我有一些包含字母数字值的字符串输出。我想从该字符串中仅获取数字。如何通过查询获取?我可以使用哪个MySQL函数？
我的查询类似于：
select DISTINCT SUBSTRING(referrerURL,71,6)
from hotshotsdblog1.annonymoustracking
where advertiserid = 10
limit 10;
输出：
100683
101313
19924&
9072&h
12368&
5888&h
10308&
100664
1&hash
101104
我希望输出结果为：
100683
101313
19924
9072
12368
5888
10308
100664
1
101104
如果字符串以数字开头，然后包含非数字字符，可以使用CAST() 函数或通过添加0来隐式将其转换为数字：
SELECT CAST('1234abc' AS UNSIGNED); -- 1234
SELECT '1234abc'+0; -- 1234
要从一个任意的字符串中提取数字，你可以添加一个自定义 函数 像是 这个：
DELIMITER $$
CREATE FUNCTION `ExtractNumber`(in_string VARCHAR(50))
RETURNS INT
NO SQL
BEGIN
DECLARE ctrNumber VARCHAR(50);
DECLARE finNumber VARCHAR(50) DEFAULT '';
DECLARE sChar VARCHAR(1);
DECLARE inti INTEGER DEFAULT 1;
IF LENGTH(in_string) > 0 THEN
WHILE(inti <= LENGTH(in_string)) DO
SET sChar = SUBSTRING(in_string, inti, 1);
SET ctrNumber = FIND_IN_SET(sChar, '0,1,2,3,4,5,6,7,8,9');
IF ctrNumber > 0 THEN
SET finNumber = CONCAT(finNumber, sChar);
END IF;
SET inti = inti + 1;
END WHILE;
RETURN CAST(finNumber AS UNSIGNED);
ELSE
RETURN 0;
END IF;
END$$
DELIMITER ;
一旦定义函数，您就可以在查询中使用它：
SELECT ExtractNumber("abc1234def") AS number; -- 1234
对于仍在寻找答案的人，请使用正则表达式：
select REGEXP_SUBSTR(name,"[0-9]+") as amount from `subscriptions`
name,"[0-9]+")) AS amount FROM subscriptions - Totte KarlssonSELECT REGEXP_SUBSTR('ab123cd45', '[0-9]+')，结果将是 123，而不是 12345。 - at54321SELECT VERSION(); 来确保。 - Rafael Augusto我使用以下函数成功了：
select REGEXP_REPLACE('abc12.34.56-ghj^-_~#@!', '[^0-9]+', '')
输出结果为：123456
简单说明一下：实际上我是在MySQL中请求将0到9区间内的所有“非数字字符”替换为空。
这是我对Eugene Yarmash的ExtractNumber函数进行的改进。
它不仅会去除非数字字符，还包括像&#[0-9];这样的HTML实体，这也应该被视为非数字Unicode字符。
以下是在纯MySQL <8上没有使用UDP的代码。
CREATE DEFINER = 'user'@'host' FUNCTION `extract_number`(
str CHAR(255)
)
RETURNS char(255) CHARSET utf8mb4 COLLATE utf8mb4_unicode_ci
DETERMINISTIC
NO SQL
SQL SECURITY DEFINER
COMMENT ''
BEGIN
DECLARE tmp VARCHAR(255);
DECLARE res VARCHAR(255) DEFAULT "";
DECLARE chr VARCHAR(1);
DECLARE len INTEGER UNSIGNED DEFAULT LENGTH(str);
DECLARE i INTEGER DEFAULT 1;
IF len > 0 THEN
WHILE i <= len DO
SET chr = SUBSTRING(str, i, 1);
/* remove &#...; */
IF "&" = chr AND "#" = SUBSTRING(str, i+1, 1) THEN
WHILE (i <= len) AND (";" != SUBSTRING(str, i, 1)) DO
SET i = i + 1;
END WHILE;
END IF;
SET tmp = FIND_IN_SET(chr, "0,1,2,3,4,5,6,7,8,9");
IF tmp > 0 THEN
SET res = CONCAT(res, chr);
END IF;
SET i = i + 1;
END WHILE;
RETURN res;
END IF;
RETURN 0;
END;
但是如果您正在使用UDP的PREG_REPLACE，您只需使用以下行：
RETURN PREG_REPLACE("/[^0-9]/", "", PREG_REPLACE("/&#[0-9]+;/", "", str));
根据 Eugene Yarmash 的回答，这里是提取两位小数的自定义函数版本。适用于价格提取。
DELIMITER $$
CREATE FUNCTION `ExtractDecimal`(in_string VARCHAR(255))
RETURNS decimal(15,2)
NO SQL
BEGIN
DECLARE ctrNumber VARCHAR(255);
DECLARE in_string_parsed VARCHAR(255);
