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sanket94
Contributor III
Contributor III

Load only the latest file from the directory

Scenario:

I receive 2 types of file a Po file and Ship file daily twice. the file names are abbreviated as 

File1: _DailyStatusPO[Date]_[time24hrs]

File2: _DailyStatusSHIP[Date]_[time24hrs]

I  want to load only the file that is latest; I do not receive these files on Holidays and weekends so load the latest in those cases. Could someone please help with this.

11ca.JPG

 

 

Labels (3)
1 Reply
martinpohl
Partner
Partner

here's an example, maybe some bugs in but as the steps to go:

for each vType in ('PO','SHIP');   // add more if available

for each vFile in filelist ([LIB//NAME/dailystatus$(vType)*.dat]  // do a loop for all files with filetype

// only load filename and date from filename

Files:

first 1 load

filebasename() as File,

date(date#(right(substring(filebasename(),'_',2),8),'MMDDYYYY'),'YYYYMMDD') as Fileday

from '$(vFile)';

next

//sort datas by date desc

Filesorder:

noconvatenate load

File,

Fileday

resident Files

order by Fileday DESC;

// write the newest (the first) value in a variable

let vLoadfile = peek('File'),0,'Filesorder');

// and then to a table with all table to load from

Loadfiles:

load

'$(vLoadfile)' as Loadfile

autogenerate (1);

drop table Filesorder;

drop table Files;

next

// when ready do a loop for all files to load

let vCountfiles = noofrows(Loadfiles)-1;

for vload = 0 to $(vCountfiles)

let vFinalfile = peek('Loadfile',$(vload),'Loadfiles');

load 

from '$(vFinalfile)';

next

drop table Finalfile;