用SELECT MySQL的INSERT根据匹配列

我想基于另一个表中的一些匹配的列在一个表中填充表的ID的，这里是我的查询：用SELECT MySQL的INSERT根据匹配列

INSERT INTO adverts (publication_id) VALUES (SELECT publications.id FROM publications WHERE publications.name LIKE CONCAT ('%', (adverts.publication), '%'));

1064 - 你必须你的SQL语法错误;检查手册中对应于你的MySQL服务器版本正确的语法使用附近“SELECT publications.id FROM出版物WHERE publications.name LIKE CONCAT（”％”在1号线

不知道在哪里我要去错在这里，任何人可以帮助

下面是我的2台结构：

广告表：

公布ications表：

编辑下面你可以找到两个表中的一些样本数据，最好是什么，我想实现的是插入来自publications表id值到publications_id的基础上，adverts表是否在advert.publication - publications.name列中有匹配项。

广告的示例数据：

出版物样本数据：样本数据的

文本格式：

"56d4085e-654c-47e1-86f0-1e7f5850bebe",,"Saturday 27th February 2016","The Daily Telegraph","188833f6-7405-11e6-96f3-0800274f2cef","2016-09-06","pdf/1/a/1a5a83.pdf","0","2016-02-29 08:59:10","2016-09-06 08:31:50" 
"56d71a59-60cc-4a9b-9170-0eb15850bebe",,"Sunday 28th February 2016","The Scottish Sunday Mail",,"2016-02-28","jpg/a/8/a8d73b.jpg","0","2016-03-02 16:52:41","2016-03-02 16:53:13" 
"56daa0f6-a2c8-4346-ac05-6f825850bebe",,"Saturday 5th March 2016","The Daily Telegraph",,"2016-03-05","pdf/9/e/9eda94.pdf","0","2016-03-05 09:03:50","2016-03-05 09:04:59" 
"56daa30b-f8c4-40aa-8197-6f555850bebe",,"Friday 12th February 2016","Daily Star",,"2016-02-12","pdf/9/2/921346.pdf","0","2016-03-05 09:12:43","2016-03-05 09:13:26" 
"56daa37c-a46c-4446-9fa5-6f825850bebe",,"Sunday 14th February 2016","Sunday Express",,"2016-02-14","pdf/3/b/3b8fc2.pdf","0","2016-03-05 09:14:36","2016-03-05 09:14:36" 
"56daa3ee-441c-4bb0-b9ff-6f555850bebe",,"Saturday 27th February 2016","Daily Express",,"2016-02-27","pdf/e/b/ebb99b.pdf","0","2016-03-05 09:16:30","2016-03-05 09:16:30" 
"56daa44f-4124-41bd-bb34-6fb95850bebe",,"Sunday 28th February 2016","Daily Star Sunday ",,"2016-02-28","pdf/1/e/1e412d.pdf","0","2016-03-05 09:18:07","2016-03-05 09:18:07"

出版物：

"000f371a-7405-11e6-96f3-0800274f2cef","The Daily Telegraph","png/6/9/69e409.png","2016-09-06 00:00:00","2016-09-06 07:43:31" 
"10895d76-7405-11e6-96f3-0800274f2cef","The Scottish Sunday Mail","jpg/4/1/41b6e6.jpg","2016-09-06 00:00:00","2016-09-06 07:44:02" 
"188833f6-7405-11e6-96f3-0800274f2cef","Daily Star","png/d/c/dcb8dd.png","2016-09-06 00:00:00","2016-09-06 07:44:22" 
"212778a5-7405-11e6-96f3-0800274f2cef","Sunday Express","gif/5/2/529a66.gif","2016-09-06 00:00:00","2016-09-06 07:45:57" 
"2fa48a98-7405-11e6-96f3-0800274f2cef","Daily Express","gif/6/c/6c5be4.gif","2016-09-06 00:00:00","2016-09-06 07:46:19" 
"37d0b036-7405-11e6-96f3-0800274f2cef","Daily Star Sunday","png/c/e/ce7c42.png","2016-09-06 00:00:00","2016-09-06 07:46:44" 
"3f78db05-7405-11e6-96f3-0800274f2cef","Scottish Daily Mail","jpg/5/c/5caf2f.jpg","2016-09-06 00:00:00","2016-09-06 07:47:10"

来源

2016-09-06 user3574492

你必须使用一个UPDATE不是INSERT声明：

UPDATE Adverts a 
JOIN Publications p ON a.publication LIKE CONCAT('%', (p.name), '%') 
SET a.publication_id = p.id

Demo here

来源

2016-09-06 08:11:03

只是尝试这样做，现在我得到'＃1054 - 'where子句'中的未知列'adverts.publication' – user3574492

@ user3574492 Ooops，我的坏。你不能在语句的SELECT部分使用'adverts'表。你究竟在努力实现什么？ –

我有一堆出版物可以有很多广告（广告属于出版物）。我试图将出版物的出版物ID插入广告所属的广告表中。如果这是有道理的？ – user3574492

正确的语法是：

INSERT INTO adverts (publication_id) 
    SELECT publications.id 
    FROM publications join adverts on <condition> 
    WHERE publications.name LIKE CONCAT ('%', (adverts.publication), '%'));

来源

2016-09-06 08:11:54 Jens

用SELECT MySQL的INSERT根据匹配列

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