I have values like this in my column: /1/0/101.00_1234.jpg
Now I want to replace the /1/0 with something else. Problem is, it can differ from row to row. It can be /h/a as well. How could I do that without any additional tools?
Thanks
I'd try the SUBSTRING_INDEX key word:
http://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_substring-index
for example:
SUBSTRING_INDEX(myfield, '/', -3)
Most likely you just need to replace first 4 chars, so use SUBSTRING and CONCAT functions:
CONCAT('/a/b', SUBSTRING(colunm_name, 4))
Write a stored function which searches for the rightmost "/" in the field content and deletes all characters before that position. Then use this stored function to update the field.
Of is this already an "additional tool" -? In this case use an inline function like (not tested)
RIGHT(fieldname,LENGTH(fieldname) - LOCATE('/', reverse(fieldname))
Related
I have a small mysql database with a column which has format of a field as following:
x_1_1,
x_1_2,
x_1_2,
x_2_1,
x_2_12,
x_3_1,
x_3_2,
x_3_11,
I want to extra the data where it matches last '_1'. So if I run a query on above sample dataset, it would return
x_1_1,
x_2_1,
x_3_1,
This should not return x_2_12 or x_3_11.
I tried like '%_1' but it returns x_2_12 and x_3_11 as well.
Thank you!
A simple method is the right() function:
select t.*
from t
where right(field, 2) = '_1';
You can use like but you need to escape the _:
where field like '%$_1' escape '$'
Or use regular expressions:
where field regexp '_1$'
The underscore character has special significance in a LIKE clause. It acts as a wildcard and represent one single character. So you would have to escape it with a backslash:
LIKE '%\_1'
RIGHT does the job too, but it requires that you provide the proper length for the string being sought and is thus less flexible.
Duh, I found the answer.
Use RIGHT (col_name, 2) = '_1'
Thank you!
Hello I have a table Gallery with a field url_immagine and I would like to use a query to replace all values that look like upload/gallery/311/ge_c1966615153f6b2fcf5d84c1e389eea8.jpg in /ge_c1966615153f6b2fcf5d84c1e389eea8.jpg
Unfortunately the a part of the string, the ID (331) is not always the same and therefore can not understand how ...
I tried the regular expression like this:
UPDATE gallery SET url_immagine = replace(url_immagine, 'upload/gallery/.*/', '/')
but it seem not to work.
Combine CONCAT and SUBSTRING_INDEX since you can use last index of "/"
UPDATE gallery
SET url_immagine = (SELECT CONCAT('/',SUBSTRING_INDEX(url_immagine, '/', -1)));
Try that to confirm it's doing what you want :
SELECT CONCAT('/',SUBSTRING_INDEX(url_immagine, '/', -1))
FROM gallery
You can see documentation for the replace function and all other string functions in the mysql manual:
https://dev.mysql.com/doc/refman/5.7/en/string-functions.html#function_replace
It does not mention that replace handles regular expressions, so we an assume it does not, and it is working verbatim and uses the your * to look for the char *.
You see also that there seem not to be a function that does the whole job for you. So you must somehow combine them. The idea of Mateo is probably the right direction.
How to reference to a group using a regex in MySQL?
I tried:
REGEXP '^(.)\1$'
but it does not work.
How to do this?
(Old question, but top search result)
For MySQL 8:
SELECT REGEXP_REPLACE('stackoverflow','(.{5})(.*)','$2$1');
-- "overflowstack"
You can create capture groups with (), and you can refer to them using $1, $2, etc.
For MariaDB, capturing is done in REGEXP_REPLACE with \\1, \\2, etc. respectively.
You can't, there is no way to reference regex capturing groups in MySql.
You can solve this problem by nesting the function calls in your query. Say you have this string in your column:
'100 SOME ST,THE VILLAGES,FL 32163,USA'
and you want to capture the city name. A Capture Group like this would work if MySQL supported it (but it doesn't):
'^[0-9A-Z\s]+,\s*([a-zA-Z\s]*)'
You CAN nest function calls to strip off the part you don't want, and then grab the part you DO want like this:
SELECT REGEXP_SUBSTR(REGEXP_REPLACE(column_name, '^[0-9\\sA-Z]+,', ''), '^[0-9\\sA-Z]+') FROM table_name;
THE VILLAGES
...
