I have a dataset that contains multiple values. I want to take those rows from that dataset from the datatable BLABLA that contains an "S" with the numbers from zero to six. Then I want to display those in a MessageBox.
My Regex is S[0-6].
Dim answer As String = ""
Dim myregex As Regex = New Regex("S[0-6]")
Dim SearchRows() As DataRow = datasetB.Tables("BLABLA").Select("Data LIKE '%myregex%'")
For k As Integer = 0 To SearchRows.Length - 1
If answer = "" Then
answer = SearchRows(k).Item("Data")
Else
answer = answer & vbNewLine & SearchRows(k).Item("Data")
End If
Next
MsgBox(answer)
Unfortunately SearchRows is empty. I couldn't find the reason by debugging.
What am I doing wrong?
The DataTable.Select method does not support regex. As the documentation states, it does allow you to pass it a filterExpression string as an argument, but just because it takes a filter expression doesn't mean that it support's regex expressions. On the contrary, it's designed to mostly support the same kinds of expressions as the WHERE clause in T-SQL. T-SQL's LIKE operator does not support regex patterns, and neither does DataTable.Select. See this documentation to learn the rules for the pattern expressions that are supported by the DataTable.Select method's LIKE operator.
The filter expressions supported by the LIKE operator are not as advanced as regex, so it's almost certainly impossible to construct a filter expression which is that specific. If there is a way to filter to digits between 0 and 6, I am unaware of it and the documentation doesn't mention it. So, if you really need to filter rows by regex, you can still do it, but you need to select all the rows and then filter them yourself:
Dim SearchRows() As DataRow = datasetB.Tables("BLABLA").Select().
Where(Function(r) myregex.IsMatch(r.Item("Data").ToString())).
ToArray()
Related
I'm trying to replace parts of a string that contains what should be dates, but which are possibly in an impermissible format. Specifically, all of the dates are in the form "mm/dd/YYYY" and they need to be in the form "YYYY-mm-dd". One caveat is that the original dates may not exactly be in the mm/dd/YYYY format; some are like "5/6/2015". For example, if
val x = "where date >= '05/06/2017'"
then
x.replaceAll("'([0-9]{1,2})/([0-9]{1,2})/([0-9]{4})'", "'$3-$1-$2'")
performs the desired replacement (returns "2017-05-06"), but for
val y = "where date >= '5/6/2017'"
this does not return the desired replacement (returns "2017-5-6" -- for me, an invalid representation). With the Joda Time wrapper nscala-time, I've tried capturing the dates and then reformatting them:
import com.github.nscala_time.time.Imports._
import org.joda.time.DateTime
val f = DateTimeFormat.forPattern("yyyy-MM-dd")
y.replaceAll("'([0-9]{1,2}/[0-9]{1,2}/[0-9]{4})'",
"'"+f.print(DateTimeFormat.forPattern("MM/dd/yyyy").parseDateTime("$1"))+"'")
But this fails with a java.lang.IllegalArgumentException: Invalid format: "$1". I've also tried using the f interpolator and padding with 0s, but it doesn't seem to like that either.
Are you not able to do additional processing on the captured groups ($1, etc.) inside the replaceAll? If not, how else can I achieve the desired result?
The $1 like backreferences can only be used inside string replacement patterns. In your code, "$1" is not a backreference any longer.
You may use a "callback" with replaceAllIn to actually get the match object and access its groups to further manipulate them:
val pattern = "'([0-9]{1,2}/[0-9]{1,2}/[0-9]{4})'".r
y = pattern replaceAllIn (y, m => "'"+f.print(DateTimeFormat.forPattern("MM/dd/yyyy").parseDateTime(m.group(1)))+"'")
Regex.replaceAllIn is overloaded and can take a Match => String.
I wish to select and add comments after certain words, e.g. “not”, “never”, “don’t” in sentences in a Word document with VBA. The Find/Replace with wildcards works fine, but “Use wildcards” cannot be selected with “Match case”. The RegEx can “IgnoreCase=True”, but the selection of the word is not reliable when there are more than one comments in a sentence. The Range.start seems to be getting modified in a way that I cannot understand.
A similar question was asked in June 2010. https://social.msdn.microsoft.com/Forums/office/en-US/f73ca32d-0af9-47cf-81fe-ce93b13ebc4d/regex-selecting-a-match-within-the-document?forum=worddev
Is there a new/different way of solving this problem?
