In my MySQL database I've some strings like this:
I'm something, Infos S. 12
The pattern I want to search for is: , Infos S., than a number (only digits) and than string end. It's case-insensitive.
How can I search for it and remove it?
I have this I'm something, Infos S. 12 and I want I'm something.
This is what I have so far:
UPDATE my_table SET title_col = REPLACE(title_col, SUBSTRING(title_col, LOCATE(', Infos S. ', title_col), LENGTH(title_col) - LOCATE(')', REVERSE(title_col)) - LOCATE(', Infos S. ', title_col) + 2), '') WHERE title_col LIKE '%(%)%';
How to do the rest?
Edit:
If there's another comma it should get ignored.
Means: I'm, something, Infos S. 12 (note the comma after I'm) should get I'm, something.
You can use regexp to check if the column has the specified pattern and then use substring to update the string with substring upto the ,.
UPDATE my_table
SET title_col = SUBSTRING(title_col,1,locate(', Infos',title_col)-1)
WHERE title_col regexp ', Infos S\\. [0-9]+$'
regexp match is case-insensitive by default. If you want to make it case-sensitive, use regexp binary.
where title_col regexp binary ', Infos S\\. [0-9]+$'
Related
I have a search module with SQL query like this:
SELECT FROM trilers WHERE title '%something%'
And when I search for keyword for example like "spiderman" it returns not found, but when I search for "spider-man" it returns my content (original row in MySQL is "spider-man").
How can I ignore all symbols like -, #, !, : and return content with "spiderman" and "spider-man" keywords at the same time?
What you can do is replace the characters you don't care about before the search takes place.
First iteration would look like this:
SELECT * FROM trilers WHERE REPLACE(title, '-', '') LIKE '%spiderman%'
This would ignore any '-'.
Next you would rap that with another REPLACE to include '#' like this:
SELECT * FROM trilers WHERE REPLACE(REPLACE(title, '-', ''), '#', '') LIKE '%spiderman%'
For all 3 ('!','-','#') you would just increase the Replace with another Replace like this:
SELECT * FROM trilers WHERE REPLACE(REPLACE(REPLACE(title, '-', ''), '#', ''),'!','') LIKE '%spiderman%'
You could try something like
SELECT * FROM trilers WHERE replace(title, '-', '') LIKE '%spiderman%'
The other answers involving using REPLACE are great, but if you don't care what characters appear between "spider" and "man" or how many characters there are between the two strings, you can use an additional wildcard in your expression:
SELECT * FROM Superheroes WHERE HeroName LIKE '%spider%man%';
If you want to match only one character, but allow any character, you can use the _ wildcard, which matches only one character:
SELECT * FROM Superheroes WHERE HeroName LIKE '%spider_man%';
This will match "spideryman" and "spideryman in la la land" but not "spiderysupereliteuberheroman".
If you have a limited number of possible symbols, a way to do it without REPLACE is to use a disjunctive expression:
SELECT * FROM Superheroes WHERE
HeroName LIKE '%spiderman%'
OR
HeroName LIKE '%spider-man%'
OR
HeroName LIKE '%spider#man%'
OR
HeroName LIKE '%spider!man%';
WHERE trilers REGEXP '[[:<:]]spider[-#!:]?man[[:>:]]'
Some discussion:
[[:<:]] -- word boundary
[-#!:] -- character set, matches any of them. ('-' must be first)
[-#!:]? -- optional -- so that 'spiderman' will still match
This, unlike the rest of the answers, will avoid matching
spidermaniac
Also, consider using FULLTEXT.
You should be able to use your search with a small update. You should be able to do something like: SELECT FROM trilers WHERE title LIKE '%spider%'. This should search for anything where spider is before or after something else like the hyphen (-)
I have a field where I save strings, like:
"one two-tree"
"one-two-tree"
"one-two tree"
"one two tree"
When I do a SELECT, I want to retrieve strings that have either "-" or " " (space). Example:
When I do:
Select name from table where name="one two tree"
I want it to bring also results where there is either space or -, in this case returning all string exemplified above.
Is there a wildcard for this?
As far as standard SQL, you must use "or", or "like". depending on what exactly you want. EXAMPLE: Select name from table where name like "one?two?tree".
