I need to select all the urls with a string pattern (With out using RLIKE). String pattern is
url needs to be ending "%news.html" SO when i use the following query to select
SELECT * FROM `search_news` WHERE `url` LIKE '%news.html'
this also gives following results which are incorrect
news01.html
news8098.html
Why does the LIKE keyword is behaving like this ? What is the best way to do this without using REGEX patterns ?
Probably you are doing something wrong. Tried your scenario, Its working fine.
create table search_news (url varchar(30));
insert into search_news (url) values
('news8098.html'),
('news01.html'),
('news.html');
SELECT * FROM `search_news` WHERE `url` LIKE '%news.html'
And the output is correct as expected
news.html
Related
I have a query like this:
SELECT * FROM mytable where description like %STRING%
The problem is: When I search for JAVAit returns me even the records with JAVAscript.
But, JAVA != JavaScript, right ? How can I work around it ?
MySQL's LIKE operator isn't really suitable to detect an exact single word inside a string. But REGEXP, which supports regular expressions, can handle this. Consider the following query:
SELECT * FROM mytable WHERE description REGEXP '[[:<:]]Java[[:>:]]';
This corresponds to matching the pattern \bJava\b, i.e. the word Java by itself.
Demo
Edit:
If you are trying to execute this query using Laravel, then whereRaw should come in handy:
$results = DB::table('mytable')
->whereRaw('description REGEXP ?', ['[[:<:]]Java[[:>:]]'])
->get();
Hello I have a table Gallery with a field url_immagine and I would like to use a query to replace all values that look like upload/gallery/311/ge_c1966615153f6b2fcf5d84c1e389eea8.jpg in /ge_c1966615153f6b2fcf5d84c1e389eea8.jpg
Unfortunately the a part of the string, the ID (331) is not always the same and therefore can not understand how ...
I tried the regular expression like this:
UPDATE gallery SET url_immagine = replace(url_immagine, 'upload/gallery/.*/', '/')
but it seem not to work.
Combine CONCAT and SUBSTRING_INDEX since you can use last index of "/"
UPDATE gallery
SET url_immagine = (SELECT CONCAT('/',SUBSTRING_INDEX(url_immagine, '/', -1)));
Try that to confirm it's doing what you want :
SELECT CONCAT('/',SUBSTRING_INDEX(url_immagine, '/', -1))
FROM gallery
You can see documentation for the replace function and all other string functions in the mysql manual:
https://dev.mysql.com/doc/refman/5.7/en/string-functions.html#function_replace
It does not mention that replace handles regular expressions, so we an assume it does not, and it is working verbatim and uses the your * to look for the char *.
You see also that there seem not to be a function that does the whole job for you. So you must somehow combine them. The idea of Mateo is probably the right direction.
I have a column in my table name as URL and it contains Multiple value like "https://www.google.com/#q=how+to+make+a+android+app"
and
http://www.bing.com/search?q=how+to+make+a+android+app&go=&qs=n&form=QBLH&pq=how+to+make+a+android+app&sc=8-15&sp=-1&sk=
I want to get data separately in output like
website = https://www.google.com
Keyword = how to make a android app.
Any Idea Plz, How can i get this in MySql.
you can use substr() function in php to cut string. In your case, you can get the characters up to the end of 'https://www.google.com'.
read more at http://ro1.php.net/substr
in case you want to do it directly in your query, you can use substr() as well. Sql syntax goes like this:
SELECT SUBSTR(column, start_position, desired_length) FROM table_name
*SELECT SUBSTR(column_name, 1, 20) FROM table_name;*
I just found a bunch of rogue data in my MYSQL db...
The only way to get to it is via one of the columns - FILE_PATH which contains a slash stripped version of a file path. There are a few rogue files in this set that I need find - they all have the file name "Thumbs.db" but they have a variety of paths
example:
F:DatasetGroupedByFormatsx-fmt-398Thumbs.db
I have a full text index on the field, however the following query doesn't give any returns:
SELECT * FROM main_small WHERE MATCH `FILE_PATH` AGAINST ('Thumbs.db')
Response:
MySQL returned an empty result set (i.e. zero rows). ( Query took 0.0382 sec )
I am unsure whether this is because I have the syntax wrong, or whether the text string needs to be isolated by whitespace/punctuation.
Surely it's
select * from main_small where FILE_PATH like '%Thumbs.db'
However, if not then does MySql Full text Search help?
The problem is that your query thinks 'Thumbs.db' is a whole word. You'll need to find some way to do wildcard searching in order to select those rows. How about:
SELECT * FROM main_small WHERE `FILE_PATH` LIKE '%Thumbs.db'
Just use LIKE:
SELECT * FROM main_small WHERE `FILE_PATH` LIKE '%Thumbs.db'
I trying to search a MySQL database using PHP, The search works fine but I'm looking for a little help with wildcards:
The data in the field (Model) I am searching is: "A4" (215 results)
My search string is:
SELECT * FROM `temp_comp`.`mvl` WHERE `Model` LIKE '%A4 Avant%'
Is there a way I can still search 'A4 Avant' but this will return any fields that contain 'A4' or 'Avant'
The search term is taken from a csv file so I wanted to try and do this without having to split the two words first and search for 'A4' and/or 'Avant', I have tried the following but get no results:
SELECT * FROM `temp_comp`.`mvl` WHERE `Model` LIKE '%A4%Avant%'
As you may have guessed this is not my normal field so any help would be very much appreciated.
SELECT * FROM temp_comp.mvl WHERE (Model LIKE '%A4%') OR (Model LIKE '%Avant%')
If you want to avoid splitting up the test you can use a regexp:
SELECT * FROM temp_comp.mvl WHERE Model REGEXP 'A4|Avant' <<-- Either '%A4% or %Avant%
SELECT * FROM temp_comp.mvl WHERE Model REGEXP 'A4*Avant' <<-- '%A4%Avant%
See the reference
MySQL doesn't have a built-in way to split strings in a query. Splitting them beforehand and then adding n predicates to the where clause is trivial.
SELECT * FROM `temp_comp`.`mvl`
WHERE `Model` LIKE '%A4%' OR `Model` LIKE '%Avant%';
Here's how you can achieve this.
Modify your query like this:
SELECT * FROM `temp_comp`.`mvl` WHERE `Model` LIKE '%A4%' AND Model LIKE '%Avant%'
The above example will look for elements of the search term in the query in such a manner that they should all be present in the record. And if you want to search for records with either of the search terms, you can replace the word AND with OR and it does the job.
And ideally, you would follow these steps to prepare for the query:
Take the input string to search.
Split it by spaces so you have an array of words
Loop through the array and build your search string.
Use the search string in your query.