I am having the following problem:
I have a table T which has a column Name with names. The names have the following structure:
A\\B\C
You can create on yourself like this:
create table T ( Name varchar(10));
insert into T values ('A\\\\B\\C');
select * from T;
Now if I do this:
select Name from T where Name = 'A\\B\C';
That doesn't work, I need to escape the \ (backslash):
select Name from T where Name = 'A\\\\B\\C';
Fine.
But how do I do this automatically to a string Name?
Something like the following won't do it:
select replace('A\\B\C', '\\', '\\\\');
I get: A\\\BC
Any suggestions?
Many thanks in advance.
You have to use "verbatim string".After using that string your Replace function will
look like this
Replace(#"\", #"\\")
I hope it will help for you.
The literal A\\B\C must be coded as A\\\\A\\C, and the parameters of replace() need escaping too:
select 'A\\\\B\\C', replace('A\\\\B\\C', '\\', '\\\\');
output (see this running on SQLFiddle):
A\\B\C A\\\\B\\C
So there is little point in using replace. These two statements are equivalent:
select Name from T where Name = replace('A\\\\B\\C', '\\', '\\\\');
select Name from T where Name = 'A\\\\B\\C';
Usage of regular expression will solve your problem.
This below query will solve the given example.
1) S\\D\B
select * from T where Name REGEXP '[A-Z]\\\\\\\\[A-Z]\\\\[A-Z]$';
if incase the given example might have more then one char
2) D\\B\ACCC
select * from T where Name REGEXP '[A-Z]{1,5}\\\\\\\\[A-Z]{1,5}\\\\[A-Z]{1,5}$';
note: i have used 5 as the max occurrence of char considering the field size is 10 as its mentioned in the create table query.
We can still generalize it.If this still has not met your expectation feel free to ask for my help.
You're confusing what's IN the database with how you represent that data in SQL statements. When a string in the database contains a special character like \, you have to type \\ to represent that character, because \ is a special character in SQL syntax. You have to do this in INSERT statements, but you also have to do it in the parameters to the REPLACE function. There are never actually any double slashes in the data, they're just part of the UI.
Why do you think you need to double the slashes in the SQL expression? If you're typing queries, you should just double the slashes in your command line. If you're generating the query in a programming language, the best solution is to use prepared statements; the API will take care of proper encoding (prepared statements usually use a binary interface, which deals with the raw data). If, for some reason, you need to perform queries by constructing strings, the language should hopefully provide a function to escape the string. For instance, in PHP you would use mysqli_real_escape_string.
But you can't do it by SQL itself -- if you try to feed the non-escaped string to SQL, data is lost and it can't reconstruct it.
You could use LIKE:
SELECT NAME FROM T WHERE NAME LIKE '%\\\\%';
Not exactly sure by what you mean but, this should work.
select replace('A\\B\C', '\', '\\');
It's basically going to replace \ whereever encountered with \\ :)
Is this what you wanted?
How can I search for "1-800-flowers" by "1800flowers" in MySQL?
I have the data "1-800-flowers", but I want to find it by "1800flowers".
You're probably best off creating a second column that you fill with 1800flowers (replacing all characters you want to ignore) and searching that. That way, you can make full use of indexing.
A quick way to convert all existing data would be
UPDATE table SET columnname_without_hyphens = REPLACE(columnname, "-", "");
If your problem is just ignoring hyphens, I may suggest using REPLACE to eliminate them, like this:
SELECT ... WHERE REPLACE(column, '-', '') ...
Otherwise, if you're looking for strings that "sound alike", you may want to have a look at the SOUNDEX function.
The use of the replace function will kill any ability to use an index on the column, but:
select *
from YourTable
where replace(YourColumn, '-', '') = '1800flowers'