERT INTO Records (Bookmark, Program, Agency)
VALUES ('$data->getBookmark()', '$data->getProgram()', '$data->getAgency()')");

I end up with Notice: Undefined property: pageData::$getBookmark in…

Notice: Undefined property: pageData::$getProgram in…

Notice: Undefined property: pageData::$getAgency in…

Using PhpMyAdmin it looks like Bookmark becomes 0 and Program becomes () and Agency is empty.

If I type

print($data->getBookmark());

it prints out the bookmark. if I type

echo $data->getBookmark();

it prints out. Why doesn’t it work when I try to insert it into the database, too?

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  1. Editorial Team
    Editorial Team
    2026-06-18T02:55:29+00:00Added an answer on June 18, 2026 at 2:55 am

    When using anything other than a normal variable in a string you should add curly brackets around the values:

    mysql_query("INSERT INTO Records (Bookmark, Program, Agency)
                 VALUES ('{$data->getBookmark()}', '{$data->getProgram()}', '{$data->getAgency()}')");
    
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