Retrive words from a string and compare

FILENAME       KEYWORD
xxx.doc      abc,caa,funny
xyz.txt      abc,cba,funnys
sdfetll.txt      ball,basketball,players,court,
d.doc      bryan,name,hobbies,male,boy

above is a datasets, fields are filename,keyword. store in a ms access table. both filename and keyword fields are string.   i.e.   'xxx.doc'  is a string,  'abc,caa,funny' is a string

if a user input "abc", the program should return filename associated with it , i.e.  xxx.doc and xyz.txt

since the datasets may be very large, the algorithim to retrive the keywords from keyword string and then compare in order to get the associated filename should be efficient.

anyone has the script or can provide some idea? like what data types., method i should use....etc

thank you for enlightening.

plovelAsked:
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mocartsCommented:
hi, plovel :)
just use SQL:
SELECT FILENAME FROM TABLE1 WHERE ','+KEYWORD+',' LIKE '%,abc,%'

delphi example:
procedure TForm1.Button1Click(Sender: TObject);
begin
  ADOQuery1.SQL.Text := 'SELECT FILENAME FROM TABLE2 WHERE '',''+KEYWORD+'','' LIKE ''%,'
    + Edit1.Text + ',%''';
  ADOQuery1.Open;
  if ADOQuery1.RecordCount = 0 then
    ShowMessage('No result found!')
  else
    ShowMessage('FileName: '+ ADOQuery1.FieldByName('FILENAME').AsString);
  ADOQuery1.Close;
end;

wbr, mo.
0
mocartsCommented:
caution - you must check for empty search string, otherwise you will get all records where KEYWORD field ends with comma.
0
ahllCommented:
I agree with mocarts, but you have to include in the KEYWORD field a comma at the beginind and at the end. If not, you will lose the first and the last word.
0
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mocartsCommented:
not actually - that can be overcome in SQL by "quoting" keyword field in commas - what is actually done in my example.
SELECT FILENAME FROM TABLE1 WHERE ','+KEYWORD+',' LIKE '%,abc,%'

also comment to my example - plovel - you should additionally implement loop (while not eof) for retrieving all selected records (filenames):
var
  sFiles: string;
begin
...
 ADOQuery1.Open;
  if ADOQuery1.RecordCount = 0 then
    ShowMessage('No result found!')
  else begin
    while not ADOQuery1.eof do begin
      sFIles := sFiles + ',' + ADOQuery1.FieldByName('FILENAME').AsString;
      ADOQuery1.Next;
    end;
    if sFiles <> '' then Delete(sFiles, 1);
    ShowMessage(sFiles);
  end;
...
0

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cerdalCommented:
You'd have much faster access if you split the list up and inserted one record for each keyword. That way you could just do a straight "select filename from Table1 where keyword = :MyKeyWord".

If the table has other fields in it, which would mean too many duplicates, add an AutoIncrement field to Table1 and fill a second table with KeyWord + Tbl1AutoIncValue in Table1's AfterPost event. You could then do:

Select TABLENAME from TABLE1, TABLE2
Where TABLE2.KEYWORD = :MyKeyWord
And TABLE1.AutoIncValue = TABLE2.Tbl1AutoIncValue

If you need more detail, just ask.

Chris.
0
mocartsCommented:
thanks, plovel :)
small comment on Chris comment..
that will be only a little bit faster, but will take more room to store all data. and if to be truly formal then there must be three tables - one for keywords, one for filenames and last one to link previous tables i.e.
Kyeword table (KT): IDKeyword, Keyword
Filename table (FT): IDFile, FileName
Link table (LT): IDKeyword, IDFile
and select would be SELECT FT.FileName FROM FT, KT, LT WHERE FT.IDFile=LT.IDFile AND KT.IDKeyword=LT.IDKeyword AND KT.Keyword = 'KeywordToSearch'

wbr, mo.
0
cerdalCommented:
reply to mo:

pragmatism is always best :-) If it's good enough, do it that way.

it's another case of horses for courses - I've never worked with Access, but some databases can be orders of magnitude slower matching parts of a field (even if it's indexed) in every record from start to finish rather than going straight to the only ones whose keyfield matches what's needed exactly.

It might be an idea for plovel to compare the 2 methods and let us know how much faster it actually is in his situation...

As for the extent of normalising that's required, I agree it's pretty rare to find a case where 100% is optimum.

Chris.
0
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