| Όνομα αρχείου | Σύνδεσμος προβολής και λήψης | Σχετικό εκπαιδευτικό υλικό από το Koha Manual |
|---|---|---|
| Room1-Session1-Week1-Day1- Εισαγωγή στο Koha, Περιβάλλον Προσωπικού, Διαχείριση Χρηστών.mp4 | https://nextcloud-ilsas.alis.uniwa.gr/index.php/s/JMFAC3MGHpqcyjy |
|
| Room1-Session1-Week2-Day1 - Εισαγωγή στο Koha, Περιβάλλον Προσωπικού, Διαχείριση Χρηστών.mp4 | https://nextcloud-ilsas.alis.uniwa.gr/index.php/s/daoZBxxXHcd9dWs | |
| Room1-Session1-Week3-Day1-Εισαγωγή στο Koha, Περιβάλλον Προσωπικού, Διαχείριση Χρηστών.mp4 | https://nextcloud-ilsas.alis.uniwa.gr/index.php/s/xggKFzEF7HTHwAk |
Εικόνα 4
Εικόνα 5
Εικόνα 6
Εικόνα 7
Εικόνα 8
Test 111
SQL Ερώτημα SELECT branchcode,categorycode,count(*) FROM borrowers WHERE branchcode LIKE <<Library|branches:all>> GROUP BY branchcode,categorycode WITH ROLLUP |
SQL Ερώτημα: SELECT COUNT(*) AS new_members FROM borrowers WHERE DATE(dateenrolled) >= DATE_SUB(DATE_FORMAT(CURDATE(), '%Y-%m-01'), INTERVAL 1 MONTH) AND DATE(dateenrolled) < DATE_FORMAT(CURDATE(), '%Y-%m-01') |
SQL Ερώτημα: SELECT cardnumber, surname, firstname, categorycode, email, branchcode FROM borrowers WHERE dateexpiry > CURDATE(); |
SQL Ερώτημα: SELECT borrowers.cardnumber, borrowers.surname, borrowers.firstname, COUNT(issues.issue_id) AS total_borrows FROM borrowers JOIN issues ON borrowers.borrowernumber = issues.borrowernumber GROUP BY borrowers.cardnumber, borrowers.surname, borrowers.firstname ORDER BY total_borrows DESC LIMIT 30; |
SQL Ερώτημα: SELECT b.cardnumber, b.surname, b.firstname, b.branchcode, b.dateenrolled FROM borrowers b LEFT JOIN issues i ON b.borrowernumber = i.borrowernumber WHERE i.issue_id IS NULL ORDER BY b.dateenrolled ASC |
SQL Ερώτημα: SELECT categorycode, COUNT(*) AS total_borrows FROM borrowers b JOIN issues i ON b.borrowernumber = i.borrowernumber GROUP BY categorycode ORDER BY total_borrows DESC; |
SQL Ερώτημα: SELECT b.authorised_by AS librarian_id, u.surname, u.firstname, COUNT(*) AS total_cataloged FROM biblio b JOIN borrowers u ON b.authorised_by = u.borrowernumber WHERE u.cardnumber = '123456789' -- Αντικαταστήστε με τον αριθμό κάρτας του βιβλιοθηκονόμου. Αλλιώς θα χτυπάει. GROUP BY b.authorised_by, u.surname, u.firstname |
SQL Ερώτημα: SELECT b.biblionumber, b.title, b.author, b.datecreated, i.homebranch FROM biblio b JOIN items i ON b.biblionumber = i.biblionumber WHERE b.datecreated >= DATE_SUB(CURDATE(), INTERVAL 1 MONTH) ORDER BY b.datecreated DESC; |
SQL Ερώτημα: SELECT homebranch, COUNT(*) FROM items GROUP BY homebranch WITH ROLLUP |
SQL Ερώτημα: SELECT i.itype AS itemtype, it.description, COUNT(*) AS total_items FROM items i LEFT JOIN itemtypes it ON i.itype = it.itemtype GROUP BY i.itype WITH ROLLUP |
SQL Ερώτημα: SELECT itemtype,COUNT(*) FROM biblioitems GROUP BY itemtype WITH ROLLUP |
SQL Ερώτημα: SELECT b.biblionumber, b.title, b.author, b.datecreated, b.frameworkcode, i.homebranch FROM biblio b LEFT JOIN items i ON b.biblionumber = i.biblionumber WHERE DATE(b.datecreated) >= DATE_SUB(CURDATE(), INTERVAL 1 MONTH) GROUP BY b.biblionumber, b.title, b.author, b.datecreated, b.frameworkcode, i.homebranch |
