I have a table structured like below (real data replaced with dummy data):
tableName
+-----------+----------------+
| code | description |
+-----------+----------------+
| A | text here |
| b | blah blah |
| c | sdfsdfsdfsdf |
| d | sdfsfkaljdaklj |
| e | asdkjasdlkjasd |
| f | adskljalkdj |
| g | asdjalkd |
| h | askdjavsd |
+-----------+----------------+
I am trying to do the following command:
SELECT * FROM tableName WHERE description LIKE '__[aeiou]%';
I expect this select statement to return all results from the table where the third character in the description is a, e, i, o, or u.
However, I am getting 0 results for the about query. I know for a fact that valid results which match this pattern (3rd character is a, e, i, o or u) exists in the table.
What is wrong with my query?
Use a regular expression instead of a SQL Server style LIKE pattern:
SELECT *
FROM tableName
WHERE description REGEXP '^..[aeiou].*$';
EDIT:
For those who don't read documentation thoroughly, the documentation says:
The other type of pattern matching provided by MySQL uses extended
regular expressions. When you test for a match for this type of
pattern, use the REGEXP and NOT REGEXP operators (or RLIKE and NOT
RLIKE, which are synonyms).
The following list describes some characteristics of extended regular
expressions:
The list of features, such as [ and ] is in this list. The standard for LIKE has only two wildcards, % and _.
In MySQL the '_' character is a not part of the extended regular expression class and can not be used as such. use RLIKE to use regular expressions
Related
I have this type of string
'160f7a4a-766a-4c23-a155-8bd3f7389f77\', \'63233bfc-b663-4c73-890b-00a48d79c4dc'
In one column and I want like
'160f7a4a-766a-4c23-a155-8bd3f7389f77','63233bfc-b663-4c73-890b-00a48d79c4dc'
This type of result in MySQL
i have to perform query like
SELECT * FROM kapp_staging.kols where `kol_id` in (select REPLACE(json_id,'\'',"'") FROM kapp_staging.news_items
where `id` = '991'))
in where in clause i have subquery and in subquery
i geting
'160f7a4a-766a-4c23-a155-8bd3f7389f77\', \'63233bfc-b663-4c73-890b-00a48d79c4dc'
this type of value
so i need to remove \ from value so my where in query work fine.
i have data like:
Kols table
| id | kol_id | name | data |
|----|---------------------------------------- |---------| ------|
| 1 |160f7a4a-766a-4c23-a155-8bd3f7389f77 | balwant | data |
| 2 |63233bfc-b663-4c73-890b-00a48d79c4dc | vikram | data |
news items
| id | json_id | data |
|----|-----------------------------------------------------------------------------------------|---------|
| 991 | {'\160f7a4a-766a-4c23-a155-8bd3f7389f77\','\160f7a4a-766a-4c23-a155-8bd3f7389f77\'} | data |
I tried many ways but didn't get this response.
Thanks in Advance : )
The backslashes aren't in the data, they're just used to escape the quotes when inserting into the table. So you don't need to remove them.
However, you can't use IN to match values in a comma-delimited list, you need to use FIND_IN_SET(); see Search with comma-separated value mysql
You also need to remove the quotes and curly braces before you can use FIND_IN_SET().
SELECT DISTINCT k.*
FROM kols AS k
JOIN news_items AS n
ON FIND_IN_SET(k.kol_id,
REPLACE(REPLACE(REPLACE(json_id, '{', ''), '}', ''), "'", ''))
DEMO
Things would be much easier if you normalized your data and put the list of IDs into a separate table with one row per ID.
I got a big data (approximately 600,000).
I want the rows with value "word's" will appear.
Special characters will be completely ignored.
TABLE:
| column_value |
| ------------- |
| word's |
| hello |
| world |
QUERY: select * from table where column_value like '%words%'
RESULTS:
| column_value |
| ------------- |
| word's |
I want the rows with special characters will appear and ignore their special characters.
Can you please help me how can we achieve it with fast runtime?
You can use replace to remove the "special" character prior the matching.
