mysql - sugarcrm 简单查询不起作用

Question

我正在尝试检索联系人所属公司的名称。关系存在于表 account_contacts 中，但是当我尝试调整查询时它会发出声音

SELECT
    accounts.`name`,
    contacts.first_name
FROM
    contacts,
    accounts
INNER JOIN accounts_contacts ON contacts.id = accounts_contacts.contact_id
AND accounts.id = accounts_contacts.account_id

我得到的错误是

[Err] 1054 - Unknown column 'contacts.id' in 'on clause'

更改后:

SELECT
accounts.`name`,
contacts.first_name,
accounts.id
FROM
    contacts
INNER JOIN accounts_contacts ON contacts.id = accounts_contacts.contact_id
JOIN accounts ON accounts.id = accounts_contacts.account_id
WHERE first_name = 'shamraiz' 

您的查询返回 2 行，其中包含我期望的结果。帐户ID不同。但是，我已经重做的查询以您的方式再次实现它不起作用。 accountid 相同，但返回 2 行。

SELECT
    contacts.id AS CONTACTID,
    accounts.id AS ACCOUNTID,
    contacts.first_name,
    contacts.last_name,
    contacts.phone_work,
    contacts.phone_fax,
    contacts.department,
    contacts.title,
    contacts.description,
    contacts.salutation,
    email_addresses.email_address,
    contacts.deleted
FROM
    contacts
INNER JOIN accounts_contacts ON contacts.id = accounts_contacts.contact_id
    JOIN accounts on accounts.id = accounts_contacts.account_id
INNER JOIN email_addr_bean_rel ON contacts.id = email_addr_bean_rel.bean_id
INNER JOIN email_addresses ON email_addresses.id = email_addr_bean_rel.email_address_id

where first_name = 'shamraiz'

下一个查询返回 3 行，但前 2 行是重复的

SELECT
    contacts.id AS CONTACTID,
    accounts.id AS ACCOUNTID,
    contacts.first_name,
    contacts.last_name,
    contacts.phone_work,
    contacts.phone_fax,
    contacts.department,
    contacts.title,
    contacts.description,
    contacts.salutation,
    email_addresses.email_address,
    contacts.deleted
FROM
    contacts
inner JOIN accounts_contacts ON contacts.id = accounts_contacts.contact_id
    left JOIN accounts on accounts.id = accounts_contacts.account_id
left JOIN email_addr_bean_rel ON contacts.id = email_addr_bean_rel.bean_id
left JOIN email_addresses ON email_addresses.id = email_addr_bean_rel.email_address_id
where first_name = 'shamraiz'

来自联系人

SELECT * FROMsugarcrm.contactswhere first_name = 'shamraiz' 返回 2 行

来自 account_contact 关系

SELECT * FROMsugarcrm.accounts_contactswhere contact_id = '17619b5e-db07-fa3b-6748-51a73ef38c5e' 返回 1 行

SELECT * FROMsugarcrm。accounts_contacts其中 contact_id = '003b0000006ZMDXAA4' 返回 1 行。

因此最终查询应该返回 2 个不同的行，因为他们是两个具有相似姓名的联系人，他们加入了 2 个不同的公司。

一个联系人可以属于 1 个公司。

更多调整:

我做了一些修改，但它返回了 1 条记录。应该返回 2。我需要它来提取电子邮件地址是否存在关系的记录。

SELECT
    contacts.id AS CONTACTID,
    accounts.id AS ACCOUNTID,
    contacts.first_name,
    contacts.last_name,
    contacts.phone_work,
    contacts.phone_fax,
    contacts.department,
    contacts.title,
    contacts.description,
    contacts.salutation,
    EM.email_address,
    contacts.deleted,
    EABR.primary_address
FROM
    contacts
LEFT JOIN accounts_contacts ON contacts.id = accounts_contacts.contact_id
JOIN accounts ON accounts.id = accounts_contacts.account_id
LEFT JOIN email_addr_bean_rel EABR ON contacts.id = EABR.bean_id
AND (
    EABR.primary_address = 1
    || (EABR.primary_address IS NOT NULL AND EABR.primary_address != 0)
) 
JOIN email_addresses EM ON EABR.email_address_id = EM.id
WHERE
    contacts.first_name = 'shamraiz'

