Comments on: SQL Puzzle: SQL Advance Query – Split the String and generate count of Each part https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/ BIGData | NoSQL | MSSQL | MySQL | PostgreSQL Wed, 12 Sep 2018 10:27:00 +0000 hourly 1 https://wordpress.org/?v=7.0.4 By: ankush g https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/#comment-6689 Wed, 12 Sep 2018 10:27:00 +0000 https://www.dbrnd.com/?p=6177#comment-6689 Select NAME, count(NAME) from (
Select DISTINCT
trim(regexp_substr(str1, ‘[^;]+’, 1, level)) Name, level
FROM tbl_SplitData
connect by regexp_substr(name, ‘[^;]+’, 1, level) is not null)foo
GROUP BY NAME

]]> By: Ankush https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/#comment-6688 Wed, 12 Sep 2018 10:26:07 +0000 https://www.dbrnd.com/?p=6177#comment-6688 Select NAME, count(NAME) from (
Select DISTINCT
trim(regexp_substr(str1, ‘[^;]+’, 1, level)) Name, level
FROM tbl_SplitData
connect by regexp_substr(name, ‘[^;]+’, 1, level) is not null)foo
GROUP BY NAME

]]> By: Ankush https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/#comment-6687 Wed, 12 Sep 2018 10:25:06 +0000 https://www.dbrnd.com/?p=6177#comment-6687 What do u say about this?

Select NAME, count(NAME) from (
Select DISTINCT
trim(regexp_substr(str1, ‘[^;]+’, 1, level)) Name, level
FROM tbl_SplitData
connect by regexp_substr(name, ‘[^;]+’, 1, level) is not null)foo
GROUP BY NAME

]]> By: Dinesh.IS https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/#comment-5953 Tue, 24 Oct 2017 04:55:50 +0000 https://www.dbrnd.com/?p=6177#comment-5953 Declare @SplitDetails Table(StringValue VarChar(5000));
Declare @StringValue VarChar(5000)=”,@Len As Int

Select @StringValue+=’;’+Str1 From (Select Str1 From TBL_SplitData ) A
Set @StringValue=STUFF(@StringValue,1,1,”)

While Len(@StringValue)>0
Begin

Set @Len=(Case CharIndex(‘;’,@StringValue) When 0 Then Len(@StringValue) Else CharIndex(‘;’,@StringValue)-1 End)

Insert Into @SplitDetails
(StringValue)
Values(SubString(@StringValue,1,@Len))

Set @StringValue=(Case Len(@StringValue)-@Len When 0 Then Null Else Right(@StringValue,Len(@StringValue)-@Len-1) End)

End

Select StringValue,Count(*) As StringCounts From @SplitDetails
Group By StringValue

]]> By: Dinesh.IS https://www.dbrnd.com/2017/10/sql-puzzle-sql-advance-query-split-the-string-and-generate-count-of-each-part/#comment-5952 Tue, 24 Oct 2017 04:55:17 +0000 https://www.dbrnd.com/?p=6177#comment-5952 Declare @SplitDetails Table(StringValue VarChar(5000));
Declare @StringValue VarChar(5000),@Len As Int
Declare Cur Cursor For Select Str1 From TBL_SplitData
Open Cur
Fetch Next From Cur Into @StringValue
While @@Fetch_Status=0
Begin

While Len(@StringValue)>0
Begin

Set @Len=(Case CharIndex(‘;’,@StringValue) When 0 Then Len(@StringValue) Else CharIndex(‘;’,@StringValue)-1 End)

Insert Into @SplitDetails
(StringValue)
Values(SubString(@StringValue,1,@Len))

Set @StringValue=(Case Len(@StringValue)-@Len When 0 Then Null Else Right(@StringValue,Len(@StringValue)-@Len-1) End)
End
Fetch Next From Cur Into @StringValue
End
CLose Cur
Deallocate Cur

Select StringValue,Count(*) As StringCounts From @SplitDetails
Group By StringValue

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