PHP can not call a functioin in veiw( $this when not in object context in )

Question

this is the class:

class abc extend def {
     public static function count_all() {
        $sql = "SELECT COUNT(*) FROM " . self::$table_name; 
        $sql= $this -> conn -> prepare($sql);
        $sql -> execute();
        return $row = $sql -> fetchAll();
    }
}

after createing on veiew

$abs = new abs();
$total_count = abc::count_all();
echo $total_count;

it should echo the total count but when i echo this it show the error

Fatal error: Using $this when not in object context in

how i can solve it regards in advance.

http://www.php.net/manual/en/language.oop5.static.php `Because static methods are callable without an instance of the object created, the pseudo-variable $this is not available inside the method declared as static.` — Royal Bg, Jun 05 '14 at 11:29
without static... http://bg2.php.net/manual/en/language.oop5.php — Royal Bg, Jun 05 '14 at 11:30
@Rangad it's a duplicate topic at all :) it shouldn't be answered I suppose :) — Royal Bg, Jun 05 '14 at 11:31

Ocean · Answer 1 · 2014-06-05T12:16:57.540

0

PHP allow you to access instance method by calling on Class (MyClass::instanceMethod) or calling class method on instance. But it is not be done in your code, it is not standard in Object oriented programming.

You are using $this in static method, change your method to this

 public function count_all() {
    /* Use $this->table_name for better design*/
    $sql = "SELECT COUNT(*) FROM " . self::table_name;
    $sql= $this -> conn -> prepare($sql);
    $sql -> execute();
    return $row = $sql -> fetchAll();
}

and then you can call like this

$abs = new abs();
$total_count = $abs->count_all();
echo $total_count;

edited Jun 05 '14 at 12:16

answered Jun 05 '14 at 11:30

Ocean

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And then another error plus the same error – Royal Bg Jun 05 '14 at 11:31
This is not the solution. You still call the method in static context which produces a error about strict standarts and the additionally the same as the op had. – Rangad Jun 05 '14 at 11:32
@MarkBaker I made mistake (cause of copy-paste) fixed – Ocean Jun 05 '14 at 11:38
@Rangad I made mistake :) fixed – Ocean Jun 05 '14 at 11:39
@MarkBaker, Thanks for attention, But where is the problem!? I run it on my computer without any error! – Ocean Jun 05 '14 at 11:50
thanks from all for the attentions, regards – SAR Jun 05 '14 at 11:51

score 0 · Accepted Answer · answered Jun 05 '14 at 11:46

i found it this way am i doing right, regarding php development steps.

public function count_all() {
        $sql = "SELECT COUNT(*) FROM " . self::$table_name; 
        $result = $this -> conn -> prepare($sql);
        $result -> execute();
        return $total_found = $result->fetchColumn();
    }

in the view of this after calling the class.

<?php echo $class->count_all(); ?>

PHP can not call a functioin in veiw( $this when not in object context in )

2 Answers2