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luegin 发表于 2017-11-26 18:19:12
感谢分享
长河 发表于 2017-11-27 17:54:52
本帖最后由 论坛小编 于 2017-11-28 22:17 编辑
欣想事成 发表于 2017-9-30 14:33
谢谢分享,支持2.3.4版本吗?


建议购买《开源IT运维管理软件--iTop实施指南》书籍,书内含有大家关心的绝大部分iTop实施和开发问题的解答。
秋韵Jenking 发表于 2018-3-12 17:53:17
致命错误, iTop 无法继续.

错误: Failed to issue SQL query: query = SELECT DISTINCT `Service_itservice`.`id` AS `Serviceid`, `Service_itservice`.`name` AS `Servicename`, `Service_itservice`.`org_id` AS `Serviceorg_id`, `Organization_org_id_itorganization`.`name` AS `Serviceorganization_name`, `Service_itservice`.`servicefamily_id` AS `Serviceservicefamily_id`, `ServiceFamily_servicefamily_id_itservicefamilly`.`name` AS `Serviceservicefamily_name`, `Service_itservice`.`description` AS `Servicedescription`, `Service_itservice`.`status` AS `Servicestatus`, CAST(CONCAT(COALESCE(`Service_itservice`.`name`, '')) AS CHAR) AS `Servicefriendlyname`, CAST(CONCAT(COALESCE(`Organization_org_id_itorganization`.`name`, '')) AS CHAR) AS `Serviceorg_id_friendlyname`, CAST(CONCAT(COALESCE(`ServiceFamily_servicefamily_id_itservicefamilly`.`name`, '')) AS CHAR) AS `Serviceservicefamily_id_friendlyname` FROM `itservice` AS `Service_itservice`INNER JOIN (`itorganization` AS `Organization_org_id_itorganization` ) ON `Service_itservice`.`org_id` = `Organization_org_id_itorganization`.`id`LEFT JOIN (`itservicefamilly` AS `ServiceFamily_servicefamily_id_itservicefamilly` ) ON `Service_itservice`.`servicefamily_id` = `ServiceFamily_servicefamily_id_itservicefamilly`.`id` WHERE 1 ORDER BY `Servicefriendlyname` ASC , mysql_error = Table 'itop.itservicefamilly' doesn't exist, mysql_errno = 1146
问一下这个问题咋解决?
cp3154 发表于 2018-8-12 16:36:59
好东西,谢谢分享
阿波罗地 发表于 2018-12-4 15:37:05
汉化方法是啥?学学汉化
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