Windows Vista ve WAMP2.0 üzerine phpFlickr (son kararlı sürümü) kullanın:

ve ben bu var:

Strict Standards: Assigning the return value of new by reference is deprecated in .../phpFlickr.php on line 91

Strict Standards: Assigning the return value of new by reference is deprecated in .....phpFlickr.php on line 330

Strict Standards: Assigning the return value of new by reference is deprecated in ...phpFlickr.php on line 399

Strict Standards: Assigning the return value of new by reference is deprecated in phpFlickr.php on line 468

Strict Standards: Assigning the return value of new by reference is deprecated in HTTP/Request.php on line 341

Strict Standards: Assigning the return value of new by reference is deprecated in ../Request.php on line 659

Strict Standards: Assigning the return value of new by reference is deprecated in ....quest.php on line 672

Strict Standards: Assigning the return value of new by reference is deprecated in ...../Request.php on line 717

Strict Standards: Assigning the return value of new by reference is deprecated in ....PEAR.php on line 569

Strict Standards: Assigning the return value of new by reference is deprecated in .....PEAR.php on line 572

Ben Symfony kullanıyorum itibariyle;, ben GUI bu uyarı mesajı göstermek istemiyorum ve ben açıklandığı gibidir uyarı mesajı yöntem sync_upload sonra gösterilen sahip bir sayfa benim isteği yönlendirmek istiyorum.

Bu benim kodudur:

require_once("../Flickr/phpFlickr.php");
//authentication
        $f = new phpFlickr("key","secret");
        if($f!=null) echo  "connexion réussie</br>";
        $f->setToken("token");

        $f->enableCache("fs", "../Flickr/cache");

        $myPerms =$f->auth("write");
        echo  "permission définie".$myPerms."</br>";
//upload
        $tags="test1,test2";
        $upload_photo_id=$f->sync_upload($filePhotoName,"www.koutchoumi.com",null,$tags);
        echo $upload_photo_id."_test";
        $photo=$f->photos_getInfo($upload_photo_id);
        print_r($photo) ;
        $url=$f->buildPhotoURL($photo);
        echo "</br>".$url."</br>"  ;