How to find certain words based on table
Hello,
I have a list of words available in one particular table, based on those records i need to match it up with one specific column on another table. Based on that match i need to update the same. Please see the example for your reference in the excel file.
Please do let me know if you need further clarifications.
https://www.dropbox.com/s/fy5oj5r4feipqrw/Certain%20word%20sample.xlsx?dl=0
Vinwin
VinWin06
One more question just out of curiosity is it possible to find out any object name or thing based on the text from table 1. Like there is no table 2 available to contain words to find the relevant records, so we have to find the words from the table 1 and
those words mainly are any objects or a thing like bolts,shaft,spanner etc.
Is it possible to do, always welcome to your suggestions.
VinWin06
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How to find certain words based on table without overwriting the old one
Hello,
I have a list of words available in one particular table, based on those records i need to match it up with one specific column on another table. Based on that match i need to update the same. below is answer for this question;
Design a new query. Select your data table and your words table. No join line between them.
Drag datatable.* to the design grid, and the field to check. In the criteria for that field write:
Like '*' & [wordstable]![wordsfield] & "*"
(of course you change this to your object names)
But now my another question is that how i can do update without replacing the existing one. Like for example " Peter is engineering degree".
In this example if i have words like ring and degree on my reference table, then as per the solution it will update the degree first then later if its find ring on the text then it will overwrite the degree on the particular column. How i can prevent
this?
VinWin06But now my another question is that how i can do update without replacing the existing one. Like for example " Peter is engineering degree".
I think what you are saying is that you want to avoid specious substring matches such as finding 'engineering' when searching for 'ring'. Right? If so use the following function rather than the LIKE operator to identify the 'word' rather than merely
a substring:
Public Function FindWord(varFindIn As Variant, varWord As Variant) As Boolean
Const PUNCLIST = """' .,?!:;(){}[]/"
Dim intPos As Integer
FindWord = False
If Not IsNull(varFindIn) And Not IsNull(varWord) Then
intPos = InStr(varFindIn, varWord)
' loop until no instances of sought substring found
Do While intPos > 0
' is it at start of string
If intPos = 1 Then
' is it whole string?
If Len(varFindIn) = Len(varWord) Then
FindWord = True
Exit Function
' is it followed by a space or punctuation mark?
ElseIf InStr(PUNCLIST, Mid(varFindIn, intPos + Len(varWord), 1)) > 0 Then
FindWord = True
Exit Function
End If
Else
' is it precedeed by a space or punctuation mark?
If InStr(PUNCLIST, Mid(varFindIn, intPos - 1, 1)) > 0 Then
' is it at end of string or followed by a space or punctuation mark?
If InStr(PUNCLIST, Mid(varFindIn, intPos + Len(varWord), 1)) > 0 Then
FindWord = True
Exit Function
End If
End If
End If
' remove characters up to end of first instance
' of sought substring before looping
varFindIn = Mid(varFindIn, intPos + 1)
intPos = InStr(varFindIn, varWord)
Loop
End If
End Function
You can supplement this with the following functions to find any or all of multiple words in the same string expression:
Public Function FindAnyWord(varFindIn, ParamArray varWordList() As Variant) As Boolean
Dim var As Variant
For Each var In varWordList
If FindWord(varFindIn, var) Then
FindAnyWord = True
Exit Function
End If
Next var
End Function
Public Function FindAllWords(varFindIn, ParamArray varWordList() As Variant) As Boolean
Dim var As Variant
For Each var In varWordList
If Not FindWord(varFindIn, var) Then
FindAllWords = False
Exit Function
Else
FindAllWords = True
End If
Next var
End Function
You can edit the PUNCLIST constant in the first function to allow for any other terminating or leading punctuation characters which may exist, in non-English text for instance.
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\[removed by moderator\]
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