vba - distinguish between two different kinds of string and extract them -


I have a column in Excel with " 01/16 14:38 ATND [notes dealer / From the distributor] JR "and" 01/16 14:14 ATND [notes from company] JR2 "and" 01/16 14:14 ATND [Notes from the company] TLO items are ordered back "

As you can see after the bracket icon, there are two letters or three letters of three letters, JR, JR2 and TLO. I wrote a program that only differentiates between JR and TLO, but is used to extract the code, for example, for example JR and JR 2 counted for example. If someone can help me with this, then I appreciate it. 

  Sub G_ExtractCodes () Long LR = Range ("A" and Rows.Count) as Long LR. End (XLEEP). Note: Route = RR (phase "O" and "I") as the range in the form of red dim I Long Dime Note (NoteCode, "JR")> gt1 = 1 then cells (i, 20) = "jr" ALSIF inset (notecodes, "JR2") & gt; = 1 then cells (i, 20) = "jr2" ALSIF inset (notecodes, "TLO") & gt; = 1 then cells (i, 20) = "TLO" end if the next end end    
 
 
  If your first condition in the statement is more restrictive than the  Alsef  section, then any instance of "JR2" will be caught by the Exam  First  ( Only gets "JR", and  ElseIf  is not evaluated). 
 
 Flip your argument, and I think it should fix it: 
 
  If InStr (NoteCode, "JR2")> = 1 then cells (i, 20 ) = "JR2" ALSIF Insert (NoteCode, "JR")> gt1 = 1 Then Cells (i, 20) = "JR" ALSIF Insert (NoteCode, "TLO") & gt; = 1 Then cells (i, 20) = "TLO" end if  
 
 Alternatively, you can parse these codes like: 
 
  String dim bracket position as a slow code value i = L for 2 to step 1 set Note code = range ("O" & I) 'Finds the position of correct bracket bracket location location = installer (notecod , ")") 'Closes Android to any character & amp; Correct square bracket codeValue = trim (including middle (notecodes, bracket location + 1), 'deletes any text that appears after code value, if any, if any instr (codeValue, "") & gt; 0 = CallWall = Left (codeValue, Instr (codeValue, "") End if the cell (i, 20) .Value = codeValue 'variable codeValue = vbNullString next

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