Question

I am selecting data from a database. The database field names are exactly the same as the class variable names. Is there a way to store this data into the class variables without specifying each one individually?

    //gets info about a specified file.
    //chosen based on a supplied $fileId
    function getFileInfo($fileId)
    {
        //select info from database
        $sql = "SELECT id, companyId, url, name, contentType, fileSize, saved, retrieved
                FROM files
                WHERE id = $fileId";
        $results = $this->mysqli->query($sql);
        $results = $results->fetch_object();

        //store info into class variables
        $this->id = $results->id;
        $this->companyId = $results->companyId;
        $this->url = $results->url;
        $this->name = $results->name;
        $this->contentType = $results->contentType;
        $this->fileSize = $results->fileSize;
        $this->saved = $results->saved;
        $this->retrieved = $results->retrieved;
    }

Answer 1

I’d propose this approach:

$nameMap = array(
   'id',
   'companyId',
   'url',
   'name',
   'contentType',
   'fileSize',
   'saved',
   'retrieved',
);

foreach( $nameMap as $attributeName ) {
   $this->$attributeName  = $results->$attributeName ;
}

While one could write

foreach($result as $var => $value) {
...
}

the outcome fully depends on backing table’s structure. If you add further attributes to the table, your code might break.

Using $nameMap, the application still works.