I'm trying to match phone numbers based on the last 6 digits. The problem is the numbers in the database are in various formats, some have whitespace within the number.
SELECT * FROM users WHERE trim(phone) LIKE '%123456'
Trim only removes the leading and trailing whitespace and doesn't find entries where clients have entered their numbers with whitespace between the numbers:
123 456, 12 34 56, etc.
So how to remove the whitespace within the search? Having the result without whitespace is not enough. Updating the database with replace is not an option either.
Use replace() to substitute all occurrences of ' ' to '' within a string.
mySQL documentation
REPLACE(str,from_str,to_str)
Returns the string str with all occurrences of the string from_str replaced by the string to_str. REPLACE() performs a case-sensitive match when searching for from_str.
SELECT * FROM users WHERE Replace(coalesce(Phone,''), ' ','') LIKE '%123456'
If you can't do it in the where clause, which seems odd to me; just nest it in a select.
Select sub.*
from (Select u.*, Replace(coalesce(Phone,''), ' ','') as ph
from users u) sub
where sub.ph LIKE '%123456'
Related
I am trying to pull a product code from a long set of string formatted like a URL address. The pattern is always 3 letters followed by 3 or 4 numbers (ex. ???### or ???####). I have tried using REGEXP and LIKE syntax, but my results are off for both/I am not sure which operators to use.
The first select statement is close to trimming the URL to show just the code, but oftentimes will show a random string of numbers it may find in the URL string.
The second select statement is more rudimentary, but I am unsure which operators to use.
Which would be the quickest solution?
SELECT columnName, SUBSTR(columnName, LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", columnName), LENGTH(columnName) - LOCATE(columnName REGEXP "[^=\-][a-zA-Z]{3}[\d]{3,4}", REVERSE(columnName))) AS extractedData FROM tableName
SELECT columnName FROM tableName WHERE columnName LIKE '%___###%' OR columnName LIKE '%___####%'
-- Will take a substring of this result as well
Example Data:
randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc
In this case, the desired string is "xyz123" and the location of said pattern is variable based on each entry.
EDIT
SELECT column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), SUBSTR(column, LOCATE(column REGEXP "([a-zA-Z]{3}[0-9]{3,4}$)", column), LENGTH(column) - LOCATE(column REGEXP "^.*[a-zA-Z]{3}[0-9]{3,4}", REVERSE(column))) AS extractData From mainTable
This expression is still not grabbing the right data, but I feel like it may get me closer.
I suggest using
REGEXP_SUBSTR(column, '(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])')
Details:
(?<=[&?]random_code=[^&#]{0,256}-) - immediately on the left, there must be & or &, random_code=, and then zero to 256 chars other than & and # followed with a - char
[a-zA-Z]{3} - three ASCII letters
[0-9]{3,4} - three to four ASCII digits
(?![^&#]) - that are followed either with &, # or end of string.
See the online demo:
WITH cte AS ( SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz123&hello_world=us&etc_etc' val
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz4567&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz89&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-xyz00000&hello_world=us&etc_etc'
UNION ALL
SELECT 'randomwebsite.com/3982356923abcd1ab?random_code=12480712_ABC_DEF_ANOTHER_CODE-aaaaa11111&hello_world=us&etc_etc')
SELECT REGEXP_SUBSTR(val,'(?<=[&?]random_code=[^&#]{0,256}-)[a-zA-Z]{3}[0-9]{3,4}(?![^&#])') output
FROM cte
Output:
I'd make use of capture groups:
(?<=[=\-\\])([a-zA-Z]{3}[\d]{3,4})(?=[&])
I assume with [^=\-] you wanted to capture string with "-","\" or "=" in front but not include those chars in the result. To do that use "positive lookbehind" (?<=.
I also added a lookahead (?= for "&".
If you'd like to fidget more with regex I recommend RegExr
I'm trying to select all rows that contain only alphanumeric characters in MySQL using:
SELECT * FROM table WHERE column REGEXP '[A-Za-z0-9]';
However, it's returning all rows, regardless of the fact that they contain non-alphanumeric characters.
