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... or if it is difficult to identify individual records, it may be more efficient to retrieve and process the data one record at a time. The frequency of updates: If the data is being updated frequently, then it may be more efficient to retrieve and process the data one record at a time in ...

... .or. ; a[ nPos2 ] == numero .or. ; a[ nPos3 ] == numero .or. ; a[ nPos4 ] == numero .or. ; a[ nPos5 ] == numero , ncount++, ) } ) it result the frequency number not the delay !!!!

the frequency run ok Function freq( aData, numero, nWheel )   local nCount   := 0   local nPos1,nPos2,nPos3,nPos4,nPos5      nPos1 := ( nWheel - 1 ) * 5 + 3    ...

Silvio, Can you show your code? LG Otto Do you saw the code ? LG ? It is not here to discuss how to do the calculation because finding the frequency of a number is quickly done. this is one of last draws https://i.postimg.cc/ZRhVp32t/h.jpg the frequency is calculated by counting ...

Otto wrote:Silvio,
Can you show your code?
LG
Otto

pls
http://forums.fivetechsupport.com/viewtopic.php?f=3&t=41773&sid=58834d69e74baf1a1e9dfff11e5ebde3&sid=58834d69e74baf1a1e9dfff11e5ebde3#p250337

Silvio,
Can you show your code?
LG
Otto

James Bott wrote:Read the database once while copying all the data into a multidimensional array.

Then calculate the frequency using the array.

I made it but the time is big

... If I make number:= 34 cFilter := "oDbf:Ba1=" +number oDbf:setFilter(cFilter) nFreq:= oDbf:keycount() It immediately gives me the frequency value in fast mode for 1 number Is it possible that if I do a search for 90 numbers it will take a long time?

Read the database once while copying all the data into a multidimensional array. Then calculate the frequency using the array. that is in practice how should I do? if I asked in the forum it means that I don't know how to do it I cannot use ascan() on all fields because ...

Read the database once while copying all the data into a multidimensional array.

Then calculate the frequency using the array.

I have to calculate the frequency of a number that is, I'll explain I have these ninety numbers https://i.postimg.cc/fyMTTWFQ/box.png For each number I must calculate the frequency that is, how many times this number is ...

... the latitude and longitude in addition to some icons and a lot of other information that it deserves to appear. The .XML file is updated with a frequency of 30 seconds therefore if it is a follow-up you should update the XML also to refresh them If Mr. Antonio allows to publish the use and use ...

... : show icon in system tray -ln <integer> : select line from text file by using of line number -fr <integer> : output audio sampling frequency in kHz (from 8 to 48) -bt <integer> : output audio bit depth (8 or 16) -ch <integer> : output audio channel mode (1 or 2) -?, ...

... Czerpak (druzus/at/priv.onet.pl) * harbour/src/vm/garbage.c + added new PRG function HB_GCSETAUTO( [nNewVal] ) - nPrevVal which allows to set the frequency of automatic GC activation. nNewVal is number of GC items (in thousands). If total number of new GC items allocated from last GC pass exceed ...

James, In adordd we cant control commit frequency, as far as I know, with :addnew method. As soon as we move or add new record the previous its auto committed because ado auto calls :update(). Being other rdd than ado in source area the loop ...