153

我有一个包含两个字段(国家和 ISO 代码)的表:

Table1

   field1 - e.g. 'Afghanistan' (without quotes)
   field2 - e.g. 'AF'(without quotes)

在某些行中,第二个字段的开头和/或结尾有空格,这会影响查询。

Table1

   field1 - e.g. 'Afghanistan' (without quotes) 
   field2 - e.g. ' AF' (without quotes but with that space in front)

有没有办法(在 SQL 中)遍历表并在 field2 中查找/替换空格?

4

10 回答 10

312

您正在寻找TRIM

UPDATE FOO set FIELD2 = TRIM(FIELD2);

似乎值得一提的是,TRIM 可以支持多种类型的空格,但一次只能支持一种,并且默认情况下会使用空格。但是,您可以嵌套TRIMs。

 TRIM(BOTH ' ' FROM TRIM(BOTH '\n' FROM column))

如果你真的想在一次调用中去掉所有的空格,你最好和符号REGEXP_REPLACE一起使用。[[:space:]]这是一个例子:

SELECT 
    -- using concat to show that the whitespace is actually removed.
    CONCAT(
         '+', 
         REGEXP_REPLACE(
             '    ha ppy    ', 
             -- This regexp matches 1 or more spaces at the beginning with ^[[:space:]]+
             -- And 1 or more spaces at the end with [[:space:]]+$
             -- By grouping them with `()` and splitting them with the `|`
             -- we match all of the expected values.
             '(^[[:space:]]+|[[:space:]]+$)', 

             -- Replace the above with nothing
             ''
         ), 
         '+') 
    as my_example;
-- outputs +ha ppy+
于 2011-07-28T11:27:34.897 回答
45

我根据您的答案和其他链接组成的一般答案对我有用,我在评论中写了它:

 UPDATE FOO set FIELD2 = TRIM(Replace(Replace(Replace(FIELD2,'\t',''),'\n',''),'\r',''));

等等

因为 trim() 不会删除所有的空格,所以最好替换所有你想要的空格而不是修剪它。

希望我能帮助你分享我的答案:)

于 2014-05-22T13:10:13.710 回答
17

为了清楚起见,TRIM 默认情况下只删除空格(不是所有空格)。这是文档:http ://dev.mysql.com/doc/refman/5.0/en/string-functions.html#function_trim

于 2012-11-19T10:00:36.430 回答
13

请在使用此解决方案之前了解用例:

进行选择查询时修剪不起作用

这有效

select replace(name , ' ','') from test;

虽然这不是

select trim(name) from test;
于 2012-09-24T07:45:13.850 回答
11

当前的答案似乎都不会真正从字符串的开头和结尾删除 100% 的空格。

正如在其他帖子中提到的,默认TRIM只删除空格——而不是制表符、换页符等TRIM。指定其他空白字符的 s 组合可能会提供有限的改进,例如TRIM(BOTH '\r' FROM TRIM(BOTH '\n' FROM TRIM(BOTH '\f' FROM TRIM(BOTH '\t' FROM TRIM(txt))))). 但是这种方法的问题是只能为特定字符指定一个TRIM字符,并且这些字符只能从开头和结尾删除。因此,如果要修剪的字符串类似于\t \t \t \t(即交替的空格和制表符),TRIM则需要更多的 s - 在一般情况下,这可能会无限期地继续下去。

对于轻量级的解决方案,应该可以编写一个简单的用户定义函数 (UDF) 来通过遍历字符串开头和结尾的字符来完成这项工作。但我不会那样做……因为我已经编写了一个更重量级的正则表达式替换器,它也可以完成这项工作 - 并且可能由于其他原因而有用,如本文所述。

