Reload automatically on a changing filename

[#1]

Hi, I'm new on this community and therefore my question may have been already asked.

The application I'm developping requires to reload automatically a file with a name that change every day. :

Today the filename would be: "20140205_two.csv", tomorrow it will be "20140206_two.csv" as you can see the name in depending on the date.

The script that I want to include should reload the correct file depending on the current day.

Thank you in advance.

[MK_QSL]

Instead of file name you can write "*_two.csv"

[giakoum]

You could use the following syntax :

for each file in filelist(????????_two.csv)

load

*

from $(file)

please check syntax, this is just an example

[Nicole-Smith · Accepted solution]

Let vFile = date(today(), 'YYYYMMDD') & '_two.csv';

YourTable:

LOAD *

FROM $(vFile);

[MK_QSL]

Example Given Below...

Instead of this

LOAD Column1,

          Column2

FROM

(ooxml, embedded labels, table is Sheet1);

[ashwanin]

Please use like this

LOAD

abc,

def

FROM

[C:\Documents and Settings\Ashwani\Desktop\2014*.csv

(ooxml, embedded labels, table is Sheet1);

See i have used  ' *'  due to this qv will load all the csv files starting with 2014. Now date change does not impact you.

If it not works then please let me know.

[#7]

Your answer is exactly what I asked for but I have another issue the real name is:
"20140204_two_201402051420114745.csv" (the last numbers can be assimilated as randomised) I pretentiously thought I could used your answer and modify it like this:

" Let vFile = date(today()-1, 'YYYYMMDD') & '_two' & '*.csv'; "

I tried to use the star without the quotes, with a "\", or after the FROM but each time I failed.

Thank you very much

[Nicole-Smith]

Try like this:

Let vFile = date(today(), 'YYYYMMDD') & '_two_';

YourTable:

LOAD *

FROM $(vFile)*.csv;

[#9]

Thank you so much!
It exactly what I was looking for.

[Nicole-Smith]

I'm glad it works for you!