ERROR

Maalesef, elimizden gelenin en iyisini yaptığımız halde: bir hata oluştu
Typedatabase
MessageYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near "AND SUBSTRING(categories_id,4,1) != "0") AS cat4, (SELECT categorie..." at line 5
QuerySELECT (SELECT categories_id FROM categories_description WHERE language_id = 12 AND categories_id = 000) AS cat1, (SELECT categories_id FROM categories_description WHERE language_id = 12 AND categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2, (SELECT categories_id FROM categories_description WHERE language_id = 12 AND categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3, (SELECT categories_id FROM categories_description WHERE language_id = 12 AND categories_id = AND SUBSTRING(categories_id,4,1) != "0") AS cat4, (SELECT categories_name FROM categories_description WHERE language_id = 12 AND categories_id = 000) AS cat1name, (SELECT categories_name FROM categories_description WHERE language_id = 12 AND categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2name, (SELECT categories_name FROM categories_description WHERE language_id = 12 AND categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3name, (SELECT categories_name FROM categories_description WHERE language_id = 12 AND categories_id = AND SUBSTRING(categories_id,4,1) != "0") AS cat4name, (SELECT url_text FROM commerce_seo_url_without_language WHERE categories_id = 000) AS cat1url, (SELECT url_text FROM commerce_seo_url_without_language WHERE categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2url, (SELECT url_text FROM commerce_seo_url_without_language WHERE categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3url, (SELECT url_text FROM commerce_seo_url_without_language WHERE categories_id = AND SUBSTRING(categories_id,4,1) != "0") AS cat4url, (SELECT wod_count_products FROM categories WHERE categories_id = 000) AS cat1cnt, (SELECT wod_count_products FROM categories WHERE categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2cnt, (SELECT wod_count_products FROM categories WHERE categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3cnt, (SELECT wod_count_products FROM categories WHERE categories_id = AND SUBSTRING(categories_id,4,1) != "0") AS cat4cnt

Arthur Conan Doyle & Rudyard Kipling 
Foundations of Fiction – Werewolves [EPUB ebook] 
The stories that gave birth to the modern genre craze

Support
Cover von Arthur Conan Doyle & Rudyard Kipling: Foundations of Fiction - Werewolves (ePUB)

A dog may be man’s best friend. It is also one of his worst nightmares. The idea that humans could suddenly present themselves as demonic werewolves can only have come from the deepest, darkest and most diabolical recesses of an author’s mind.In this volume we present a roll-call of classic authors including Rudyard Kipling, Saki, Arthur Conan Doyle, Catherine Crowe and many others who short story by short story establish the building blocks of this horrific yet thrilling genre. Here all manner of characters and narratives weave together to bring a unique yet intricate account of the beginnings of this most troubling of literary genres.

€4.48
Zahlungsmethoden
Dieses Ebook kaufen – und ein weitere GRATIS erhalten!
Sprache Englisch ● Format EPUB ● Seiten 67 ● ISBN 9781803546445 ● Verlag Copyright Group ● Erscheinungsjahr 2023 ● herunterladbar 3 mal ● Währung EUR ● ID 8818378 ● Kopierschutz Adobe DRM
erfordert DRM-fähige Lesetechnologie

Ebooks vom selben Autor / Herausgeber

Arthur Conan DoyleRudyard KiplingSaki

20.968 Ebooks in dieser Kategorie