演示

Rextester 在线演示。特别是,最后一行显示其他方法失败但正则表达式方法成功。

功能

-- ------------------------------------------------------------------------------------
-- USAGE
-- ------------------------------------------------------------------------------------
-- SELECT reg_replace(<subject>,
--                    <pattern>,
--                    <replacement>,
--                    <greedy>,
--                    <minMatchLen>,
--                    <maxMatchLen>);
-- where:
-- <subject> is the string to look in for doing the replacements
-- <pattern> is the regular expression to match against
-- <replacement> is the replacement string
-- <greedy> is TRUE for greedy matching or FALSE for non-greedy matching
-- <minMatchLen> specifies the minimum match length
-- <maxMatchLen> specifies the maximum match length
-- (minMatchLen and maxMatchLen are used to improve efficiency but are
--  optional and can be set to 0 or NULL if not known/required)
-- Example:
-- SELECT reg_replace(txt, '^[Tt][^ ]* ', 'a', TRUE, 2, 0) FROM tbl;
DROP FUNCTION IF EXISTS reg_replace;
CREATE FUNCTION reg_replace(subject VARCHAR(21845), pattern VARCHAR(21845),
  replacement VARCHAR(21845), greedy BOOLEAN, minMatchLen INT, maxMatchLen INT)
RETURNS VARCHAR(21845) DETERMINISTIC BEGIN 
  DECLARE result, subStr, usePattern VARCHAR(21845); 
  DECLARE startPos, prevStartPos, startInc, len, lenInc INT;
  IF subject REGEXP pattern THEN
    SET result = '';
    -- Sanitize input parameter values
    SET minMatchLen = IF(minMatchLen < 1, 1, minMatchLen);
    SET maxMatchLen = IF(maxMatchLen < 1 OR maxMatchLen > CHAR_LENGTH(subject),
                         CHAR_LENGTH(subject), maxMatchLen);
    -- Set the pattern to use to match an entire string rather than part of a string
    SET usePattern = IF (LEFT(pattern, 1) = '^', pattern, CONCAT('^', pattern));
    SET usePattern = IF (RIGHT(pattern, 1) = '$', usePattern, CONCAT(usePattern, '$'));
    -- Set start position to 1 if pattern starts with ^ or doesn't end with $.
    IF LEFT(pattern, 1) = '^' OR RIGHT(pattern, 1) <> '$' THEN
      SET startPos = 1, startInc = 1;
    -- Otherwise (i.e. pattern ends with $ but doesn't start with ^): Set start position
    -- to the min or max match length from the end (depending on "greedy" flag).
    ELSEIF greedy THEN
      SET startPos = CHAR_LENGTH(subject) - maxMatchLen + 1, startInc = 1;
    ELSE
      SET startPos = CHAR_LENGTH(subject) - minMatchLen + 1, startInc = -1;
    END IF;
    WHILE startPos >= 1 AND startPos <= CHAR_LENGTH(subject)
      AND startPos + minMatchLen - 1 <= CHAR_LENGTH(subject)
      AND !(LEFT(pattern, 1) = '^' AND startPos <> 1)
      AND !(RIGHT(pattern, 1) = '$'
            AND startPos + maxMatchLen - 1 < CHAR_LENGTH(subject)) DO
      -- Set start length to maximum if matching greedily or pattern ends with $.
      -- Otherwise set starting length to the minimum match length.
      IF greedy OR RIGHT(pattern, 1) = '$' THEN
        SET len = LEAST(CHAR_LENGTH(subject) - startPos + 1, maxMatchLen), lenInc = -1;
      ELSE
        SET len = minMatchLen, lenInc = 1;
      END IF;
      SET prevStartPos = startPos;
      lenLoop: WHILE len >= 1 AND len <= maxMatchLen
                 AND startPos + len - 1 <= CHAR_LENGTH(subject)
                 AND !(RIGHT(pattern, 1) = '$' 
                       AND startPos + len - 1 <> CHAR_LENGTH(subject)) DO
        SET subStr = SUBSTRING(subject, startPos, len);
        IF subStr REGEXP usePattern THEN
          SET result = IF(startInc = 1,
                          CONCAT(result, replacement), CONCAT(replacement, result));
          SET startPos = startPos + startInc * len;
          LEAVE lenLoop;
        END IF;
        SET len = len + lenInc;
      END WHILE;
      IF (startPos = prevStartPos) THEN
        SET result = IF(startInc = 1, CONCAT(result, SUBSTRING(subject, startPos, 1)),
                        CONCAT(SUBSTRING(subject, startPos, 1), result));
        SET startPos = startPos + startInc;
      END IF;
    END WHILE;
    IF startInc = 1 AND startPos <= CHAR_LENGTH(subject) THEN
      SET result = CONCAT(result, RIGHT(subject, CHAR_LENGTH(subject) + 1 - startPos));
    ELSEIF startInc = -1 AND startPos >= 1 THEN
      SET result = CONCAT(LEFT(subject, startPos), result);
    END IF;
  ELSE
    SET result = subject;
  END IF;
  RETURN result;
END;

