ERROR

Нам шкода - хоча ми зробили все можливе: сталася помилка
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 = 10 AND categories_id = 000) AS cat1, (SELECT categories_id FROM categories_description WHERE language_id = 10 AND categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2, (SELECT categories_id FROM categories_description WHERE language_id = 10 AND categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3, (SELECT categories_id FROM categories_description WHERE language_id = 10 AND categories_id = AND SUBSTRING(categories_id,4,1) != "0") AS cat4, (SELECT categories_name FROM categories_description WHERE language_id = 10 AND categories_id = 000) AS cat1name, (SELECT categories_name FROM categories_description WHERE language_id = 10 AND categories_id = 00 AND SUBSTRING(categories_id,2,1) != "0") AS cat2name, (SELECT categories_name FROM categories_description WHERE language_id = 10 AND categories_id = 0 AND SUBSTRING(categories_id,3,1) != "0") AS cat3name, (SELECT categories_name FROM categories_description WHERE language_id = 10 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

Tác giả: Karlheinz Keppler

Ủng hộ
Dr. Karlheinz Keppler, M.A., Jahrgang 1951, ist einer der bekanntesten Gefängnisärzte Deutschlands und seit mehr als 20 Jahren in einem großen Frauengefängnis tätig. Für die Insassinnen ist er nicht nur Arzt, sondern häufig auch der erste Ansprechpartner bei Drogenproblemen, untereinander verübten Übergriffen, existenziellen Ängsten und emotionalen Dramen. Er ist verheiratet und lebt in Lohne, Niedersachsen.




3 Ebooks bởi Karlheinz Keppler

Bìa của Karlheinz Keppler & Heino Stöver: Gefängnismedizin
Karlheinz Keppler & Heino Stöver: Gefängnismedizin
Bessere Gesundheitsversorgung in Haft! Alles zur besonderen Situation des medizinischen Personals in Haftanstalten – der Arzt im Justizvollzug als Gutachter; Haftfähigkeitsprüfung; Problematik der Ag …
PDF
tiếng Đức
€34.99
Bìa của Karlheinz Keppler: Frauenknast
Karlheinz Keppler: Frauenknast
Raue Sitten, starke Gefühle, bewegende Schicksale – hautnah erzählt Kriminell sind Frauen auch – aber anders. Warum sie in Haft kommen und wie es im Frauenknast zugeht, darüber wissen wir so gut wie …
EPUB
tiếng Đức
€11.99
Bìa của Karlheinz Keppler & Wolfgang Lesting: Medizin in Haft
Karlheinz Keppler & Wolfgang Lesting: Medizin in Haft
Die gesundheitliche Versorgung von Gefangenen bzw. Patienten im Justiz- und Maßregelvollzug wird zunehmend Gegenstand fachlicher Auseinandersetzungen, sowohl auf den Gebieten der Medizin, der Pflege, …
PDF
tiếng Đức
€109.99