如果main_table的{{1}}空值，我似乎无法获得正确的语法，以便将数据从一个表插入到另一个表中。如果email的{{1}}电子邮件为空，那么我会尝试从main_table插入id的{{1}}：
email在此示例中，id中secondary_table号码MariaDB> SELECT * FROM `main_table`;
+------+---------------------+----------------------------------+------+
| Id | Email | Other | More |
+------+---------------------+----------------------------------+------+
| 1 | user1@somewhere.com | blah | A |
| 2 | | needs email from secondary_table | B |
| 3 | user3@someplace.com | blah | C |
+------+---------------------+----------------------------------+------+
3 rows in set (0.09 sec)
MariaDB> SELECT * FROM `secondary_table`;
+------+---------------------+-------+
| Id | Email | Info |
+------+---------------------+-------+
| 1 | user1@somewhere.com | blah |
| 1 | | blank |
| 2 | user2@something.com | blah |
| 2 | | blank |
| 3 | user3@someplace.com | blah |
| 3 | | blank |
+------+---------------------+-------+
6 rows in set (0.09 sec)
的{{1}}为空id。我试图将2从main_table插入到email id。我试过了：
email还有其他各种各样的失败......也许它很简单，但是我被卡住了！
答案 0 :(得分：1)
查看您的示例似乎您需要更新（使用加入）而不是插入
（假设main_table中的电子邮件字段为空）
update main_table
inner join secondary_table ON main_table.id = secondary_table.id
set main_table.Email = secondary_table.Email
where main_table.Email is null
and secondary_table.email <> ''
或假设在main_table中的电子邮件字段为=＆＃39;＆＃39;
update main_table
inner join secondary_table ON main_table.id = secondary_table.id
set main_table.Email = secondary_table.Email
where main_table.Email = ''
and secondary_table.email <> ''