Question

我不确定原因：

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

给了我这个错误：

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

是这样的：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。如果要运行条件INSERT语句，可以使用带有伪表的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

要做得更简单（也更快）

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

不存在代码时出错 - mariaDB

2 个答案: