Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Parsing a string SQL

I have this very long string in MSSQL which I require the I value.

I represents Invoice, A represents Amount, D represents Date

I=940;A=29.5;D=20090901|I=941;A=62.54;D=20090910|I=942;A=58.99;D=20091005|I=954;A=93.45;D=20091201|I=944;A=96.76;D=20091101|I=946;A=52.5;D=20091101|I=943;A=28.32;D=20091101|I=945;A=52.5;D=20091101|I=955;A=79.81;D=20091201|I=950;A=25.2;D=20091124|I=948;A=31.86;D=20091110|I=949;A=28.32;D=20091120|I=947;A=25.2;D=20091109|I=951;A=242.54;D=20091124|I=952;A=28.32;D=20091129|I=956;A=38.94;D=20091210|I=957;A=107.39;D=20091215|I=958;A=32.55;D=20091228|I=959;A=27.3;D=20091228|I=960;A=24.79;D=20091230|I=1117;A=28.32;D=20100131|I=1115;A=272.58;D=20100131|I=1116;A=159.6;D=20100209

This is one of the scariest cases.

Each of these are invoice numbers which have relevant values which i will use to link to another transaction. I would really appreciate it if someone could explain the best manner to go about this without making a app if possible

like image 861
Enzero Avatar asked Jun 10 '11 14:06

Enzero


People also ask

Can you PARSE strings in SQL?

String parsing is a common task for data analysts and data engineers working with raw data. With the growth of unstructured qualitative data, parsing strings efficiently has become increasingly important for fast analysis.

Can you PARSE data in SQL?

Simple SQL operations like LOAD, ALTER, INSERT, and UPDATE can turn parsing data from a chore into an efficient and mistake-free task.

What is parsing in SQL?

The parsing stage involves separating the pieces of a SQL statement into a data structure that other routines can process. The database parses a statement when instructed by the application, which means that only the application, and not the database itself, can reduce the number of parses.

How fetch part of a string in SQL?

Use the SUBSTRING() function. The first argument is the string or the column name. The second argument is the index of the character at which the substring should begin. The third argument is the length of the substring.


1 Answers

declare @s varchar(max) = 'I=940;A=29.5;D=20090901|I=941;A=62.54;D=20090910|I=942;A=58.99;D=20091005|I=954;A=93.45;D=20091201|I=944;A=96.76;D=20091101|I=946;A=52.5;D=20091101|I=943;A=28.32;D=20091101|I=945;A=52.5;D=20091101|I=955;A=79.81;D=20091201|I=950;A=25.2;D=20091124|I=948;A=31.86;D=20091110|I=949;A=28.32;D=20091120|I=947;A=25.2;D=20091109|I=951;A=242.54;D=20091124|I=952;A=28.32;D=20091129|I=956;A=38.94;D=20091210|I=957;A=107.39;D=20091215|I=958;A=32.55;D=20091228|I=959;A=27.3;D=20091228|I=960;A=24.79;D=20091230|I=1117;A=28.32;D=20100131|I=1115;A=272.58;D=20100131|I=1116;A=159.6;D=20100209'
declare @xml xml

select @xml = '<item><value>'+replace(replace(@s, ';','</value><value>'), '|','</value></item><item><value>')+'</value></item>'

select N.value('substring(value[1],3)', 'int') as Invoice,
       N.value('substring(value[2],3)', 'money') as Amount,
       N.value('substring(value[3],3)', 'date') as [Date]
from @xml.nodes('item') as T(N)

Result:

Invoice     Amount                Date
----------- --------------------- ----------
940         29,50                 2009-09-01
941         62,54                 2009-09-10
942         58,99                 2009-10-05
954         93,45                 2009-12-01
944         96,76                 2009-11-01
946         52,50                 2009-11-01
943         28,32                 2009-11-01
945         52,50                 2009-11-01
955         79,81                 2009-12-01
950         25,20                 2009-11-24
948         31,86                 2009-11-10
949         28,32                 2009-11-20
947         25,20                 2009-11-09
951         242,54                2009-11-24
952         28,32                 2009-11-29
956         38,94                 2009-12-10
957         107,39                2009-12-15
958         32,55                 2009-12-28
959         27,30                 2009-12-28
960         24,79                 2009-12-30
1117        28,32                 2010-01-31
1115        272,58                2010-01-31
1116        159,60                2010-02-09

For SQL Server 2005 you need to use datetime instead of date

select N.value('substring(value[1],3)', 'int') as Invoice,
       N.value('substring(value[2],3)', 'money') as Amount,
       N.value('substring(value[3],3)', 'datetime') as [Date]
from @xml.nodes('item') as T(N)

To read from a table you need to do it like this.

declare @s varchar(max) = 'I=940;A=29.5;D=20090901|I=941;A=62.54;D=20090910|I=942;A=58.99;D=20091005|I=954;A=93.45;D=20091201|I=944;A=96.76;D=20091101|I=946;A=52.5;D=20091101|I=943;A=28.32;D=20091101|I=945;A=52.5;D=20091101|I=955;A=79.81;D=20091201|I=950;A=25.2;D=20091124|I=948;A=31.86;D=20091110|I=949;A=28.32;D=20091120|I=947;A=25.2;D=20091109|I=951;A=242.54;D=20091124|I=952;A=28.32;D=20091129|I=956;A=38.94;D=20091210|I=957;A=107.39;D=20091215|I=958;A=32.55;D=20091228|I=959;A=27.3;D=20091228|I=960;A=24.79;D=20091230|I=1117;A=28.32;D=20100131|I=1115;A=272.58;D=20100131|I=1116;A=159.6;D=20100209'

declare @YourTable table(ID int, s varchar(max))
insert into @YourTable values
(1, @s),
(2, @s)

select Y.ID,
       T.N.value('substring(value[1],3)', 'int') as Invoice,
       T.N.value('substring(value[2],3)', 'money') as Amount,
       T.N.value('substring(value[3],3)', 'date') as [Date]
from @YourTable as Y
  cross apply (select cast('<item><value>'+replace(replace(Y.s, ';','</value><value>'), '|','</value></item><item><value>')+'</value></item>' as xml)) as X(XMLCol)
  cross apply X.XMLCol.nodes('item') as T(N)
like image 149
Mikael Eriksson Avatar answered Oct 26 '22 22:10

Mikael Eriksson