DECLARE digitsAndDotsNumber VARCHAR(255) DEFAULT '';
DECLARE finalNumber VARCHAR(255) DEFAULT '';
DECLARE sChar VARCHAR(1);
DECLARE inti INTEGER DEFAULT 1;
DECLARE digitSequenceStarted boolean DEFAULT false;
DECLARE negativeNumber boolean DEFAULT false;
-- FIX FIND_IN_SET cannot find a comma ","
SET in_string_parsed = replace(in_string,',','.');
IF LENGTH(in_string_parsed) > 0 THEN
-- extract digits and dots
WHILE(inti <= LENGTH(in_string_parsed)) DO
SET sChar = SUBSTRING(in_string_parsed, inti, 1);
SET ctrNumber = FIND_IN_SET(sChar, '0,1,2,3,4,5,6,7,8,9,.');
IF ctrNumber > 0 AND (sChar != '.' OR LENGTH(digitsAndDotsNumber) > 0) THEN
-- add first minus if needed
IF digitSequenceStarted = false AND inti > 1 AND SUBSTRING(in_string_parsed, inti-1, 1) = '-' THEN
SET negativeNumber = true;
END IF;
SET digitSequenceStarted = true;
SET digitsAndDotsNumber = CONCAT(digitsAndDotsNumber, sChar);
ELSEIF digitSequenceStarted = true THEN
SET inti = LENGTH(in_string_parsed);
END IF;
SET inti = inti + 1;
END WHILE;
-- remove dots from the end of number list
SET inti = LENGTH(digitsAndDotsNumber);
WHILE(inti > 0) DO
IF(SUBSTRING(digitsAndDotsNumber, inti, 1) = '.') THEN
SET digitsAndDotsNumber = SUBSTRING(digitsAndDotsNumber, 1, inti-1);
SET inti = inti - 1;
ELSE
SET inti = 0;
END IF;
END WHILE;
-- extract decimal
SET inti = 1;
WHILE(inti <= LENGTH(digitsAndDotsNumber)-3) DO
SET sChar = SUBSTRING(digitsAndDotsNumber, inti, 1);
SET ctrNumber = FIND_IN_SET(sChar, '0,1,2,3,4,5,6,7,8,9');
IF ctrNumber > 0 THEN
SET finalNumber = CONCAT(finalNumber, sChar);
END IF;
SET inti = inti + 1;
END WHILE;
SET finalNumber = CONCAT(finalNumber, RIGHT(digitsAndDotsNumber, 3));
IF negativeNumber = true AND LENGTH(finalNumber) > 0 THEN
SET finalNumber = CONCAT('-', finalNumber);
END IF;
IF LENGTH(finalNumber) = 0 THEN
RETURN 0;
END IF;
RETURN CAST(finalNumber AS decimal(15,2));
ELSE
RETURN 0;
END IF;
END$$
DELIMITER ;
测试:
select ExtractDecimal("1234"); -- 1234.00
select ExtractDecimal("12.34"); -- 12.34
select ExtractDecimal("1.234"); -- 1234.00
select ExtractDecimal("1,234"); -- 1234.00
select ExtractDecimal("1,111,234"); -- 11111234.00
select ExtractDecimal("11,112,34"); -- 11112.34
select ExtractDecimal("11,112,34 and 123123"); -- 11112.34
select ExtractDecimal("-1"); -- -1.00
select ExtractDecimal("hello. price is 123"); -- 123.00
select ExtractDecimal("123,45,-"); -- 123.45
根据Eugene Yarmash和Martins Balodis的答案。
在我的情况下，我不知道源字符串是否包含点作为小数分隔符。虽然，我知道应该如何处理特定的列。例如，如果值出现为“10,00”小时而不是“1.00”，我们知道最后一个分隔符字符应该被视为点小数分隔符。为此，我们可以依靠一个次要的布尔参数来指定最后一个逗号分隔符应该如何行为。
DELIMITER $$
CREATE FUNCTION EXTRACT_DECIMAL(
inString VARCHAR(255)
, treatLastCommaAsDot BOOLEAN
) RETURNS varchar(255) CHARSET utf8mb4
NO SQL
DETERMINISTIC
BEGIN
DECLARE ctrNumber VARCHAR(255);