SQL Ερώτημα: SELECT itemnumber, barcode, biblionumber, dateaccessioned, homebranch, location, itemcallnumber FROM items WHERE DATE(dateaccessioned) >= DATE_SUB(CURDATE(), INTERVAL 1 MONTH) |
SQL Ερώτημα: SELECT homebranch, COUNT(*) AS book_count FROM items GROUP BY homebranch; |
SQL Ερώτημα: SELECT biblio.title, biblio.author, COUNT(issues.issue_id) AS borrow_count FROM issues JOIN items ON issues.itemnumber = items.itemnumber JOIN biblio ON items.biblionumber = biblio.biblionumber GROUP BY biblio.title, biblio.author ORDER BY borrow_count DESC LIMIT 50; |
SQL Ερώτημα: SELECT biblio.title, borrowers.surname, borrowers.firstname, issues.date_due FROM issues JOIN borrowers ON issues.borrowernumber = borrowers.borrowernumber JOIN items ON issues.itemnumber = items.itemnumber JOIN biblio ON items.biblionumber = biblio.biblionumber WHERE issues.date_due < DATE_SUB(CURDATE(), INTERVAL 6 MONTH) ORDER BY issues.date_due ASC |
SQL Ερώτημα: SELECT biblio.title, biblio.author, items.barcode, issues.date_due, borrowers.surname, borrowers.firstname FROM issues JOIN items ON issues.itemnumber = items.itemnumber JOIN biblio ON items.biblionumber = biblio.biblionumber JOIN borrowers ON issues.borrowernumber = borrowers.borrowernumber ORDER BY issues.date_due; |
SQL Ερώτημα: SELECT i.itemnumber, i.borrowernumber, b.cardnumber, b.surname, b.firstname, i.date_due, i.issuedate FROM issues i JOIN borrowers b ON i.borrowernumber = b.borrowernumber WHERE i.date_due < CURDATE() AND i.date_due >= DATE_SUB(CURDATE(), INTERVAL 1 MONTH) ORDER BY i.date_due ASC |
SQL Ερώτημα: SELECT b.cardnumber, b.surname, b.firstname, i.itemnumber, it.barcode, bi.title, bi.author, i.issuedate, i.date_due FROM borrowers b JOIN issues i ON b.borrowernumber = i.borrowernumber JOIN items it ON i.itemnumber = it.itemnumber JOIN biblio bi ON it.biblionumber = bi.biblionumber WHERE i.date_due < CURDATE() |
SQL Ερώτημα: SELECT i.homebranch, COUNT(*) AS total_loans FROM issues iss JOIN items i ON iss.itemnumber = i.itemnumber WHERE DATE(iss.issuedate) >= DATE_SUB(CURDATE(), INTERVAL 1 MONTH) GROUP BY i.homebranch ORDER BY total_loans DESC |
SQL Ερώτημα: SELECT author COUNT(datetime) FROM statistics LEFT JOIN items USING(itemnumber) LEFT JOIN biblio USING(biblionumber) WHERE type='issue' AND author is NOT NULL GROUP BY biblionumber ORDER BY COUNT(datetime) DESC LIMIT 0,10 ; |
SQL Ερώτημα: SELECT borrowers.cardnumber AS 'Αριθμός κάρτας', borrowers.surname AS 'Επώνυμο', borrowers.firstname AS 'Όνομα', accountlines.date AS 'Ημερομηνία χρέωσης', accountlines.amountoutstanding AS 'Οφειλόμενο ποσό', accountlines.amount AS 'Αρχικό ποσό', accountlines.description AS 'Περιγραφή' FROM accountlines JOIN borrowers ON accountlines.borrowernumber = borrowers.borrowernumber WHERE accountlines.debit_type_code = 'F' AND accountlines.date >= DATE_SUB(DATE_FORMAT(NOW(), '%Y-%m-01'), INTERVAL 1 MONTH) AND accountlines.date < DATE_FORMAT(NOW(), '%Y-%m-01') ORDER BY accountlines.date |
όνειρο θερινής νυκτός
όνειρο θερινής ημέρας