Any suggestion will be appreciated.
The code using RegEx follows:
Function zRegExCommentor(zPhrase As String, tComment As String) As Long
Dim sTheseSentences As Sentences
Dim rThisSentenceToSearch As Word.Range, rThisSentenceResult As Word.Range
Dim myRegExp As RegExp
Dim myMatches As MatchCollection
Options.CommentsColor = wdByAuthor
Set myRegExp = New RegExp
With myRegExp
.IgnoreCase = True
.Global = False
.Pattern = zPhrase
End With
Set sTheseSentences = ActiveDocument.Sentences
For Each rThisSentenceToSearch In sTheseSentences
Set rThisSentenceResult = rThisSentenceToSearch.Duplicate
rThisSentenceResult.Select
Do
DoEvents
Set myMatches = myRegExp.Execute(rThisSentenceResult)
If myMatches.Count > 0 Then
rThisSentenceResult.Start = rThisSentenceResult.Start + myMatches(0).FirstIndex
rThisSentenceResult.End = rThisSentenceResult.Start + myMatches(0).Length
rThisSentenceResult.Select
Selection.Comments.Add Range:=Selection.Range
Selection.TypeText Text:=tComment & "{" & zPhrase & "}"
rThisSentenceResult.Start = rThisSentenceResult.Start + 1 'so as not to find the same phrase again and again
rThisSentenceResult.End = rThisSentenceToSearch.End
rThisSentenceResult.Select
End If 'If myMatches.Count > 0 Then
Loop While myMatches.Count > 0
Next 'For Each rThisSentenceToSearch In sTheseSentences
End Function
Relying on Range.Start or Range.End for position in a Word document is not reliable due to how Word stores non-printing information in the text flow. For some kinds of things you can work around it using Range.TextRetrievalMode, but the non-printing characters inserted by Comments aren't affected by these settings.
I must admit I don't understand why Word's built-in Find with wildcards won't work for you - no case matching shouldn't be a problem. For instance, based on the example: "Never has there been, never, NEVER, a total drought.":
FindText:="[n,N][e,E][v,V][e,E][r,R]"
Will find all instances of n-e-v-e-r regardless of the capitalization. The brackets let you define a range of values, in this case the combination of lower and upper case for each letter in the search term.
The workarounds described in my MSDN post you link to are pretty much all you can if you insist on RegEx:
Using the Office Open XML (or possibly Word 2003 XML) file format will let you use RegEx and standard XML processing tools to find the information, add comment "tags" into the Word XML, close it all up... And when the user sees the document it will all be there.
If you need to be doing this in the Word UI a slightly different approach should work (assuming you're targeting Word 2003 or later): Work through the document on a range-by-range basis (by paragraph, perhaps). Read the XML representation of the text into memory using the Range.WordOpenXML property, perform the RegEx search, add comments as WordOpenXML, then write the WordOpenXML back into the document using the InserXml method, replacing the original range (paragraph). Since you'd be working with the Paragraph object Range.Start won't be a factor.
I have a data column that has a heading value with multiple levels, where I only want the first three levels, but I cannot figure out how to get the parsed value?
I was reading this and it shows how to use create a function to return a boolean for the condition, but how would I create a function that would return a parsed value?
This is the Regular Expression that I think I need.
^(\d.\d.\d)
I'm looking for something that would change 1.2.3.4.5. to 1.2.3 and similar for any other header I have that has more than three levels.
Ideally, I'd like to be able to put it into my Query Design as a Field Expression, but I'm not sure how I would do that.
I assumed your input values could have more than one digit between the dots. In other words, I think you want this ...
? RegExpGetMatch("1.2.3.4.5.", "^(\d+\.\d+\.\d+).*", 1)
1.2.3
? RegExpGetMatch("1.27.3.4.5.", "^(\d+\.\d+\.\d+).*", 1)
1.27.3
If that is the correct behavior, here is the function I used.
Public Function RegExpGetMatch(ByVal pSource As String, _
ByVal pPattern As String, _
ByVal pGroup As Long) As String
'requires reference to Microsoft VBScript Regular Expressions
'Dim re As RegExp
'Set re = New RegExp
'late binding; no reference needed
Dim re As Object
Set re = CreateObject("VBScript.RegExp")
re.Global = True
re.Pattern = pPattern
RegExpGetMatch = re.Replace(pSource, "$" & pGroup)
Set re = Nothing
End Function
See also this answer by KazJaw. His answer taught me how to select the match group with RegExp.Replace.
In a query run within an Access session, you could use the function like this:
SELECT
RegExpGetMatch([Data Column], "^(\d+\.\d+\.\d+).*", 1) AS parsed_value
FROM YourTable;
Note however a custom VBA function is not usable for queries run from outside an Access session.
Try changing your RegEx to ^(\d\.\d\.\d). You need to escape the . since it has a special meaning in RegExp.
I have a string
IsNull(VSK1_DVal.RuntimeSUM,0),
I need to remove IsNull part, so the result would be
VSK1_DVal.RuntimeSUM,
I'm absolute new to RegEx, but it wouldn't be a problem, if not one thing :
VSK1 is dynamic part, can be any combination of A-Z,0-9 and any length. How to replace strings with RegEx? I use MSSQL 2k5, i think it uses general set of RegEx rules.
EDIT : I forgot to say, that I'm doing replacement in SSMS Query window's Replace Box (^H) - not building RegEx query
br
marius
here's a regex that should work:
[^(]+\(([^,]+),[^)]\)
Then use $1 capture group to extract the part that you need.
I did a sanity check in ruby:
orig = "IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /[^(]*\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => VSK1_DVal.RuntimeSUM,
It gets trickier if you have a prefix that you want to retain. Like if you have this:
"somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
In this case, you need someway to identify the start of the pattern. Maybe you can use '=' to identify the start of the pattern? If so, this should work:
orig = "somestuff = IsNull(VSK1_DVal.RuntimeSUM,0),"
regex = /=\s*\w+\(([^,]+),[^)]\)/
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
But then the case where you don't have an equals sign will fail. Maybe you can use 'IsNull' to identify the start of the pattern? If so, try this (note the '/i' representing case insensitive matching):
orig = "somestuff = isnull(VSK1_DVal.RuntimeSUM,0),"
regex = /IsNull\(([^,]+),[^)]\)/i
result = orig.sub(regex){$1} # result => somestuff = VSK1_DVal.RuntimeSUM,
/IsNULL\((A-Z0-9+),0\)/
Then pick group match number 1.
Here's a very useful site: http://www.regexlib.com/RETester.aspx
They have a tester and a cheatsheet that are very useful for quick testing of this sort.
I tested the solution by Dave and it works fine except it also removes the trailing comma you wanted retained. Minor thing to fix.
Try this:
IsNULL\((.*,)0\)
You say in your question
I use MSSQL 2k5, i think it uses
general set of RegEx rules.
This is not true unless you enable CLR and compile and install an assembly. You can use its native pattern matching syntax and LIKE for this as below.
WITH T(C) AS
(
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,0),' UNION ALL
SELECT 'IsNull(VSK1_DVal.RuntimeSUM,123465),' UNION ALL
SELECT 'No Match'
)
SELECT SUBSTRING(C,8,1+LEN(C)-8-CHARINDEX(',',REVERSE(C),2))
FROM T
WHERE C LIKE 'IsNull(%,_%),'
my string style like this
expression1/field1+expression2*expression3+expression4/field2*expression5*expression6/field3
a real style mybe like this:
computer/(100)+web*mail+explorer/(200)*bbs*solution/(300)
"+" and "*" represent operator
"computer","web"...represent expression
(100),(200) represent field num . field num may not exist.
I want process the string to this:
<computer>/(100)+web*<mail>+explorer/(200)*bbs*<solution>/(300)
rules like this
if expression length is more than 3 and its field is not (200), then add brackets to it.
My recommendation is to mix regex with other language features. The complication arises from the fact that field appears before expressions, and lookbehind is usually more limited than lookforward.
In pseudo-Java-code, I recommend doing something like this:
String[] parts = input.split("/");
for (int i = 0; i < parts.length; i++) {
if (!parts[i].startsWith("(200)"))
parts[i] = parts[i].replaceAll("(?=[a-z]{4})([a-z]+)", "<$1>");
}
String output = parts.join("/");
I would not use just regular expression.
You say "if expression length is more than 3 and its field is not (200), then add brackets to it"
I think a normal conditional statement is the best and clearest solutoion for this.
I think regular expressions are sometimes overused. Regexes are hard to read, and when a couple of conditional statements can do the same but more clearly, then I'd say the code quality is higher.