i need to know how substring_index can only return all rows that match exactly the number of delimiters. In this case the .
For example this query:
SELECT
SUBSTRING_INDEX(ABC, '.', 4)
FROM xxx
only should output when the row is exactly something like this (with 4 words):
aaa.bbb.ccc.ddd
The problem is that: this row is also showed .
aaa.bbb
This will return anything where ABC has 3 . delimiters.
select *
from xxx
where char_length(replace(ABC, '.', '')) + 3 = char_length(ABC)
You would need to multiply 3 by your delimiter length if you had a multi-character string for your delimiter.
Related
Length will be dynamic and i want to find the data before last occurrence of a character in a string in MYSQL
Like strrchr in php
To get last occurrence of _ (underscore) I need to pass length. and here it's 3
mysql> SELECT SUBSTRING_INDEX ('this_is_something_here', '_', 3);
+----------------------------------------------------+
| SUBSTRING_INDEX ('this_is_something_here', '_', 3) |
+----------------------------------------------------+
| this_is_something |
+----------------------------------------------------+
And here, to get last occurrence of _ (underscore) i need to pass length. and here it's 6
mysql> SELECT SUBSTRING_INDEX ('and_this_may_go_like_this_too', '_', 6);
+-----------------------------------------------------------+
| SUBSTRING_INDEX ('and_this_may_go_like_this_too', '_', 6) |
+-----------------------------------------------------------+
| and_this_may_go_like_this |
+-----------------------------------------------------------+
i want data string before last occurrence of _ (underscore) just shown in above example but without passing length.
Note : from above example i want before data of "_here" and "_too"
last occurrence of _ (underscore)
Is there any built-in functionality to achieve this in MySQL?
Thanks in advance amigos.
I didn't quite get your examples, but I think what you want is to pass -1 as the length and prepend the substring prior.
Compare
strrchr('and_this_may_go_like_this_too', '_'); // Returns _too
SELECT SUBSTRING_INDEX('and_this_may_go_like_this_too', '_', -1);
-- Returns too, just need to concatenate `_` so...
SELECT CONCAT('_', SUBSTRING_INDEX('and_this_may_go_like_this_too', '_', -1));
-- Returns _too
If you're looking for the part of the string before and up to the needle, and not from the needle to the end of the string, you can use:
SET #FULL_STRING = 'this_is_something_here';
SELECT LEFT(#FULL_STRING, LENGTH(#FULL_STRING) - LOCATE('_', REVERSE(#FULL_STRING)));
-- Returns this_is_something
Note that the second statement is not what strrchr does.
select reverse(substr(reverse('this_is_something_here'), 1+locate('_', reverse('this_is_something_here'))));
Use reverse, locate, right then replace without using length
Set #str = 'there_is_something';
Select replace(#str,right(#str,locate('_',reverse(#str))),'');
You can write query like this
SELECT SUBSTRING_INDEX('and_this_may_go_like_this_too','_',(LENGTH('and_this_may_go_like_this_too')-LENGTH(REPLACE('and_this_may_go_like_this_too' ,'_',''))) - 1);
I have a database with some codes seperated by / or -, I want to show the left side only, this is an example of the data:
45/84
12/753
68-53
15742-845
2/556
So, i want to get this:
45
12
68
15742
2
I tried using LEFT(), but this search for 1 character only, and returns a warning if the character is not found, this is what LEFT(field,'/') returns.
45
12
(WARNING)
(WARNING)
2
So, what about a REGEXP?
an IF?
any way to ignore from the first non numeric character?
I dont' have more ideas...
Thank you!
Try this:
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(col, '-', 1), '/', 1)
FROM mytable
Demo here
You can do it with this statement. Replace the string '15742/845' with your fieldname
SELECT SUBSTRING_INDEX( REPLACE('15742/845','/','-'), '-', 1)
I am trying to come up with a query that will select a row if a column has at least X words in it. For example:
SELECT * FROM TABLE IF COLUMN <has >= 3 words>
Coming up empty though. Any ideas?
3 words need at least 2 spaces. You can count the spaces in your column
select * from your_table
where length(your_column) - length(replace(your_column, ' ', '')) > 1
I am writing a SQL query to select row, where a field with space separated numbers contains a single number, in this example the 1.