DECLARE inStringParsed VARCHAR(255);
DECLARE digitsAndDotsNumber VARCHAR(255) DEFAULT '';
DECLARE digitsBeforeDotNumber VARCHAR(255) DEFAULT '';
DECLARE digitsAfterDotNumber VARCHAR(255) DEFAULT '';
DECLARE finalNumber VARCHAR(255) DEFAULT '';
DECLARE separatorChar VARCHAR(1) DEFAULT '_';
DECLARE iterChar VARCHAR(1);
DECLARE inti INT DEFAULT 1;
DECLARE digitSequenceStarted BOOLEAN DEFAULT false;
DECLARE negativeNumber BOOLEAN DEFAULT false;
-- FIX FIND_IN_SET cannot find a comma ","
-- We need to separate entered dot from another delimiter characters.
SET inStringParsed = TRIM(REPLACE(REPLACE(inString, ',', separatorChar), ' ', ''));
IF LENGTH(inStringParsed) > 0 THEN
-- Extract digits, dots and delimiter character.
WHILE(inti <= LENGTH(inStringParsed)) DO
-- Might contain MINUS as the first character.
SET iterChar = SUBSTRING(inStringParsed, inti, 1);
SET ctrNumber = FIND_IN_SET(iterChar, CONCAT('0,1,2,3,4,5,6,7,8,9,.,', separatorChar));
-- In case the first extracted character is not '.' and `digitsAndDotsNumber` is set.
IF ctrNumber > 0 AND (iterChar != '.' OR LENGTH(digitsAndDotsNumber) > 0) THEN
-- Add first minus if needed. Note: `inti` at this point will be higher than 1.
IF digitSequenceStarted = FALSE AND inti > 1 AND SUBSTRING(inStringParsed, inti - 1, 1) = '-' THEN
SET negativeNumber = TRUE;
END IF;
SET digitSequenceStarted = TRUE;
SET digitsAndDotsNumber = CONCAT(digitsAndDotsNumber, iterChar);
ELSEIF digitSequenceStarted = true THEN
SET inti = LENGTH(inStringParsed);
END IF;
SET inti = inti + 1;
END WHILE;
-- Search the left part of string until the separator.
-- https://dev59.com/qGct5IYBdhLWcg3wHJzF#43699586
IF (
-- Calculates the amount of delimiter characters.
CHAR_LENGTH(digitsAndDotsNumber)
- CHAR_LENGTH(REPLACE(digitsAndDotsNumber, separatorChar, SPACE(LENGTH(separatorChar)-1)))
) + (
-- Calculates the amount of dot characters.
CHAR_LENGTH(digitsAndDotsNumber)
- CHAR_LENGTH(REPLACE(digitsAndDotsNumber, '.', SPACE(LENGTH(separatorChar)-1)))
) > 0 THEN
-- If dot is present in the string. It doesn't matter for the other characters.
IF LOCATE('.', digitsAndDotsNumber) != FALSE THEN
-- Replace all special characters before the dot.
SET inti = LOCATE('.', digitsAndDotsNumber) - 1;
-- Return the first half of numbers before the last dot.
SET digitsBeforeDotNumber = SUBSTRING(digitsAndDotsNumber, 1, inti);
SET digitsBeforeDotNumber = REPLACE(digitsBeforeDotNumber, separatorChar, '');
SET digitsAfterDotNumber = SUBSTRING(digitsAndDotsNumber, inti + 2, LENGTH(digitsAndDotsNumber) - LENGTH(digitsBeforeDotNumber));
SET digitsAndDotsNumber = CONCAT(digitsBeforeDotNumber, '.', digitsAfterDotNumber);
ELSE
IF treatLastCommaAsDot = TRUE THEN
-- Find occurence of the last delimiter within the string.
