这个查询语句怎么写呀？求助 ~~o(>_<)o ~~

[puzhichen]

CREATE TABLE #phoneNote
(
ID INT ,
PhoneNumber VARCHAR(11)
)

INSERT INTO #phoneNote(ID,[PhoneNumber])
(SELECT 1,'13800000001' UNION
SELECT 2,'13800000001' UNION
SELECT 3,'13800000002' UNION
SELECT 4,'13800000002' UNION
SELECT 5,'13800000002' UNION
SELECT 6,'13800000003' UNION
SELECT 7,'13800000004' UNION
SELECT 8,'13800000005')


希望得到的结果：
phoneNumber 总数
----------- -----------
13800000001 2
13800000002 3
13800000003 1
13800000001 2
13800000000 0

(5 行受影响)

[Vidor]

declare @str varchar(8000) = '13800000001,13800000002,13800000003,13800000000'
set @str = replace(@str, ',', ''' p union all select ''')
exec('select p, cn=isnull(cn,0) from (select '''+@str + ''') a left join (select pn=PhoneNumber, cn=count(1) from #phoneNote group by PhoneNumber) b on p=pn')
/*
phoneNumber 总数
----------- -----------
13800000001 2
13800000002 3
13800000003 1
13800000001 2
13800000000 0
*/

[puzhichen]

[Quote=引用 1 楼 vidor 的回复:]

SQL code
declare @str varchar(8000) = '13800000001,13800000002,13800000003,13800000000'
set @str = replace(@str, ',', ''' p union all select ''')
exec('select p, cn=isnull(cn,0) from (select '''+@st……
[/Quote]

@str 有长度限制哦
只能输入几百个手机号码...

[开着拖拉机泡妞]

中午试试写一个，感觉动态的实现不怎么好
[Quote=引用 2 楼 puzhichen 的回复:]

引用 1 楼 vidor 的回复:

SQL code
declare @str varchar(8000) = '13800000001,13800000002,13800000003,13800000000'
set @str = replace(@str, ',', ''' p union all select ''')
exec('select p, cn=isnull(cn,……
[/Quote]

[SQL777]

[Quote=引用 3 楼 travylee 的回复:]
中午试试写一个，感觉动态的实现不怎么好

引用 2 楼 puzhichen 的回复:

引用 1 楼 vidor 的回复:

SQL code
declare @str varchar(8000) = '13800000001,13800000002,13800000003,13800000000'
set @str = replace(@str, ',', ''' p union……
[/Quote]
不动态的实现也可以用拆分成一个表。再关联。思路差不多都这样

[Vidor]

[Quote=引用 2 楼 puzhichen 的回复:]

引用 1 楼 vidor 的回复:

@str 有长度限制哦
只能输入几百个手机号码...
[/Quote]

我说个方法而已，事实上你都是11位的，用数字表substring(@str,number*12+1,11)辅助拆分不就可以了。
where number between 0 and len(@str)/12

总之输入字符串是要拆成行集。

[Vidor]

declare @str varchar(8000) = '13800000001,13800000002,13800000003,13800000000'
select substring(@str, number*12+1, 11) from master..spt_values where type='P' and number between 0 and len(@str)/12
/*
----------------------
13800000001
13800000002
13800000003
13800000000
*/

[q806294478]

declare @sql varchar(max)='13800000001,13800000002,13800000003,13800000001,13800000000'
declare @str xml
set @str=convert(xml,'<root><v>'+REPLACE(@sql,',','</v><v>')+'</v></root>')
select a.num,COUNT(b.PhoneNumber) from
(select ROW_NUMBER() over(order by getdate()) id,num=N.v.value('.','varchar(100)')  from @str.nodes('/root/v')N(v)) a 
left join #phoneNote b on a.num=b.PhoneNumber group by a.id,a.num
结果显示：
13800000000	0
13800000001	2
13800000001	2
13800000002	3
13800000003	1

[五更琉璃]

CREATE TABLE #phoneNote
(
ID INT ,
PhoneNumber VARCHAR(11)
)

INSERT INTO #phoneNote(ID,[PhoneNumber])
(SELECT 1,'13800000001' UNION
SELECT 2,'13800000001' UNION
SELECT 3,'13800000002' UNION
SELECT 4,'13800000002' UNION
SELECT 5,'13800000002' UNION
SELECT 6,'13800000003' UNION
SELECT 7,'13800000004' UNION
SELECT 8,'13800000005')


CREATE TABLE #phoneNoteSearch
(
PhoneNumber VARCHAR(11)
)
INSERT INTO #phoneNoteSearch(PhoneNumber)
select '13800000001' union all
select '13800000002' union all
select '13800000003' union all
select '13800000001' union all
select '13800000000'


select PhoneNumber,(select count(*) from  #phoneNote where #phoneNote.PhoneNumber=#phoneNoteSearch.PhoneNumber) as 总数 from #phoneNoteSearch 

13800000001	2
13800000002	3
13800000003	1
13800000001	2
13800000000	0

[开着拖拉机泡妞]

[Quote=引用 7 楼 q806294478 的回复:]

SQL code
declare @sql varchar(max)='13800000001,13800000002,13800000003,13800000001,13800000000'
declare @str xml
set @str=convert(xml,'<root><v>'+REPLACE(@sql,',','</v><v>')+'</v></root>')
select a……
[/Quote]

这个好，XML，学习了