Example fields:
"1 2 3 4 5 11 12 21" - match, contains number one
"2 3 4 6 11 101" - no match, does not contain number one
The best query so far is:
$sql = "SELECT * from " . $table . " WHERE brands REGEXP '[/^1$/]' ORDER BY name ASC;";
Problem is that this REGEXP also finds 11 a match
I read many suggestions on other post, for instance [\d]{1}, but the result always is the same.
Is it possible to accomplish what I want, and how?
You don't need regex: You can use LIKE if you add a space to the front and back of the column:
SELECT * from $table
WHERE CONCAT(' ', brands, ' ') LIKE '% 1 %'
ORDER BY name
Try:
WHERE brands REGEXP '[[:<:]]1[[:>:]]'
[[:<:]] and [[:>:]] match word boundaries before and after a word.
Why not FIND_IN_SET() + REPLACE() ?
SELECT
*
FROM
`table`
WHERE
FIND_IN_SET(1, REPLACE(`brands`, ' ', ','))
ORDER BY
`name` ASC;
I have 2 columns in a table and I would like to roughly report on the total number of words.
Is it possible to run a MySQL query and find out the total number of words down a column.
It would basically be any text separated by a space or multiple space.
Doesn't need to be 100% accurate as its just a general guide.
Is this possible?
Try something like this:
SELECT COUNT(LENGTH(column) - LENGTH(REPLACE(column, ' ', '')) + 1)
FROM table
This will count the number of caracters in your column, and substracts the number of caracters in your column removing all the spaces. Hereby you know how many spaces you have in your row and hereby know how many words there are (roughly because you can also type in a double space, this wil count as two words but you say you want it roughly so this should suffice).
Count simply gives you the number of found rows. You need to use SUM instead.
SELECT SUM(LENGTH(column) - LENGTH(REPLACE(column, ' ', '')) + 1) FROM table
A less rough count:
SELECT LENGTH(column) - LENGTH(REPLACE(column, SPACE(1), ''))
FROM
( SELECT CONCAT(TRIM(column), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(2), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(3), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(5), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(9), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(17), SPACE(1)) AS column
FROM
( SELECT REPLACE(column, SPACE(33), SPACE(1)) AS column
FROM tableX
) AS x
) AS x
) AS x
) AS x
) AS x
) AS x
) AS x
I stumbled upon this post while I was looking for an answer myself and truthfully I've tested all of the answers here and the closest one was #fikre's answer. However, I have concern over data that have leading spaces and/or extra spaces between the words (trailing spaces doesn't seem to have effect to fikre's query during my testing). So, I'm looking for a way to identify any spaces in between words and remove them. While I found a few answers using advanced function (which is beyond my skill set), I did find a very simple way to do it.
tl;dr > #fikre's answer is the only one working for me but I did a minor tweak to ensure that I'll get the most accurate word count.
Query 1 -- This will return 5 "Word Count"
SELECT SUM(LENGTH(input) - LENGTH(REPLACE(input, ' ', '')) + 1) AS "Word Count" FROM
(SELECT TRIM(REPLACE(REPLACE(REPLACE(input,' ','<>'),'><',''),'<>',' ')) AS input
FROM (SELECT ' too late to the party ' AS input) i) r;
Query 2 -- This will return 13 "Word Count"
SELECT SUM(LENGTH(input) - LENGTH(REPLACE(input, ' ', '')) + 1) AS "Word Count"
FROM (SELECT ' too late to the party ' AS input) i;
-- breakdown ' too late to the party '
1 leading space= 1 word count
2 spaces after the first space from the word 'too'= 2 word count
1 space after the first space from the word 'late'= 1 word count
4 spaces after the first space from the word 'the'= 4 word count
trailing space(s) wasn't counted at all.
Total spaces > 1+2+1+4=8 + 5 word count = 13
So, basically if the data row contains even a million spaces in between (disclaimer: an assumption. I've only tested 336,896 spaces), Query 1 will still return Word count=5.
Note: The mid part REPLACE(REPLACE(REPLACE(input,' ','<>'),'><',''),'<>',' ') I took from this answer https://stackoverflow.com/a/55476224/10910692