DROP FUNCTION IF EXISTS format_result;
CREATE FUNCTION format_result(result VARCHAR(21845))
RETURNS VARCHAR(21845) DETERMINISTIC BEGIN
  RETURN CONCAT(CONCAT('|', REPLACE(REPLACE(REPLACE(REPLACE(result, '\t', '\\t'), CHAR(12), '\\f'), '\r', '\\r'), '\n', '\\n')), '|');
END;

DROP TABLE IF EXISTS tbl;
CREATE TABLE tbl
AS
SELECT 'Afghanistan' AS txt
UNION ALL
SELECT ' AF' AS txt
UNION ALL
SELECT ' Cayman Islands  ' AS txt
UNION ALL
SELECT CONCAT(CONCAT(CONCAT('\t \t ', CHAR(12)), ' \r\n\t British Virgin Islands \t \t  ', CHAR(12)), ' \r\n') AS txt;     

SELECT format_result(txt) AS txt,
       format_result(TRIM(txt)) AS trim,
       format_result(TRIM(BOTH '\r' FROM TRIM(BOTH '\n' FROM TRIM(BOTH '\f' FROM TRIM(BOTH '\t' FROM TRIM(txt))))))
         AS `trim spaces, tabs, formfeeds and line endings`,
       format_result(reg_replace(reg_replace(txt, '^[[:space:]]+', '', TRUE, 1, 0), '[[:space:]]+$', '', TRUE, 1, 0))
         AS `reg_replace`
FROM tbl;

用法:

SELECT reg_replace(
         reg_replace(txt,
                     '^[[:space:]]+',
                     '',
                     TRUE,
                     1,
                     0),
         '[[:space:]]+$',
         '',
         TRUE,
         1,
         0) AS `trimmed txt`
FROM tbl;
于 2016-08-04T12:21:33.790 回答
4

此语句将删除和更新数据库的字段内容

删除字段值左侧的空格

更新表 SET field1 = LTRIM(field1);

前任。更新成员 SET firstName = LTRIM(firstName);

删除字段值右侧的空格

更新表 SETfield1 = RTRIM(field1);

前任。更新成员 SET firstName = RTRIM(firstName);

于 2013-03-05T06:58:48.400 回答
2

我需要修剪具有名字和姓氏的主键列中的值,所以我不想修剪所有空格,因为这会删除我需要保留的名字和姓氏之间的空格。对我有用的是...

UPDATE `TABLE` SET `FIELD`= TRIM(FIELD);

或者

UPDATE 'TABLE' SET 'FIELD' = RTRIM(FIELD);

或者

UPDATE 'TABLE' SET 'FIELD' = LTRIM(FIELD);

请注意,FIELD 的第一个实例是单引号,而第二个则根本不是引号。我必须这样做,否则它给了我一个语法错误,说当我在引号中时它是一个重复的主键。

于 2017-08-01T08:40:15.077 回答
1

如果需要在select查询中使用trim,也可以使用正则表达式

SELECT * FROM table_name WHERE field RLIKE ' * query-string *'

返回具有类似“查询字符串”字段的行

于 2013-10-09T17:43:33.247 回答
0

您可以使用 ltrim 或 rtrim 清除右侧或左侧或字符串的空格。

于 2013-02-03T19:45:50.560 回答
-5

我知道它已经被接受了,但是对于像我这样寻找“删除所有空格”的人(不仅仅是在字符串的开头和结尾):

select SUBSTRING_INDEX('1234 243', ' ', 1);
// returns '1234'

编辑 2019/6/20:是的,那不好。该函数返回从“第一次出现字符空间时”开始的字符串部分。所以,我猜这样说会删除前导和尾随空格并返回第一个单词:

select SUBSTRING_INDEX(TRIM(' 1234 243'), ' ', 1);
于 2012-12-18T16:44:27.180 回答