SET inti = CHAR_LENGTH(digitsAndDotsNumber) - LOCATE(separatorChar, REVERSE(digitsAndDotsNumber));
-- Break the string into left part until the last occurrence of separator character.
SET digitsBeforeDotNumber = SUBSTRING(digitsAndDotsNumber, 1, inti);
SET digitsBeforeDotNumber = REPLACE(digitsBeforeDotNumber, separatorChar, '');
SET digitsAfterDotNumber = SUBSTRING(digitsAndDotsNumber, inti + 2, LENGTH(digitsAndDotsNumber) - LENGTH(digitsBeforeDotNumber));
-- Remove any dot occurence from the right part.
SET digitsAndDotsNumber = CONCAT(digitsBeforeDotNumber, '.', REPLACE(digitsAfterDotNumber, '.', ''));
ELSE
SET digitsAndDotsNumber = REPLACE(digitsAndDotsNumber, separatorChar, '');
END IF;
END IF;
END IF;
SET finalNumber = digitsAndDotsNumber;
IF negativeNumber = TRUE AND LENGTH(finalNumber) > 0 THEN
SET finalNumber = CONCAT('-', finalNumber);
END IF;
IF LENGTH(finalNumber) = 0 THEN
RETURN 0;
END IF;
RETURN CAST(finalNumber AS DECIMAL(25,5));
ELSE
RETURN 0;
END IF;
END$$
DELIMITER ;
--
-- SELECT EXTRACT_DECIMAL('-711,712,34 and 123123', FALSE); -- -71171234.00000
-- SELECT EXTRACT_DECIMAL('1.234', FALSE); -- 1.23400
-- SELECT EXTRACT_DECIMAL('1,234.00', FALSE); -- 1234.00000
-- SELECT EXTRACT_DECIMAL('14 9999,99', FALSE); -- 14999999.00000
-- SELECT EXTRACT_DECIMAL('-149,999.99', FALSE); -- -149999.99000
-- SELECT EXTRACT_DECIMAL('3 536 500.53', TRUE); -- 3536500.53000
-- SELECT EXTRACT_DECIMAL('3,536,500,53', TRUE); -- 3536500.53000
-- SELECT EXTRACT_DECIMAL("-1"); -- -1.00000
-- SELECT EXTRACT_DECIMAL('2,233,536,50053', TRUE); -- 2233536.50053
-- SELECT EXTRACT_DECIMAL('13.01666667', TRUE); -- 13.01667
-- SELECT EXTRACT_DECIMAL('1,00000000', FALSE); -- 100000000.00000
-- SELECT EXTRACT_DECIMAL('1000', FALSE); -- 1000.00000
-- ==================================================================================
DROP FUNCTION IF EXISTS GetNumeric;
DELIMITER //
CREATE FUNCTION GetNumeric(str CHAR(255)) RETURNS CHAR(255) AS
DECLARE i SMALLINT = 1;
DECLARE len SMALLINT = 1;
DECLARE ret CHAR(255) = '';
DECLARE c CHAR(1);
BEGIN
IF str IS NULL
THEN
RETURN "";
END IF;
WHILE i < CHAR_LENGTH( str ) + 1 LOOP
BEGIN
c = SUBSTRING( str, i, 1 );
IF c BETWEEN '0' AND '9' THEN
ret = CONCAT(ret,c);
END IF;
i = i + 1;
END;
END LOOP;
RETURN ret;
END //
DELIMITER ;
SELECT GetNumeric('abc123def456xyz789') as test;
对于任何有类似需求的新手，这将是您所需的。
select DISTINCT CONVERT(SUBSTRING(referrerURL,71,6), SIGNED) as `foo`
from hotshotsdblog1.annonymoustracking
where advertiserid = 10
limit 10;
可以用PHP来实现。
foreach ($query_result as &$row) {
$row['column_with_numbers'] = (int) filter_var($query_result['column_with_numbers'], FILTER_SANITIZE_NUMBER_INT);
}
尝试，查询级别，
SELECT CAST('1&hash' AS UNSIGNED);
对于PHP，
echo intval('13213&hash');