Try this code:
SELECT * FROM table WHERE column REGEXP '^[A-Za-z0-9]+$'
This makes sure that all characters match.
Your statement matches any string that contains a letter or digit anywhere, even if it contains other non-alphanumeric characters. Try this:
SELECT * FROM table WHERE column REGEXP '^[A-Za-z0-9]+$';
^ and $ require the entire string to match rather than just any portion of it, and + looks for 1 or more alphanumberic characters.
You could also use a named character class if you prefer:
SELECT * FROM table WHERE column REGEXP '^[[:alnum:]]+$';
Try this:
REGEXP '^[a-z0-9]+$'
As regexp is not case sensitive except for binary fields.
There is also this:
select m from table where not regexp_like(m, '^[0-9]\d+$')
which selects the rows that contains characters from the column you want (which is m in the example but you can change).
Most of the combinations don't work properly in Oracle platforms but this does. Sharing for future reference.
Try this
select count(*) from table where cast(col as double) is null;
Change the REGEXP to Like
SELECT * FROM table_name WHERE column_name like '%[^a-zA-Z0-9]%'
this one works fine
How do I remove all superfluous full-stop . and semi-colon ; characters from end of last name field values in SQL?
One way to check of the last character is a "full stop" or "semicolon" is to use a substring function to get the last character, and compare that to the characters you are looking for. (There are several ways to do this, for example, using LIKE or REGEXP operator.
If that last character matches, then lop off that last character. One way to do that is to use a substring function. (Use the CHAR_LENGTH function to return the number of characters in the string.)
For example, something like this:
UPDATE mytable t
SET t.last_name = SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1)
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
But, I'd strongly recommend that you test those expressions using a SELECT statement, before running an UPDATE statement.
SELECT t.last_name AS old_val
, SUBSTR(t.last_name,1,CHAR_LENGTH(t.last_name)-1) AS new_val
FROM mytable t
WHERE SUBSTRING(t.last_name,CHAR_LENGTH(t.last_name),1) IN ('.',';')
Substring rows that have a semi-colon or dot :
update emp
set ename = substring(ename, 1, char_length(ename) - 1)
where ename REGEXP '[.;]$';
I need a SELECT query in MYSQL that will retrieve all rows in one table witch field values contain "?" char with one condition: the char is not the last character
Example:
ID Field
1 123??see
2 12?
3 45??78??
Returning rows would then be those from ID 1 and 3 that match the condition given
The only statement I have is:
SELECT *
FROM table
WHERE Field LIKE '%?%'
But, the MySQL query does not solve my problem..
The LIKE expressions also support a wildcard "_" which matches exactly one character.
So you can write an expression like the example below, and know that your "?" will not be the last character in the string. There must be at least one more character.
WHERE intrebare LIKE '%?_%'
Re comment from #JohnRuddell,
Yes, that's true, this will match the string "??" because a "?" exists in a position that is not the last character.
It depends whether the OP means for that to be a match or not. The OP says the string "45??78??" is a match, but it's not clear if they would intend that "4578??" to be a match.
An alternative is to use a regular expression, but this is a little more tricky because you have to escape a literal "?", so it won't be interpreted as a regexp metacharacter. Then also escape the escape character.
WHERE intrebare REGEXP '\\?[^?]'
you can just add an additional where where the last character is not a ?
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND intrebare NOT LIKE '%?'
you could also do it like this
SELECT *
FROM intrebari
WHERE intrebare LIKE '%?%' AND RIGHT(intrebare,1) <> '?'
DEMO
I've been trying to write this query, I need to select the rows where a column has only letters (a-z) and a full stop.
I tried this but it's not working:
SELECT * FROM table WHERE (c1 REGEXP '[^a-zA-Z\.]') = 0
This one would usually work in PHP.
Try:
SELECT * FROM table WHERE c1 REGEXP '^[a-zA-Z.]+$'
The anchor ^ and $ ensure that you are matching the entire string and not part of it. Next the character class [a-zA-Z.] matches a single upper/lower case letter or a period. The + is the quantifier for one or more repetitions of the previous sub-regex, so in this case it allows us to match one or more of either a period or a upper/lower case letter.
More info on regex usage in MySQL