Count 0's between 1's - SQL - sql-server

Question

I need a query or function to count the 0's between 1's in a string.
For example:
String1 = '10101101' -> Result=3
String2 = '11111001101' -> Result=1
String3 = '01111111111' -> Result=1
I only need to search for 101 pattern or 01 pattern if its at the beginning of the string.

Answer 1

You may try to decompose the input strings using SUBTRING() and a number table:
SELECT
String, COUNT(*) AS [101Count]
FROM (
SELECT
v.String,
SUBSTRING(v.String, t.No - 1, 1) AS PreviousChar,
SUBSTRING(v.String, t.No, 1) AS CurrentChar,
SUBSTRING(v.String, t.No + 1, 1) AS NextChar
FROM (VALUES
('10101101'),
('11111001101'),
('01111111111')
) v (String)
CROSS APPLY (VALUES (1), (2), (3), (4), (5), (6), (7), (8), (9), (10)) t (No)
) cte
WHERE
CASE WHEN PreviousChar = '' THEN '1' ELSE PreviousChar END = '1' AND
CurrentChar = '0' AND
NextChar = '1'
GROUP BY String
Result:
String 101Count
10101101 3
11111001101 1
01111111111 1
Notes:
The table with alias v is the source table, the table with alias t is the number table. If the input strings have more than 10 characters, use an appropriate number (tally) table.

Answer 2

-- This converts "111101101010111" in "01101010" and "011101000" in "01110"
regexp_replace(field, '^1*(.*)1*0*$', '\1')
-- This converts "01101010" in "0000"
regexp_replace(field, '1', '')
-- This counts the string length, returning 4 for '0000':
LENGTH(field)
-- Put all together:
LENGTH(
regexp_replace(
regexp_replace(field, '^1*(.*)1*0*$', '\1')
, '1', '')
)
Different or more complicated cases require a modification of the regular expression.
Update
For "zeros between 1s" I see now you mean "101" sequences. This is more complicated because of the possibility of having "10101". Suppose you want to count this as 2:
replace 101 with 11011. Now 10101 will become either 1101101 or 1101111011. In either case, you have the "101" sequence well apart and still only have two of them.
replace all 101s with 'X'. You now have 1X11X1
replace [01] with the empty string. You now have XX.
use LENGTH to count the X's.
Any extra special sequence like "01" at the beginning you can convert as first thing with "X1" ("10" at the end would become "1X"), which will then neatly fold back in the above workflow.

Answer 3

By using the LIKE operator with % you can decide how to search a specific string. In this SQL query I am saying that I want every record that starts as 101 or 01.
SELECT ColumnsYouWant FROM TableYouWant
WHERE ColumnYouWant LIKE '101%' OR '01%';
You can simple COUNT the ColumnYouWant, like this:
SELECT COUNT(ColumnYouWant) FROM TableYouWant
WHERE ColumnYouWant LIKE '101%' OR '01%';
Or you can use a method of your backend language to count the results that the first query returns. This count method will depend on the language you are working with.
SQL Documentation for LIKE: https://www.w3schools.com/sql/sql_like.asp
SQL Documentation for COUNT; https://www.w3schools.com/sql/sql_count_avg_sum.asp

Answer 4

The other solutions do not account for all of the characters (max of 11, of the examples shown)
Data
drop table if exists #tTEST;
go
select * INTO #tTEST from (values
(1, '10101101'),
(2, '11111001101'),
(3, '01111111111')) V(id, string);
Query
;with
split_cte as (
select id, n, substring(t.string, v.n, 1) subchar
from #tTEST t
cross apply (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10),
(11),(12),(13),(14),(15),(16),(17),(18),(19),(20)) v(n)
where v.n<=len(t.string)),
lead_lag_cte as (
select id, n, lead(subchar, 1, 9) over (partition by id order by n) lead_c, subchar,
lag(subchar, 1, 9) over (partition by id order by n) lag_c
from split_cte)
select id, sum(case when (lead_c=1 and lag_c=9) then 1 else
case when (lead_c=1 and lag_c=1) then 1 else 0 end end) zero_count
from lead_lag_cte
where subchar=0
group by id;
Results
id zero_count
1 3
2 1
3 1

Answer 5

Another way, perhasp quicker:
DECLARE #T TABLE (ID INT, STRING VARCHAR(32));
INSERT INTO #T
VALUES (1, '10101101'),
(2, '11111001101'),
(3, '01111111111');
SELECT *, LEN(STRING) - LEN(REPLACE(STRING, '0', '')) AS NUMBER_OF_ZERO
FROM #T
Result:
ID STRING NUMBER_OF_ZERO
----------- -------------------------------- --------------
1 10101101 3
2 11111001101 3
3 01111111111 1

Answer 6

select (len(replace('1'+x, '101', '11011')) - len(replace(replace('1'+x, '101', '11011'), '101', '')))/3
from
(
values
('10101101'),
('11111001101'),
('01111111111'),
('01010101010101010101')
) v(x);