已解决的答案:

CREATE ALGORITHM=UNDEFINED DEFINER=`root`@`%` SQL SECURITY DEFINER VIEW `view_contacts_sugar_hdb`
AS
select
    `hdb`.`contacts`.`CONTACTID` AS `CONTACTID`,
    `hdb`.`contacts`.`CLIENTID` AS `CLIENTID`,
     concat(`hdb`.`contacts`.`FIRSTNAME`,_utf8' ',coalesce(`hdb`.`contacts`.`INITIALS`,_utf8'')) AS `FIRSTNAME`,
    `hdb`.`contacts`.`LASTNAME` AS `LASTNAME`,
    `hdb`.`contacts`.`PHONE` AS `PHONE`,
    `hdb`.`contacts`.`FAX` AS `FAX`,
    `hdb`.`contacts`.`DEPARTMENT` AS `DEPARTMENT`,
    `hdb`.`contacts`.`TITLE` AS `TITLE`,
    `hdb`.`contacts`.`INFO` AS `INFO`,
    `hdb`.`contacts`.`SALUTATION` AS `SALUTATION`,
    `hdb`.`contacts`.`EMAIL` AS `EMAIL`,
    CASE
        WHEN `hdb`.`contacts`.`ACTIVE` != 0 THEN 0
        ELSE 1
    END DELETED,
    'paradox' AS `SOURCEDATABASE`
    from `hdb`.`contacts`
    union
        SELECT
        contacts.id AS CONTACTID,
        accounts_contacts.account_id AS CLIENTID,
        contacts.first_name AS FIRSTNAME,
        contacts.last_name AS LASTNAME,
        contacts.phone_work AS PHONE,
        contacts.phone_fax AS FAX,
        contacts.department AS DEPARTMENT,
        contacts.title AS TITLE,
        contacts.description AS INFO,
        contacts.salutation AS SALUTATION,
        email_addresses.email_address AS EMAIL,
        contacts.deleted AS DELETED,
        'sugar' AS SOURCEDATABASE
    FROM
        (
            (
                (
                    sugarcrm.contacts
                    LEFT JOIN sugarcrm.email_addr_bean_rel ON (
                        (
                            contacts.id = email_addr_bean_rel.bean_id
                        )
                    )
                    AND (
                        email_addr_bean_rel.primary_address = 1 || (
                            email_addr_bean_rel.primary_address IS NOT NULL
                            AND email_addr_bean_rel.primary_address != 0
                        )
                    )
                )
                LEFT JOIN sugarcrm.accounts_contacts ON (
                    (
                        contacts.id = accounts_contacts.contact_id
                    )
                )
            )
            JOIN sugarcrm.email_addresses ON (
                (
                    email_addr_bean_rel.email_address_id = email_addresses.id
                )
            )
        )
    LEFT JOIN sugarcrm.accounts ON accounts.id = accounts_contacts.account_id
ORDER BY
    `LASTNAME`,
    `FIRSTNAME`;

Answer 1

SELECT
       contacts.id AS CONTACTID,
       accounts.id AS ACCOUNTID,
       contacts.first_name,
       contacts.last_name,
       contacts.phone_work,
       contacts.phone_fax,
       contacts.department,
       contacts.title,
       contacts.description,
       contacts.salutation,
       email_addresses.email_address,
       contacts.deleted
   FROM
      contacts
         INNER JOIN accounts_contacts 
            ON contacts.id = accounts_contacts.contact_id
            JOIN accounts
               ON accounts.id = accounts_contacts.account_id
         INNER JOIN email_addr_bean_rel EABR
            ON contacts.id = EABR.bean_id
            INNER JOIN email_addresses EM
               ON EABR.email_address_id = EM.id
   WHERE 
      contacts.first_name = 'shamraiz' 

就像我帮助你解决的其他问题一样......

一次列出一个表，INNER JOIN(或 LEFT JOIN)到下一个表“ON”，无论这两个表相关的标准是什么......然后，INNER JOIN(或 LEFT JOIN)到关系中的下一个表层次结构。

如果您有同一个人的多个联系人记录，例如不同的帐户和/或电子邮件，您将获得多个记录。