February 16, 2015 at 1:35 am
Hello
I have a date file with no delimiter like bellow
0080970393102312072981103378000004329392643958
0080970393102312072981103378000004329392643958
I just know 5 first number in a line is for example "ID of bank"
or 6th and 7th number in a line is for example "ID of employee"
Could you help me how can I create a XML format file?
thanks alot
February 16, 2015 at 7:32 am
bkshn (2/16/2015)
HelloI have a date file with no delimiter like bellow
0080970393102312072981103378000004329392643958
0080970393102312072981103378000004329392643958
I just know 5 first number in a line is for example "ID of bank"
or 6th and 7th number in a line is for example "ID of employee"
Could you help me how can I create a XML format file?
thanks alot
Since your data is fixed width just use substring to get each data element. Then you can do whatever you want with it.
For example.
declare @Something varchar(100) = '0080970393102312072981103378000004329392643958'
select LEFT(@Something, 5) as BankID
, SUBSTRING(@Something, 6, 2) as EmpID
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February 16, 2015 at 7:55 am
Check the following page from BOL: https://msdn.microsoft.com/en-us/library/ms187833.aspx
In the Sample XML Format Files section (bottom of the page):
F. Importing fixed-length or fixed-width fields
The following example describes fixed fields of 10 or 6 characters each. The format file represents these field lengths/widths as LENGTH="10" and LENGTH="6", respectively. Every row of the data files ends with a carriage return-line feed combination, {CR}{LF}, which the format file represents as TERMINATOR="\r".
<?xml version="1.0"?>
<BCPFORMAT
xmlns="http://schemas.microsoft.com/sqlserver/2004/bulkload/format"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<RECORD>
<FIELD ID="1" xsi:type="CharFixed" LENGTH="10"/>
<FIELD ID="2" xsi:type="CharFixed" LENGTH="6"/>
<FIELD ID="3" xsi:type="CharTerm" TERMINATOR="\r"
</RECORD>
<ROW>
<COLUMN SOURCE="1" NAME="C1" xsi:type="SQLINT" />
<COLUMN SOURCE="2" NAME="C2" xsi:type="SQLINT" />
</ROW>
</BCPFORMAT>
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