我有一些包含字母数字值的字符串输出。我想从该字符串中只获取数字。我如何通过查询获取此信息？我可以使用哪些MySql功能？
我的查询如下：
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
我希望输出如下：
100683
101313
19924
9072
12368
5888
10308
100664
1
101104
答案 0 :(得分：21)
如果字符串以数字开头，则包含非数字字符，您可以使用CAST()函数或通过添加0隐式将其转换为数字：
SELECT CAST('1234abc' AS UNSIGNED); -- 1234
SELECT '1234abc'+0; -- 1234
要从任意字符串中提取数字，您可以添加function之类的自定义this：
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 ;
定义函数后，您可以在查询中使用它：
SELECT ExtractNumber("abc1234def") AS number; -- 1234
答案 1 :(得分：13)
对于仍在寻找的人，请使用正则表达式：
select REGEXP_SUBSTR(name,"[0-9]+") as amount from `subscriptions`
答案 2 :(得分：2)
基于Eugene Yarmash答案。这是自定义函数的版本，该函数提取一个带两个小数位的小数。有利于价格提取。
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
答案 3 :(得分：1)
在php中试试这个
$string = '9072&h';
echo preg_replace("/[^0-9]/", '', $string);// output: 9072
或点击此链接在MySql中执行此操作 Refer the link
答案 4 :(得分：1)
这是我对Eugene Yarmash对ExtractNumber函数的改进。
它不仅会去除非数字字符，还会去除&#[0-9];之类的HTML实体，它们也应被视为非数字unicode字符。
这是纯MySQL <8上没有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，如果您使用的是UDP的PREG_REPLACE，则可以使用以下行：
RETURN PREG_REPLACE("/[^0-9]/", "", PREG_REPLACE("/&#[0-9]+;/", "", str));
答案 5 :(得分：1)
在这里我成功使用了这个功能：
select REGEXP_REPLACE('abc12.34.56-ghj^-_~#@!', '[^0-9]+', '')
输出：123456
解释：基本上我是在要求 mysql 将 0 到 9 之间的所有“非数字”替换为“”。
答案 6 :(得分：0)
尝试， 查询级别，
SELECT CAST('1&hash' AS UNSIGNED);
for PHP，
echo intval('13213&hash');
答案 7 :(得分：0)
对于任何有类似要求的新人来说，这应该是你所需要的。
select DISTINCT CONVERT(SUBSTRING(referrerURL,71,6), SIGNED) as `foo`
from hotshotsdblog1.annonymoustracking
where advertiserid = 10
limit 10;
答案 8 :(得分：0)
我已将其重写为MemSQL语法：
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;
答案 9 :(得分：0)
我建议使用数据透视表（例如，仅包含从1到至少字符串长度的有序数字向量的表），然后执行以下操作：
SELECT group_concat(c.elem separator '')
from (
select b.elem
from
(
select substr('PAUL123f3211',iter.pos,1) as elem
from (select id as pos from t10) as iter
where iter.pos <= LENGTH('PAUL123f3211')
) b
where b.elem REGEXP '^[0-9]+$') c
答案 10 :(得分：0)
可以用 PHP 代替。
foreach ($query_result as &$row) {
$row['column_with_numbers'] = (int) filter_var($query_result['column_with_numbers'], FILTER_SANITIZE_NUMBER_INT);
}
答案 11 :(得分：0)
基于 Eugene Yarmash 和 Martins Balodis 的回答。
就我而言，我不知道源字符串是否包含点作为小数点分隔符。虽然，我知道应该如何处理特定的列。例如。如果 value 出现为“10,00”小时而不是“1.00”，我们知道最后一个分隔符应该被视为点小数点分隔符。为此，我们可以依靠辅助布尔参数来指定最后一个逗号分隔符的行为方式。
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://stackoverflow.com/a/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 ;
以下是一些用法示例：
--
-- 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
-- ==================================================================================