使用SELECT的结果在表中插入数据

Question

我正在尝试使用select的结果填充表，但我总是收到以下错误：

  

重复条目'0'表示密钥'PRIMARY'SQL = INSERT INTO url(r'^feed/basic/(?P<category>[0-9a-zA-Z]+)/$', BasicFeed(), name='basic_feed'),（fab_bloquer_jours，id，day）VALUES（''，'4'， '2017-12-04 14:55:49'）

我的表fab_bloquer_jours是：

date

我的查询是：

1   id Primary  int(11)
2   day int(13)
3   date    datetime

我通过下拉列表获得$day=$data['fab_bloquer_jours___day']; $query = $db->getQuery(true); $query = "INSERT INTO fab_bloquer_jours (day) SELECT * FROM ( SELECT DATE_ADD('2018-01-01', INTERVAL n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) AS DATE FROM ( SELECT 0 AS num UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 ) AS n1, ( SELECT 0 AS num UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 ) AS n2, ( SELECT 0 AS num UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 ) AS n3, ( SELECT 0 AS num UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL SELECT 9 ) AS n4 ) AS a WHERE DATE >= '2018-01-01' AND DATE < '2019-01-01' AND WEEKDAY(DATE) = '$day'"; $db->setQuery($query); $query=$db->execute(); 值。

我会在这里感谢一些帮助，因为我无法弄清楚如何使它发挥作用。

Answer 1

删除ID及其空值，请参阅下文：

INSERT INTO `fab_bloquer_jours` (`day`,`date`) VALUES ('4','2017-12-04 14:55:49');

在第二个查询中，select语句错误。您正在尝试插入一列，但您的select语句表示全选。您为select语句所做的任何操作都必须在括号内并作为一列。 e.g。

INSERT INTO fab_bloquer_jours (day) VALUES (your SELECT COLUMNNAME FROM TABLE)

Answer 2

尝试将表格中的id列设置为自动递增。然后不要在insert语句中包含它。

因此，对于您的示例，请在将id字段更新为自动增量后尝试此查询：

INSERT INTO fab_bloquer_jours (day,date) VALUES ('4','2017-12-04 14:55:49');