accessing objects, static variables in PHP

Question

Here's my code:

class Photograph extends DatabaseObject {

    protected static $table_name="photographs";
    protected static $db_fields=array('id', 'filename', 'type', 'size', 'caption','album_id');
    public $id;
    public $filename;
    public $type;
    public $size;
    public $caption;
    //public $album_id;
    protected static $album_id;

    private $temp_path;
    protected $upload_dir="images";

Now, when I use this function below on another page like '$photos = Photograph::find_by_album();'

I get an sql error that says: 'Database query failed: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '=' at line 1'

public static function find_by_album($album_id='') 
{
return self::find_by_sql("SELECT * FROM ".self::$table_name."WHERE album_id = ".self::$album_id."");
}

Basically, what I would to happen is to get all values saved in the database from the $table_name where $album_id entered by the user is equivalent to the album_id found in the database. You may find this problem simple, but unfortunately, I am not able to find a solution. Any ideas please? Thanks in advance. :)

EDIT:

Following Mr. Elias Ootegem,

I have modified the code and it now looks like

public static function find_by_album($album_id='') 
{ return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".$album_id.""); 
}

However, I still get the same error. I tried using this code:

public static function find_by_album() 
{ return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".$album_id.""); 
}

Now another error comes up which says: Undefined variable: album_id

Any other ideas?

put a space after table name SELECT * FROM photographsWHERE album_id= — Waygood
– Waygood, Commented Aug 6, 2012 at 10:19
Output the SQL query like it looks like, not only the error message. Then you can actually learn what is wrong (and others, too). Also you are abusing static to the extreme so my feelings for you are a little bit limited. You should learn about inheritance first, not static function stacking (or use a Super::$tatic to provide the database if you love static that much). — hakre
– hakre, Commented Aug 6, 2012 at 10:21

Elias Van Ootegem · Accepted Answer · 2012-08-06 13:52:08Z

3

First off, it's extremely bad practice to make your db functions static, because you'll either connect and disconnect for each query, and you effectively render all safety features utterly useless. You can't, for example, roll-back inserts, because the connection has been lost once an error is thrown, terminating your session and committing potentially corrupted data.
After posting this answer I read hakre's comment. Though he comes across a bit blunt, he does have a point. You might want to read up on inheritance and design patterns, that'll probably make you ease up on the statics

Now, for your actual question: have you tried dumping the query you're trying to execute? If you do that, you'll probably see that your query looks something like

SELECT * FROM photographsWHERE album_id = NULL

Fix: add a space before WHERE, and drop the self:: from self::$album_id. As I gather, you want to use the argument passed to the method. This argument might have the same name as a static property, but it isn't assigned to it by default.

In response to your update:

public static function find_by_album($album_id='') 
{
    return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".$album_id.""); 
}

This method could work, provided you call it like so:

Photograph::find_by_album(123);//or some variable that holds a valid id

If you fail to pass an argument to this method, it'll fall back to the default value, which is an empty string. The result is a query like:

SELECT * FROM photographs WHERE album_id =

This won't be regarded as valid SQL for obvious reasons. So make sure the method is invoked correctly. Another thing that might be the cause of your headaches is the absence of a Database name. If your connection does not have a default db selected, edit your query as follows:

SELECT * FROM yourDb.theTable WHERE album_id = 123

On the variation, the undefined variable error is 100% expected, you've removed the $album_id argument from the method's definition, yet you refer to the variable all the same. So what value is filled in? An undefined variable will trigger an error or silently declare itself (or gets declared) as null. If you expected PHP to fall back to the static variable you declared (but didn't initialize in your snippet), you do need to specify the scope to self::$album_id, and -of course- initialize it to a value that makes sence.

public static function find_by_album() 
{
    self::$album_id = (self::$album_id ? self::$album_id : 1);//if is set use the value, else use 1
    return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".self::$album_id.""); 
}

If none of the above resolves your issue, try dumping whatever object holds the db connection in the find_by_sql method. Chances are there you haven't made a connection.
Remember: statics don't require an instance, your constructor isn't called, so everything you do there (which in db objects is, generally speaking, connecting to the database) does NOT get executed.

edited Aug 6, 2012 at 13:52

answered Aug 6, 2012 at 10:26

Elias Van Ootegem

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2 Comments

user1242749 Over a year ago

I have modified the code and it now looks like public static function find_by_album($album_id='') { return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".$album_id.""); } However, I still get the same error. I tried using this code: public static function find_by_album() { return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".$album_id.""); } Now another error comes up which says: Undefined variable: album_id Any ideas

Elias Van Ootegem Over a year ago

The latter is easy, you've removed the parameter, so there is no variable called $album_id anymore, hence the error undefined variable. In the first case, you should check to see if the $album_id has an actual value: set it's default to 1 or something, if no value is passed the resulting query would be SELECT * FROM photographsWHERE album_id = , which is syntactically incorrect, which could explain the first problem. var_dump or echo the query string, and post it here, if possible

Michael · Accepted Answer · 2012-08-06 10:20:42Z

0

Lack of a single space?

return self::find_by_sql("SELECT * FROM ".self::$table_name." WHERE album_id = ".self::$album_id."");

answered Aug 6, 2012 at 10:20

Michael

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Comments

Waygood · Accepted Answer · 2012-08-06 10:21:25Z

0

add a space and introduce some secuity to the album id with intval() or mysql_real_escape_string()

return self::find_by_sql("SELECT * FROM `".self::$table_name."` WHERE `album_id` = ".self::$album_id);

answered Aug 6, 2012 at 10:21

Waygood

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