Getting name from a list of csv files

[#1]

Hi,

I need to load a list of csv files to qvds.

So I am doing it with a loop.

The issue is that every qvd has to have csv name in an appropriate order.

For example, I have ra_daily_2012-10-01.csv and ra_daily_2012_10_02.csv.

So I have to get  ra_daily_2012-10-01.qvd and  ra_daily_2012_10_02.qvd .

So how do I do an appropriate variable?

Thanks.

[erichshiino]

Hi,

Is there a special requirement to read them in this specific order and in a loop?

You can avoid the loop using wild cards to read your .csv.

The syntax would be something like this:

Table:

Load *

from FOLDER\*.csv (txt) ;

Hope it helps,

Erich

[#3]

Hi,

I am doing something like :

SET FilePath='*.csv';

set vFileName = 'file';



for each File in filelist ('C:\Users\innas\Desktop\Qvd\QV_new\daily\' & FilePath)

TRACE File = $(File);

    SET sFile = '$(File)';

    Directory;

   NetworkDomainDailyData:

    LOAD *

    FROM

    [$(sFile)]

    (txt, codepage is 1252, embedded labels, delimiter is ',', msq);



store * from NetworkDomainDailyData into C:\Users\innas\Desktop\Qvd\Sql_Data\$(vFileName).qvd(qvd);

drop table NetworkDomainDailyData;



    next File

So what has to be in 

set vFileName = ?

Thanks!!!

[erichshiino · Accepted solution]

Hi,

You can use this insted of set vFileName, use both lines below:

let	vFileBaseName = SubField( File,'\',SubStringCount(File,'\')+1);
let	vFileName = left ( vFileBaseName,FindOneOf(vFileBaseName, '.', SubStringCount(vFileBaseName, '.'))-1);

File contains full path and filename

vFileBaseName will contain the full file name ( 2012-10-01.csv)

vFileName will remove the file extension.

(the expression is a bit long to make sure that only the string after the last '.' is removed)

Hope it helps,

Erich

[#5]

Hi,

you really helped me!

Thanks a lot.

My model works:)

[Anonymous]

Hi Erich,

Can You Please expalin how the FIle Variable will load file path and File Name.

I have more than 100 Excel files in one folder.