parsing - VBA parse string Place values to new columns -


i have sheets rows of data like.

nom(lsl,usl)=207.3980(206.1990,208.5970)    nom(lsl,usl)=207.3980(206.1990,208.5970)    nom(lsl,usl)=18.8200(18.4400,19.2100)

i grab values , place them in own cells like

207.3980    207.3980    18.8200 206.1990    206.1990    18.4400 208.5970    208.5970    19.2100

i continue recieve "byref argument mismatch" errors. believe relating how defining reference cell.

sub parse_replace()         dim double         dim ws worksheet             set ws = thisworkbook.activesheet         dim col range         dim rlastcell range             set rlastcell = ws.cells.find(what:="*", after:=ws.cells(1, 1), lookin:=xlformulas, lookat:= _             xlpart, searchorder:=xlbycolumns, searchdirection:=xlprevious, matchcase:=false)     = rlastcell.column 1 step -1         col = collett(rlastcell.column)         columns(i).cells(4) = splitstring(col3, ",", 4)         columns(i).cells(5) = splitstring(col3, ",", 5)         columns(i).cells(6) = splitstring(col3, ",", 6)     next      end sub  function collett(col integer) string      if col > 26         collett = collett((col - (col mod 26)) / 26) + chr(col mod 26 + 64)     else         collett = chr(col + 64)     end if      end function  function splitstring(pvalue string, pchar string, pindex integer) variant     dim ystring variant          ystring = replace(replace(replace(replace(pvalue, " ", ""), "=", ""), "(", ","), ")", ",")         splitstring = split(ystring, pchar)(pindex - 1)      end function

process

  • establish number of columns data

  • loop through each column

    • convert column index column collett

    • set cell value splitstring

  • loop

thank you

edit : replaced splitstring value inteded.

you declare col range here:

dim col range

you try set col string here:

col = collett(rlastcell.column)

when set range have set range. furthermore, have use set keyword so:

set col = <a range>

when set col set rlastcell.column repeatedly in each loop of for. if need column letter last column, before entering loop.

all of pointless anyway. @ no point use column letter went through trouble retrieving in function. , really, doing don't need column letter. column letters humans; column number important in vba how.


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