代码不存在的错误 - mariaDB

Question

我不确定为什么：

INSERT INTO $db.further_assess (taskid) VALUES ('id')
            WHERE NOT EXISTS (SELECT * FROM $db.further_assess where taskid='$id')

给我这个错误：

You have an error in your SQL syntax; check the manual that corresponds to 
your MariaDB server version for the right syntax to use near 'WHERE NOT 
EXISTS (SELECT 1 FROM risk_assessment.further_assess where taskid='222' at line 2

关注了这个：Sql insert if row does not exist

更新：

我的，现在正确，查询：

        INSERT INTO $db.further_assess (taskid, reportid)
         SELECT '$id', '$report_id'
         FROM (SELECT 1) as dummytable
         WHERE NOT EXISTS (SELECT * FROM $db.further_assess where taskid='$id');

Answer 1

INSERT 语句没有 WHERE 子句。如果你想运行条件 INSERT 语句，你可以使用带有虚拟 table:

的 INSERT-SELECT 语句

INSERT INTO $db.further_assess (taskid) 
   SELECT 'id'
   FROM (SELECT 1) as dummytable
   WHERE NOT EXISTS (SELECT * FROM $db.further_assess where taskid='$id')

Answer 2

做起来更简单（也更快）

INSERT IGNORE  INTO $db.further_assess
    (taskid)
    VALUES
    ('$id')

代码不存在的错误 - mariaDB

Error with not exists code - mariaDB

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