SELECT *
FROM table
WHERE replace(column_value, '''', '') LIKE '%words%';
Nest the replace() calls for other characters.
Or you try it with regular expressions.
SELECT *
FROM table
WHERE column_value REGEXP 'w[^a-zA-Z]*o[^a-zA-Z]*r[^a-zA-Z]*d[^a-zA-Z]*s';
[^a-zA-Z]* matches optional characters, that are not a, ..., y and z and not A, ..., Y and Z, so this matches your search word also with any non alphas between the letters.
Or you have a look at the options full text search brings with it. Maybe that can help too.
You must add an index on your column_value.
MySQL doc
I have a table "story" as follows:
+++++++++++++++++++++++++++++++++++++++++++
| id | keywords |
+++++++++++++++++++++++++++++++++++++++++++
| 1 | romance,movie,drama |
| 2 | newmovie,horor,comedy |
| 3 | movie,scifi |
| 4 | newmovie,romance,drama,asia |
| 5 | kids,movie |
+++++++++++++++++++++++++++++++++++++++++++
I try a query to search 'movie' in keywords field as below:
SELECT id FROM story WHERE keywords LIKE '%movie%'
and the result is
1,2,3,4,5
but in this case I wanted the result is 1,3,5 (field value with newmovie not include). Can someone help me how the query to do it?
Thank you for your help..
You want to use find_in_set like this:
SELECT id FROM story WHERE find_in_set('movie', keywords) > 0;
Though you should really consider normalizing your table structure.
In this case, you could've stored one single keyword in one row, then the query would be simply like:
select id from story where keyword = 'movie';
and that would've been the end of it. No heavy string functions needed.
You could have structure like this:
keywords(id, name);
story(story_id,. . ., keyword_id);
then, you could simply join the two like this:
select s.*
from story s
inner join keywords k on s.keyword_id = k.id
where k.name = 'movie';
Your problem is that "newmovie" can be found by "%movie%" you need only search "movie".
I have a SQL table with the following values:
+---------+----------+
| post_id | path |
+---------+----------+
| 1 | 1/ |
| 2 | 1/2/ |
| 3 | 1/2/3/ |
| 4 | 1/2/3/4/ |
| 5 | 1/2/5/ |
+---------+----------+
How can I create a query that would get the path with the exact number of values separated by slashes?
For example, if I wanted all post_ids where the path is exactly 1/%/%/ (where each % represents a single number), meaning return anything of the form 1/2/3/, 1/2/5/, but not 1/2/3/4/.
Here's one option using regexp:
select *
from yourtable
where path regexp '1/[0-9]/[0-9]/$'
SQL Fiddle Demo
There are several ways to do that:
MySQL LIKE operator.
The LIKE operator provides two wildcard characters, the percentage % ( match any string of zero or more characters), and underscore _ ( match any single character ).
SELECT * FROM `table` WHERE `path` LIKE '1/_/_/'
SELECT * FROM `table` WHERE `path` LIKE '1/%/%/'
MySQL Regular Expressions.
SELECT * FROM `table` WHERE `path` regexp '^1/[0-9]/[0-9]/$'
Hierarchical Data in MySQL
Since this structure involves hierarchical data maybe you should consider to change the table structure to something that represents actual hierarchy. http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/ has an excellent tutorial about the subject.
I have a bunch of regular expressions in a MySQL table. I want to know whether a given string matches a part of any regular expression or not.
Eg:
+----+--------------------------------+
| id | regexps |
+----+--------------------------------+
| 1 | foo-[0-9]*\.example\.com |
| 2 | (bar|tux)-[0-9]*\.example\.com |
+----+--------------------------------+
(The regexps attribute is of VARCHAR type)
foo-11.example.com matches the first regexp.
I want a MySQL query that returns the first row with the given string as foo-11
This should do it on MySql:
select * from table t where 'foo-11.example.com' rlike t.data;
There are other ways in PostGreSQL. Here's the link from where I have referenced this:
http://www.tutorialspoint.com/mysql/mysql-regexps.htm
Match a Query to a Regular Expression in SQL?
PS: Using * is tricky though!