I think it means you’re trying to use the IDENTITY ID generator with a table column which is not properly configured. It should be an auto_increment column for the IDENTITY generator to work. Else, the database doesn’t return any generated ID.
@minaz answer was good, but I want to append a little bit more. The problem was caused by –databases keyword. If you omit the keyword, it will not contain any database creation contents. So, Dump without –databases keyword. mysqldump -u username -p database_name > dump.sql And restore it with the target database name. mysql -u … Read more
You ask if you can have more than one primary key field and you most certainly can. You can have only one primary key, but that can consist of as many columns as you need to uniquely identify your rows. Use something like this when you are creating your table: CONSTRAINT pk_PersonID PRIMARY KEY (P_Id,LastName) … Read more
For MySql, Check in the EXTRA column: SELECT * FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = ‘my_table’ AND COLUMN_NAME = ‘my_column’ AND DATA_TYPE = ‘int’ AND COLUMN_DEFAULT IS NULL AND IS_NULLABLE = ‘NO’ AND EXTRA like ‘%auto_increment%’ For Sql Server, use sys.columns and the is_identity column: SELECT is_identity FROM sys.columns WHERE object_id = object_id(‘my_table’) AND name=”my_column”
I think that you are after this: SELECT image_id FROM list WHERE (style_id, style_value) IN ((24,’red’),(25,’big’),(27,’round’)) GROUP BY image_id HAVING count(distinct style_id, style_value)=3 You can’t use AND, because values can’t be 24 red and 25 big and 27 round at the same time in the same row, but you need to check the presence of … Read more
The approach is to have a nested query that has one line per duplicate, and an outer query returning just the count of the results of the inner query. SELECT count(*) AS duplicate_count FROM ( SELECT name FROM tbl GROUP BY name HAVING COUNT(name) > 1 ) AS t
I had to combine some of the previous answers , plus a little more hackery for my data set . This is what works for me , it returns the domain and any sub-domains: SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(target_url, “https://stackoverflow.com/”, 3), ‘://’, -1), “https://stackoverflow.com/”, 1), ‘?’, 1) AS domain Explanation ( cause non-trivial SQL rarely makes sense ): SUBSTRING_INDEX(target_url, … Read more
I’m not sure exactly how you’d check, but if you just want to create it if it doesn’t exist, then you can do CREATE DATABASE IF NOT EXISTS DBname
INSERT INTO `table` (`primary_key`) VALUES (NULL);