i'm trying to to look for all strings that start with the letter D. in my mysql query i have this.
REGEXP '^[D]{4}$'
Problem is it returns everything what stars with D and End with D too.
example:
DDD 123 - true
D 123 - true
DDDD 1234 - true
SSS 123 D - returns true, but should be false.
Any idea what i'm missing?
"...for all strings that start with the letter D"
You can simply do it by using LIKE
WHERE columnName LIKE 'D%'
or this pattern in regex.
WHERE columnName REGEXP '^D.*'
SQLFiddle Demo (for both queries)
select * from tablename where columnname RLIKE '^D'
or
select * from tablename where columnname REGEXP '^D'
Related
I am basic on SQL queries and I need some help.
I have to select all string values which contains number e.g. 7 only on specific position in that string.
For example:
I have string: 987654321 and if on position 3 I will have number 7, then it should be selected.
So in example this string will be selected, because on 3rd position I have number 7.
Is there any SQL function for that, or something which could help me?
EDIT:
Example table
TABLE
Numbers Value
987654321 1
123456789 2
789009871 3
654321092 4
847949372 5
Output:
TABLE
Numbers Value
987654321 1
847949372 5
Statement:
SELECT table.numbers
FROM TABLE
WHERE substr(table.numbers,3,1)='7' <--- what to do here? --->
Many thanks in advance.
For a regex option, you may use MySQL's REGEXP operator:
SELECT *
FROM yourTable
WHERE num REGEXP '^[0-9]{2}7';
On Oracle, you could use REGEXP_LIKE:
SELECT *
FROM yourTable
WHERE REGEXP_LIKE(num, '^[0-9]{2}7');
You should use case statement.
select case when substr(stringcol, 3,1) = '7' then stringcol else "not valid" end as stringcol from <Table Name>
We're trying to use a REGEX expression inside MySQL.
Say we have a 2-column table with 5 rows as follow:
1 marketing
2 marketing1
3 marketing12
4 office5
5 marketing44Tomorrow
I'd like to have a SELECT statement that returns: marketing, marketing1, marketing12. Meaning a string (marketing) followed by nothing or by a number only.
This statement:
select * from ddd
where column_name2 REGEXP 'marketing[0-9]'
doesn't work as it does not return "marketing" alone and it will return "marketing44Tomorrow".
You can use : marketing([0-9]+)?[[:>:]]
`marketing` - any word start with **marketing**
`([0-9]+)` - any digit where....
1. `?` - Maybe there may there not
2. `[[:>:]]` - Must be the last
Result:
SELECT * FROM ddd WHERE column_name2 REGEXP 'marketing([0-9]+)?[[:>:]]'
try this,
select * from ddd where column_name2 REGEXP 'marketing[0-9]$'
As a conclusion, the perfect answer to my question in the MySQL context is:
SELECT * FROM ddd WHERE column_name2 REGEXP 'marketing([0-9]+)?[[:>:]]'
"MJN Belief" got it almost right up here.
This is the example of my current MySQL my_table...
id name code
1 111 XXXX123456XXXXXXXXXXXXXX
2 222 XXXX133456XXXXXXX5XXXXXX
3 333 XXXX123454XXX11XXXXXXABC
Code is a 24 character hexadecimal value where X is the wildcard.
I need to write a query that will return the NAME based on the CODE without a wildcard value X.
The given CODE value will be exact but it should compare the string in place, X could match any character.
For example:
SELECT name FROM my_table where code = '012312345611111111111111';
name
111
SELECT name FROM my_table where code = '000013345622222225123456';
name
222
SELECT name FROM my_table where code = '000123454ABC11234567FABC';
name
333
You can use like for this. Are you aware of the _ wildcard?
select t.*
from t
where #YourCode like replace(t.code, 'X', '_');
Of course, you can use regular expressions too. The regular expression would be: concat('^', replace(t.code, 'X', '.'), '$').
ABC:123 UVW XYZ NN-000
What is the best method to get the value after the last space using substr()? In this case I want to get NN-000 but also be able to get that last value in the case that it's greater than or less than 6 characters.
In Oracle, use SUBSTR and INSTR functions
SELECT SUBSTR('ABC:123 UVW XYZ NN-000', INSTR('ABC:123 UVW XYZ NN-000', ' ', -1))
AS LASTOCCUR
FROM DUAL
RESULT:
| LASTOCCUR |
-------------
| NN-000 |
Refer LIVE DEMO
In MySQL you could use reverse and substring_index:
select data,
rv,
reverse(substring_index(rv, ' ', 1)) yd
from
(
select data,
reverse(data) rv
from yt
) d;
See SQL Fiddle with Demo
In Oracle you could use reverse, substr and instr:
select data,
reverse(substr(rv, 0, instr(rv, ' '))) rv
from
(
select data, reverse(data) rv
from yt
) d
See SQL Fiddle with Demo
Combine the powers of RIGHT(),REVERSE() AND LOCATE()
SELECT RIGHT('ABC:123 UVW XYZ NN-000',LOCATE(' ',REVERSE('ABC:123 UVW XYZ NN-000'))-1)
EDIT: Locate in MYSQL, not CHARINDEX
REVERSE() reverses the string, so that the 'first' space it finds is really the last one. You could use SUBSTRING instead of RIGHT, but if what you're after is at the end of the string, might as well use RIGHT.
I'm using this query to getresults from my database:
MATCH(`Text2`) AGAINST ('$s')
I want to get only results when there is a full match of the string, like when on google when you search between quotes "".
How can I do this with Match/MySQL?
EG: Query is "ab cd"
ID | Text
1 ab cd
2 aab cda
3 aab a cd
Row 1 and 2 should be returned
SELECT FROM your_table WHERE Text2 LIKE '%yourstring%';
Try this::
If you need the wild search irrespective to the cases ::
SELECT FROM LOWER(mytable) WHERE LOWER(Text2) LIKE LOWER('%yourstring%');
You can use REGEXP for this purpose too.
SELECT *
FROM `tableName`
WHERE `columnName` REGEXP 'ab cd';