Thread esskar in der SQL welt (27 answers)
Opened by esskar at 2006-09-26 10:47

renee
 2006-09-26 12:06
#34711 #34711
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Wie ist die inquiry_time dargestellt? Als Integer-Zeitstempel?

Dann kannst Du mit
Code: (dl )
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SELECT
contacts.id AS 'contacts_id',
contacts.sex AS 'contacts_sex',
contacts.givenname AS 'contacts_givenname',
contacts.surname AS 'contacts_surname',
contacts.company AS 'contacts_company',
contacts.email AS 'contacts_email',
contacts.phone AS 'contacts_phone',
contacts.fax AS 'contacts_fax',
contacts.ticket AS 'contacts_ticket',
contacts.position_id AS 'contacts_position_id',
contacts.numofemployees_id AS 'contacts_numofemployees_id',
contacts.contacttype_id AS 'contacts_contacttype_id',
conversation.id AS 'conversation_id',
conversation.contact_id AS 'conversation_contact_id',
conversation.agent_id AS 'conversation_agent_id',
conversation.inquiry AS 'conversation_inquiry',
conversation.inquiry_time AS 'conversation_inquiry_time',
conversation.inquiry_notification_sent AS 'conversation_inquiry_notification_sent',
conversation.answer AS 'conversation_answer',
conversation.answer_time AS 'conversation_answer_time',
conversation.answer_notification_sent AS 'conversation_answer_notification_sent',
jm_contact_positions.value,
jm_contact_numofemployees.value,
jm_contact_contacttypes.value,
MIN(jm_contact_conversation.inquiry_time),
MAX(jm_contact_conversation.inquiry_time),
FROM
jm_contact_contacts contacts
LEFT JOIN
jm_contact_conversation conversation
ON
(contacts.id = conversation.contact_id)
JOIN jm_contact_positions ON (contacts.position_id = jm_contact_positions.id)
JOIN jm_contact_numofemployees ON (contacts.numofemployees_id = jm_contact_numofemployees.id)
JOIN jm_contact_contacttypes ON (jm_contact_contacttypes.id = contacts.contacttype_id)
GROUP BY
jm_contact_contacts.contact_id
ORDER BY
conversation.inquiry_time DESC


Jedenfalls so ungefähr... Mit MIN() und MAX() sollte es funktionieren...
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