SugarCRM的简单查询不工作(sugarcrm simple query not working

我试图检索联系人属于COMPNAY的名称。 关系船舶存在于表account_contacts ，当我尝试怎么过囤它吠叫查询

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

错误我得到的是

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

前后发生了改变：

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行的结果我的期望。 帐户ID是不同的。 然而，查询我已经重做再次实现它自己的方式不起作用。 该ACCOUNTID是相同的，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'

接下来的查询返回3行，但前2 dupplicated

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'

从通讯录

SELECT * FROM SugarCRM的. 接触where first_name = 'shamraiz'返回2行

从account_contact关系

SELECT * FROM SugarCRM的. accounts_contacts where contact_id = '17619b5e-db07-fa3b-6748-51a73ef38c5e'返回1行

SELECT * FROM SugarCRM的. accounts_contacts where contact_id = '003b0000006ZMDXAA4'返回1行。

所以最终的查询应该返回2个不同行，因为它们具有类似名称的两个触点连接到2家不同的公司。

一个联系人可以属于1家公司。

更多的调整：

我已经做了一些修改，但它返回1分的纪录。 应该返回2.我需要它拔出是否存在对电子邮件地址的关系记录。

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'

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' 

就像我帮助你在其他问题...

列出一个表的时间，INNER JOIN（或LEFT JOIN）到下表中“ON”不管标准，这两个表相关......那么，INNER JOIN（或LEFT JOIN）在关系层次结构中的下一个表。

如果您有同一人多个联系人记录，如不同的帐户和/或电子邮件，你会得到多条记录。

Answer 2:

如果您使用SugarCRM的框架，你知道你要找的联系人的ID。 你能避免所有SQL。

$contact = BeanFactory::getBean('Contacts', $id);
$contact->account_name;

希望所有的联系人？

$contact = BeanFactory::getBean('Contacts');
$all = $contact->get_full_list();
foreach ($all as $contact) {
  echo "{$contact->name} {$contact->account_name} \n <br>";
}

Answer 3:

我想你可以使用一个INNER JOIN查询，并使用别名表

SELECT a.`name`, c.first_name
FROM contacts c
INNER JOIN accounts_contacts ac
ON c.id = ac.contact_id
INNER JOIN accounts a
ON ac.contact_id = a.id

前面加上定别名到所有的列将使数据库明白表他们在执行查询。

文章来源: sugarcrm simple query not working