[英]Counting the number of occurrences of a substring within a string in PostgreSQL
How can I count the number of occurrences of a substring within a string in PostgreSQL?如何计算 PostgreSQL 中字符串中子字符串的出现次数?
Example:例子:
I have a table我有一张桌子
CREATE TABLE test."user"
(
uid integer NOT NULL,
name text,
result integer,
CONSTRAINT pkey PRIMARY KEY (uid)
)
I want to write a query so that the result
contains column how many occurrences of the substring o
the column name
contains.我想写一个查询,以便
result
包含列怎样子的许多事件o
列name
包含的内容。 For instance, if in one row, name
is hello world
, the column result
should contain 2
, since there are two o
in the string hello world
.例如,如果在一行中,
name
是hello world
,则列result
应包含2
,因为字符串hello world
有两个o
。
In other words, I'm trying to write a query that would take as input:换句话说,我正在尝试编写一个作为输入的查询:
and update the result
column:并更新
result
列:
I am aware of the function regexp_matches
and its g
option, which indicates that the full ( g
= global) string needs to be scanned for the presence of all occurrences of the substring).我知道函数
regexp_matches
及其g
选项,这表明需要扫描完整的( g
= global)字符串以查找所有出现的子字符串)。
Example:例子:
SELECT * FROM regexp_matches('hello world', 'o', 'g');
returns回报
{o}
{o}
and和
SELECT COUNT(*) FROM regexp_matches('hello world', 'o', 'g');
returns回报
2
But I don't see how to write an UPDATE
query that would update the result
column in such a way that it would contain how many occurrences of the substring o the column name
contains.但是我不知道如何编写一个
UPDATE
查询来更新result
列,以便它包含列name
包含的子字符串的出现次数。
A common solution is based on this logic: replace the search string with an empty string and divide the difference between old and new length by the length of the search string 一个常见的解决方案是基于这样的逻辑: 用空字符串替换搜索字符串,并将新旧长度之间的差异除以搜索字符串的长度
(CHAR_LENGTH(name) - CHAR_LENGTH(REPLACE(name, 'substring', '')))
/ CHAR_LENGTH('substring')
Hence: 因此:
UPDATE test."user"
SET result =
(CHAR_LENGTH(name) - CHAR_LENGTH(REPLACE(name, 'o', '')))
/ CHAR_LENGTH('o');
A Postgres'y way of doing this converts the string to an array and counts the length of the array (and then subtracts 1): Postgres的做法是将字符串转换为数组并计算数组的长度(然后减去1):
select array_length(string_to_array(name, 'o'), 1) - 1
Note that this works with longer substrings as well. 请注意,这也适用于较长的子串。
Hence: 因此:
update test."user"
set result = array_length(string_to_array(name, 'o'), 1) - 1;
Return count of character, 返回字符数,
SELECT (LENGTH('1.1.1.1') - LENGTH(REPLACE('1.1.1.1','.',''))) AS count
--RETURN COUNT OF CHARACTER '.'
另一种方式:
UPDATE test."user" SET result = length(regexp_replace(name, '[^o]', '', 'g'));
Occcurence_Count = LENGTH(REPLACE(string_to_search,string_to_find,'~'))-LENGTH(REPLACE(string_to_search,string_to_find,''))
This solution is a bit cleaner than many that I have seen, especially with no divisor. 这个解决方案比我见过的许多解决方案更清晰,特别是没有除数。 You can turn this into a function or use within a Select.
您可以将其转换为函数或在Select中使用。
No variables required. 无需变量。 I use tilde as a replacement character, but any character that is not in the dataset will work.
我使用tilde作为替换字符,但任何不在数据集中的字符都可以使用。
SELECT array_length(string_to_array('a long name here', 'o'),1)
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