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Object could not be converted to string?

Tags:

php

Why am I getting this error:

Catchable fatal error: Object of class Card could not be converted to string in /f5/debate/public/Card.php on line 79

Here is the code:

public function insert()
{
    $mysql = new DB(debate);

    $this->initializeInsert();

    $query = "INSERT INTO cards
            VALUES('$this->$type','$this->$tag','$this->$author->$last','$this->$author->$first',
            '$this->$author->$qualifications','$this->$date->$year','$this->$date->$month',
            '$this->$date->$day','$this->$title', '$this->$source', '$this->$text')";
            $mysql->execute($query);
}

(Line 79 is the $query and the function is part of class Card)

All the declarations of Card:

public $type;

public $tag;
public $title;
public $source;
public $text;

public function __construct() {
    $this->date = new Date;
    $this->author = new Author;
}

After changing line 79 to this:

$query = "INSERT INTO cards
    VALUES('$this->type','$this->tag','$this->author->last','$this->author->first',
    '$this-$author->qualifications','$this->date->year','$this->date->month','$this->date->day',
    '$this->title', '$this->source', '$this->text')";

I now get this error:

Catchable fatal error: Object of class Author could not be converted to string in /f5/debate/public/Card.php on line 79

like image 433
cactusbin Avatar asked Jul 08 '10 05:07

cactusbin


3 Answers

Read about string parsing, you have to enclose the variables with brackets {}:

$query = "INSERT INTO cards VALUES('$this->type','$this->tag','{$this->author->last}',"

Whenever you want to access multidimensional arrays or properties of a property in string, you have to enclose this access with {}. Otherwise PHP will only parse the variable up to the first [i] or ->property.

So with "$this->author->last" instead of "{$this->author->last}", PHP will only parse and evaluate $this->author which gives you the error as author is an object.

like image 152
Felix Kling Avatar answered Nov 04 '22 13:11

Felix Kling


I don't think you need the $ sign when using arrow operator.

like image 27
uvgroovy Avatar answered Nov 04 '22 13:11

uvgroovy


you shouldn't put $ before property names when you access them:

public function insert() {
        $mysql = new DB(debate);
        $this->initializeInsert();
        $query = "INSERT INTO cards VALUES('$this->type','$this->tag','$this->author->last','$this->author->first','$this-$author->qualifications','$this->date->year','$this->date->month','$this->date->day','$this->title', '$this->source', '$this->text')";
        $mysql->execute($query);
    }
like image 3
Sergey Eremin Avatar answered Nov 04 '22 13:11

Sergey Eremin