Question

我正在进行数据验证，我正在尝试计算字符串中的空格数。我的问题是，当我计算空格时，文本之间有多个空格的任何sting或任何带尾随空格的字符串都不计算在内没有运气，我尝试了以下代码。每个代码给出不同的结果，但不是所需的输出

DECLARE @MyTbl TABLE (ID INT, Name VARCHAR(300))
INSERT INTO @MyTBL VALUES
(1, 'Alfreds Futterkiste'), -- 1 space
(2,'Mike James Ray  '),     -- 4 spaces 1 space between each text and 2 spaces after text
(3,'Hanari  Carnes'),       -- 2 spaces between text
(4,'James Michael')

-- 1
SELECT ID,
LEN(Name)-LEN(REPLACE(Name, ' ', '')) AS Count_Of_Spaces 
FROM  @MyTBL
-- 2
SELECT ID,
LEN(Name + ';')-LEN(REPLACE(Name,' ','')) AS Count_Of_Spaces2 
FROM  @MyTBL
-- 3
SELECT ID,
LEN(Name)-LEN(REPLACE(Name,' ', '')) AS Count_Of_Spaces3  
FROM  @MyTBL

基于第一个查询的当前输出

 ID Count_Of_Spaces
  1    1
  2    2
  3    2
  4    1

期望的输出

 ID Count_Of_Spaces
 1     1
 2     4
 3     2
 4     1

Answer 1

您可以使用DATALENGTH：

SELECT ID,
DATALENGTH(Name)-LEN(REPLACE(Name,' ', '')) AS Count_Of_Spaces
FROM  @MyTBL;

<强> DBFiddle Demo

LEN不计算尾随空格。

如果NVARCHAR则需要除以2.

DECLARE @MyTbl TABLE (ID INT, Name NVARCHAR(300))
INSERT INTO @MyTBL VALUES
(1, 'Alfreds Futterkiste'), -- 1 space
(2,'Mike James Ray  '),     -- 4 spaces 1 space between 
                            -- each text and 2 spaces after text
(3,'Hanari  Carnes'),       -- 2 spaces between text
(4,'James Michael');

SELECT ID,
DATALENGTH(Name)/2-LEN(REPLACE(Name,' ', '')) AS Count_Of_Spaces
FROM  @MyTBL;

<强> DBFiddle Demo2

Answer 2

你在尝试＃2时得到了答案。可能只是没有意识到在查询的第二部分（REPLACE）中进行追加

DECLARE @MyTbl TABLE (ID INT, Name VARCHAR(300))
INSERT INTO @MyTBL VALUES
(1, 'Alfreds Futterkiste'), -- 1 space
(2,'Mike James Ray  '),     -- 4 spaces 1 space between each text and 2 spaces after text
(3,'Hanari  Carnes'),       -- 2 spaces between text
(4,'James Michael')


-- 2
SELECT ID,
LEN(';' + Name + ';')-LEN(REPLACE(';' + Name + ';',' ','')) AS Count_Of_Spaces2 
FROM  @MyTBL

计算字符串中的空格数

2 个答案: