I need a MySQL pattern to match a number, followed by a question mark.
I need something like
... like '%[0-9]?%'
but I have no idea how to create this regular expression.
http://dev.mysql.com/doc/refman/5.0/en/pattern-matching.html does not help.
Thanks!
you could try this:
SELECT * FROM YourTable WHERE YourField REGEXP '[0-9]\\?'
That will return rows where YourField contains a number followed by a ? anywhere in the value.
If you want it to only match if the whole field is a number followed by a ?. I.e. 9? then you could use this regex instead:
^[0-9]\\?$
I guess you're looking for something like this:
select * from table
where field rlike '[0-9]\\?'
Remember to escape the question mark. Otherwise, it will make the number optional.
Source.
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!
Quick questions, I bet the answer is so simple and I am just being blind.
I want to select from the database all the names that only start with "test_1_1_".
I would guess that I do this with;
SELECT * FROM my_table WHERE names LIKE "test_1_1_%";
This doesn't seem to work.
The results keep showing up as
test_1_1_1
test_1_1_2
test_1_11_1
test_1_11_2
test_1_12_1
test_1_12_2
How can I select with MySQL only the results that start with "test_1_1_"?
Thank you in advance.
Wesley
Underscore has a special meaning in LIKE and needs to be escaped:
SELECT * FROM my_table WHERE names LIKE 'test\_1\_1\_%';
You could also use REGEXP here, and avoid the escaping problem:
SELECT * FROM my_table WHERE names REGEXP '^test_1_1_';
When you use _ inside LIKE, it will mean replace that space with any character.
For example, searching for something like field LIKE 'a_' will result in any field with 2 characters starting with "a".
If you really want to search for the underscore characters, you need to escape the value with \ and your query will look like this: LIKE 'test\_1\_1\_%';
The underscore is a single character wildcard when use with LIKE. So to specifically locate an underscore it needs to be "escaped" as follows:
select
*
from mytable
where names like 'test\_1\_1\_%' escape '\'
To test for literal instances of a wildcard character, precede it by
the escape character. If you do not specify the ESCAPE character, \ is
assumed.
https://dev.mysql.com/doc/refman/8.0/en/string-comparison-functions.html
Can do this:
SELECT * FROM my_table WHERE names LIKE "test\_1\_\1\_%";
A few min ago I found out that mysql accepts regex, and is great becouse I think it can solve my problem, but I don't know how to write it. So, I need something like this:
SELECT name FROM products WHERE name REGEXP 'regex code'
To give a little more explanation, the name must be in this format: 123425HT and not string99-123425HT. The 123425HT and string99-123425HT is taken arbitrary
Please Help, thanks!
Something like that:
SELECT * FROM t2 WHERE str REGEXP "^([0-9]+)([a-zA-Z]{2})$";
This regexp will found strings which starting from any digits and two small or big characters, for example:
123123hd
12345435MF
6572Sg
If you want use only 6 digits change from [0-9]+ to [0-9]{6}
I want to remove the names which may be registered with fake names.
As the developer forgot to put validation on form registration.
Now i want to remove the fake names.
And for checking if that name is fake or not, I am checking if the name content any numbers or not ?
This is my query which i have written but its not working...
SELECT registration.regi_id, student.first_name,
student.cont_no, student.email_id,
registration.college,
registration.event_name,
registration.accomodation
FROM student, registration
WHERE student.stud_id = registration.stud_id
AND student.first_name NOT RLIKE '%[0-9]%'
How to fix this problem ?
Sorry for my language issues,
P.S.
There are many names in "first_name" field like "asdfasdf12323", i don't want that kind of names to be shown on list.
Your column may contain Alphanumeric characters also.YOu need to filter Numbers and Alphanumeric characters both
For Alphanumeric characters Try REGEXP '^[A-Za-z0-9]+$'
For numbers Try REGEXP '[0-9]'
Well as far as the regex is involved, your expression is only looking for a single number. Also, your 'NOT RLIKE' isn't using regex but is doing a basic string search for the literal '[0-9]' I believe. MySql has support for regex, and your last clause would look like so: AND student.first_name NOT REGEXP '[0-9]*'
let's say I have a string in which the words are separated by 1 or more spaces and I want to use that string in and SQL LIKE condition. How do I make my SQL and tell it to match 1 or more blank space character in my string? Is there an SQL wildcard that I can use to do that?
Let me know
If you're just looking to get anything with atleast one blank / whitespace then you can do something like the following WHERE myField LIKE '% %'
If your dialect allows it, use SIMILAR TO, which allows for more flexible matching, including the normal regular expression quantifiers '?', '*' and '+', with grouping indicated by '()'
where entry SIMILAR TO 'hello +there'
will match 'hello there' with any number of spaces between the two words.
I guess in MySQL this is
where entry RLIKE 'hello +there'
I know this is late, but I never found a solution to this in relation to a LIKE question.
There is no way to do what you're wanting within a SQL LIKE. What you would have to do is use REGEXP and [[:space:]] inside your expression.
So to find one or more spaces between two words..
WHERE col REGEXP 'firstword[[:space:]]+secondword'
Another way to match for one or more space would be to use [].
It's done like this.
LIKE '%[ ]%'
This will match one or more spaces.
you can't do this using LIKE but what you can do, if you know this condition can exist in your data, is as you're inserting the data into the table, use regular expression matching to detect it up front and set a flag in a different column created for this purpose.
I just replace the whitespace chars with '%'. Lets say I want to do a LIKE query on a string like this 'I want to query this string with a LIKE'
#search_string = 'I want to query this string with a LIKE'
#search_string = ("%"+#search_string+"%").tr(" ", "%")
#my_query = MyTable.find(:all, :conditions => ['my_column LIKE ?', #search_string])
first I add the '%' to the start and end of string with
("%"+#search_string+"%")
and then replace other remaining whitespace chars with '%' like so
.tr(" ", "%")
http://www.techonthenet.com/sql/like.php
The patterns that you can choose from are:
% allows you to match any string of any length (including zero length)
_ allows you to match on a single character
I think that the question is not asking to match any spaces but to match two strings one a pattern and the other with wrong number of spaces because of typos.
In my case I have to check two fields from different tables one preloaded and the other filled typed by users so sometimes they don't respect 100% the pattern.
The solution was to use LIKE in the join
Select table1.field
from table1
left join table2 on table1.field like('%' + replace(table2.field,' ','%')+'%')
if the condition:
WHERE myField LIKE '%Hello world%'
doesn't work try
WHERE myField LIKE '%Hello%'
and
WHERE myField LIKE '%world%'
this approach is helpful in a few specific use cases, hope this helps.