我有一个包含3列(First_ID,Second_ID,Third_ID
)的表,所有列都是int列。
现在我有3个值,第一个和第三个值是int值(1和0),第二个值是逗号分隔的字符串('188,189,190,191,192,193,194'
)
我应该如何填充表格,如下所示:
1 188 0
1 189 0
1 190 0
1 191 0
1 192 0
1 193 0
1 194 0
我尝试了不同的方法,但无法按照我的意愿使用它。
提前致谢
答案 0 :(得分:17)
最好使用XML,
imgThumbnail.setTag(<your image url>);
new DownloadImage().execute(imgThumbnail);
答案 1 :(得分:13)
使用您在评论中提到的Split()
功能,
-- Variable holding comma separated values
DECLARE @Var VARCHAR(4000);
SET @Var = '188,189,190,191,192,193,194'
-- Test Target Table
DECLARE @Target_Table TABLE (First_ID INT,Second_ID INT,Third_ID INT)
-- Insert statement
INSERT INTO @Target_Table
SELECT 1, CAST(Items AS INT) , 0
FROM dbo.Split(@Var, ',')
-- Test Select
SELECT * FROM @Target_Table
结果集
╔══════════╦═══════════╦══════════╗
║ First_ID ║ Second_ID ║ Third_ID ║
╠══════════╬═══════════╬══════════╣
║ 1 ║ 188 ║ 0 ║
║ 1 ║ 189 ║ 0 ║
║ 1 ║ 190 ║ 0 ║
║ 1 ║ 191 ║ 0 ║
║ 1 ║ 192 ║ 0 ║
║ 1 ║ 193 ║ 0 ║
║ 1 ║ 194 ║ 0 ║
╚══════════╩═══════════╩══════════╝
答案 2 :(得分:2)
从SQL Server 2016开始,您可以使用此功能string_split
DECLARE @Var VARCHAR(4000);
SET @Var = '188,189,190,191,192,193,194'
SELECT 1 as First_ID, value as Second_ID ,0 as Third_ID FROM string_split(@Var,',')
答案 3 :(得分:0)
DECLARE @Table VARCHAR(100)='AAA,BBB,CCC,DDD'
IF OBJECT_ID('[Comma_Split]') IS NOT NULL
DROP TABLE [Comma_Split]
CREATE TABLE [Comma_Split](ID INT IDENTITY(1,1),COL VARCHAR(100))
while LEN(@Table)>0
BEGIN
DECLARE @COMMA INT= CHARINDEX(',', @Table)
IF @COMMA=0 SET @COMMA=LEN(@Table)+1
INSERT INTO [Comma_Split]
SELECT SUBSTRING(@Table,1,@COMMA-1)
SET @COMMA=@COMMA+1
SET @Table=SUBSTRING(@Table,@COMMA,LEN(@Table))
END