However, mySQL supports a REGEX extension that will give you what you want:
http://dev.mysql.com/doc/refman/5.7/en/pattern-matching.html
One option is to use replace:
select name
from yourtable
where replace(name, '-', ' ') = 'one two tree'
There is, but it is slow: you can use REGEXP:
SELECT name
FROM table
WHERE name REGEXP 'one[- ]two[- ]tree'
or you can use replacements:
SELECT name
FROM table
WHERE REPLACE(name, '-', ' ') = 'one two three'
but your best bet is to make an additional column where you will have a normalised name (with dashes always replaced with spaces, for example) so you can take advantage of indices.
You can use the LIKE condition with '%' (wildcard operator) for this.
e.g.
SELECT name from table_name WHERE name LIKE '%-%' OR name LIKE '% %'
-- will return all names that have `-` or ` `.
Field X in table may contain special characters e.g hello!World and I would like to know if there is a way to match that with HelloWorld (Ignore case and special characters).
SELECT * FROM table WHERE X='Helloworld'
http://sqlfiddle.com/#!9/2afa1/1
if you need exaclty match of string:
SELECT *
FROM table1
WHERE x REGEXP '^hello[[:punct:],[:space:]]world$';
And if hello world could be a part of larger string:
SELECT *
FROM table1
WHERE x REGEXP 'hello[[:punct:],[:space:]]world';
What you can do is to replace all special characters like this:
SELECT * FROM table WHERE LOWER(REPLACE(X, '!', '')) = LOWER('HelloWorld');
Chain those replacements if you have to replace more:
SELECT * FROM table WHERE LOWER(REPLACE(REPLACE(X, '!', ''), '?', '')) = LOWER('HelloWorld');
If I understood your question right, you need to filter out non-ASCII characters? Please confirm whether this is true. In order to do that, have a look at REGEXP matching as in the comment link and this question.
Try something like
SELECT * FROM `table ` WHERE `X` REGEXP 'Helloworld';
REGEXP 'hello[^[:alpha:]]*world'
Notes:
This finds the string in the middle of other stuff; add ^ and $ to anchor to ends.
This assumes the non-alpha character(s) are between hello and world, not some other spot in the string.
This relies on the relevant collation to do (or not do) case folding.
How do I remove all superfluous full-stop . and semi-colon ; characters from end of last name field values in SQL?
One way to check of the last character is a "full stop" or "semicolon" is to use a substring function to get the last character, and compare that to the characters you are looking for. (There are several ways to do this, for example, using LIKE or REGEXP operator.
If that last character matches, then lop off that last character. One way to do that is to use a substring function. (Use the CHAR_LENGTH function to return the number of characters in the string.)
For example, something like this:
UPDATE mytable t
SET t.last_name = SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1)
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
But, I'd strongly recommend that you test those expressions using a SELECT statement, before running an UPDATE statement.
SELECT t.last_name AS old_val
, SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1) AS new_val
FROM mytable t
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
Substring rows that have a semi-colon or dot :
update emp
set ename = substring(ename, 1, char_length(ename) - 1)
where ename REGEXP '[.;]$';
I have a column named streetaddress that contains
<Street Number> <Street Name>
for example:
15 rue Gontier-Patin
4968 Hillcrest Circle
how can i remove the numbers from the beginning of the row purely in sql?
How about something like this - trim off everything up to and including the first space in strings which start with a number
UPDATE mytable
SET addresscol=SUBSTRING(addresscol, LOCATE(' ', addresscol)+1)
WHERE addresscol REGEXP '^[0-9]';
This is based on #Paul Dixon above but is just for returning results without updating the data for things like sorting:
SELECT IF(address REGEXP '^[0-9]', SUBSTRING(address, LOCATE(' ', address)+1), address) AS addrSort FROM table;
MySQL does not have regexp replace functions so it depends on the values in your column. If the string always begins with a number and then a space, you can do it with the SQl query Paul Dixon posted (doh, he was faster than me :D ).
If you need regular expressions to solve it, you have to do it in application code outside the database.
I think this should do the job on the basis that each entry has it's street address seperated from the house number by a space (" ");
UPDATE table
SET streetaddress = MID(streetaddress, LOCATE(' ', streetaddress) + 1);
I've tested this and it works fine.
You can also use the following if you'd like to extract the house number to a new column previous to the update;
UPDATE table
SET housenumber = MID(streetaddress, 1, LOCATE(' ', streetaddress) - 1);